Course notes · Andrew Park · Professor Xavier Mela · Duke University, Fall 2025 · LaTeX source
Simple Interest: Only the principal is reinvested at the end of each year.
The discount factor is \(\frac{1}{1+\tau r}\).
Banker's Rule: Use exact time (the number of days from \(t_0\) to \(t_0 + \tau\)) and ordinary interest (\(1 \text{ year} = 360 \text{ days}\), \(1 \text{ month} = 30 \text{ days}\)).
Compound Interest: Divide each year into \(k\) equal time periods, and apply simple interest at the end of each period to the previous balance.
The return rate over \(\tau\) years is:
Example (APR vs APY): A credit card company advertises a \(10.99\%\) APR (annual percentage rate). How much interest did you pay after \(1\) year given \(F_0 = \$2\,500\) (assuming no payment is made and no late fees incurred)?
If \(x\) is a nonnegative real number, the future value after \(x\) interest periods is:
\[ F_x = (1+\frac{r}{k})^x F_0 \]Claim: For a fixed time \(\tau\), the future value \(F_{\tau k}\) is an increasing function of \(k\).
Claim: The limit of \(F_{\tau k}\) as \(k \to \infty\) is called the future value under continuous compounding: \(F_{cts} = e^{\tau r}F_0\).
Suppose that you are considering a new investment opportunity requiring an initial capital of \(C_0\) to generate future net cash flows:
\[ C_1, C_2, \dots, C_n \]at future times \(t_1, t_2, \dots, t_n\).
The present value of this sequence of cash flows under \(k\)-periodic compounding and discount rate of \(r\) is:
\[ PV(r) = \frac{C_1}{(1+\frac{r}{k})^{(t_1 - t_0)k}} + \frac{C_2}{(1+\frac{r}{k})^{(t_2 - t_0)k}} + \dots + \frac{C_n}{(1+\frac{r}{k})^{(t_n - t_0)k}} \]The net present value of the net cash flows is:
\[ NPV(r) = PV(r) - C_0 \]The \(r_{RRR}\) is the mean compounding (annual) growth rate from investing in an alternative opportunity in the marketplace with business profile and risk similar to the new investment opportunity.
Example: Decide whether to invest \(\$250\,000\) in a startup given that:
An IRR of the new investment is a positive solution, \(r = r_{IRR}\), of the equation \(NPV(r)=0 \Leftrightarrow PV(r) = C_0\).
\[ C_0 = \frac{C_1}{(1+\frac{r}{k})^{(t_1-t_0)k}} + \dots + \frac{C_n}{(1+\frac{r}{k})^{(t_n-t_0)k}} \]Solve for \(r\) above.
In the previous example:
\[ 250000 = \frac{155000}{(1+\frac{r}{12})^{12}} + \frac{215000}{(1+\frac{r}{12})^{18}} \Rightarrow r \approx 29.42\% \]Definition: An annuity is a series of payments made at equal time periods with interest.
1. Future Value (\(S_n\))
The Future Value (FV) is the total accrued amount at the end of the term. Let \(S_i\) be the total accrued at the end of period \(i\):
Using the geometric sum formula (\(1+x+\dots+x^{n-1} = \frac{x^n-1}{x-1}\)):
\[ \boxed{S_n = P \cdot \frac{(1+\frac{r}{k})^n - 1}{r/k}} \]2. Present Value (\(A_n\))
The Present Value (PV) is the amount needed today to generate the cash flow \(P\) for \(n\) periods. We sum the PV of each individual payment:
To relate this to Future Value, multiply by \((1+\frac{r}{k})^n\):
\[ \left(1+\frac{r}{k}\right)^n A_n = S_n \implies A_n = S_n \left(1+\frac{r}{k}\right)^{-n} \]Substituting the formula for \(S_n\):
\[ \boxed{A_n = P \cdot \frac{1 - (1+\frac{r}{k})^{-n}}{r/k}} \]Definition: A sequence of cash flows that continues indefinitely (\(n \to \infty\)).
A loan is essentially an annuity where the Loan Amount equals the Present Value (\(A_n\)).
Key Formulas (Rearrangements of \(A_n\))
Notice: Doubling the monthly payment will reduce the term of the loan by more than half (e.g., a 30-year term might drop to less than 10 years).
Definition: Amortization is the reducing of a given loan amount (the principal) through a series of payments over a fixed time span. One portion of each payment goes towards the principal and the other goes towards the interest.
Assume \(k\)-periodic compounding (e.g., \(k=12\)) with payment \(P\) and loan amount \(A_n\). Let \(y = 1 + \frac{r}{k}\).
The remaining balance \(B_i\) after \(i\) periods follows the recursion:
\[ \begin{aligned} B_0 &= A_n \\ B_1 &= y B_0 - P = y A_n - P \\ B_2 &= y B_1 - P = y(y A_n - P) - P = y^2 A_n - (1+y)P \\ &\vdots \\ B_i &= y^i A_n - (1 + y + \dots + y^{i-1})P \end{aligned} \]Using the geometric sum formula for the payment term:
\[ B_i = y^i A_n - \frac{y^i - 1}{y - 1} P \]Recall from the Annuity derivation that \(P = A_n \frac{y^n(y-1)}{y^n - 1}\). Substituting this into the equation for \(B_i\) and simplifying yields:
\[ \boxed{B_i = A_n \frac{(1+\frac{r}{k})^n - (1+\frac{r}{k})^i}{(1+\frac{r}{k})^n - 1}} \]Note: At the end of the loan (\(i=n\)), \(B_n = 0\).
Decomposition of Payment (\(P = \mathcal{I}_i + \mathcal{B}_i\)) Each payment \(P\) is split into an interest portion (\(\mathcal{I}_i\)) and a principal repayment portion (\(\mathcal{B}_i\)). Note that \(\mathcal{B}_i \neq B_i\).
Interest Payment (\(\mathcal{I}_i\)) Interest is calculated on the remaining balance from the previous period (\(B_{i-1}\)).
\[ \mathcal{I}_i = \frac{r}{k} B_{i-1} \]Substituting the formula for \(B_{i-1}\) (using index \(i-1\)):
\[ \boxed{\mathcal{I}_i = A_n \frac{r}{k} \frac{(1+\frac{r}{k})^n - (1+\frac{r}{k})^{i-1}}{(1+\frac{r}{k})^n - 1}} \]Principal Payment (\(\mathcal{B}_i\)) The remainder of the payment goes toward reducing the loan balance.
\[ \mathcal{B}_i = P - \mathcal{I}_i \]After algebraic simplification,:
\[ \boxed{\mathcal{B}_i = A_n \frac{r}{k} \frac{(1+\frac{r}{k})^{i-1}}{(1+\frac{r}{k})^n - 1}} \]Example (Equity in a House): A couple bought a house \(11\) years ago for \(\$225\,000\). They put down \(10\%\) and took out a \(15\)-year mortgage at \(5.75\%\) (with monthly compounding) for the remaining balance. Suppose that the current net market value of their house (its current market value minus selling costs) is now \(\$350\,000\).
Solution: The loan is \(A_n = 225000-22500 = \$202\,500\).
Hypotheses:
If you hold the stock in perpetuity (indefinitely) rather than for \(n\) years, then there is no terminal price, and the stock's present share price becomes:
Example: Suppose that a preferred stock has a fixed total annual cash dividend per share of \(\$2.50\). Assume an annual required return of \(13\%\) for the stock. How much should you pay for the preferred stock according to the DDM?
Example: Now, suppose the total cash dividend of a common stock last year was \(\$2.75\) per share and dividends are expected to increase at \(3\%\) per annum. If the annual required return rate is \(10\%\), find the share price of the stock today according to the DDM.
Proposition:
Example: A \(10\)-year bond has maturity value \(\$1\,000\) and coupon rate \(6\%\). Find the market value of the bond \(6\) years after it was issued when \(\text{YTM}=10\%, 5\%, 6\%\).
Example: Suppose that a \(30\)-year bond with \(3\%\) coupon rate was issued by the US Treasury today. If in \(10\) years the Fed raises interest rates and another \(30\)-year bond with an annual \(6\%\) coupon rate is issued, no investors will buy a bond with \(3\%\) annual yield. So the current yield of the first bond will be forced to approach \(6\%\) on the issue date of the second bond under the law of supply and demand.
Consider a portfolio of \(2\) securities with unit prices \(S_1(t)\) and \(S_2(t)\). The return rates \(R_i\) over \([t_0, t_f]\) are random variables defined by:
\[ R_i = \underbrace{\frac{S_i(t_f) - S_i(t_0)}{S_i(t_0)}}_{\text{capital gain}} + \underbrace{\frac{D_i}{S_i(t_0)}}_{\text{dividend yield}} \quad (i = 1, 2) \]Properties:
Portfolio Weights: \(V_P(t_0)\) is the total investment, with the weights for each security being:
\[ w_1 = \frac{n_1S_1(t_0)}{V_P(t_0)}, \quad w_2 = \frac{n_2S_2(t_0)}{V_P(t_0)}, \quad \text{with } w_1 + w_2 = 1. \]Note: We allow short selling, meaning weights can be negative (\(w_i < 0\)) or greater than 1.
To simplify, let \(w = w_1\) and \(1-w = w_2\). The portfolio properties are:
\[ \begin{aligned} \text{Return: } R_P(w) &= wR_1 + (1-w)R_2 \\ \text{Exp. Return: } \mu_P(w) &= w\mu_1 + (1-w)\mu_2 \\ \text{Variance: } \sigma_P(w)^2 &= w^2\sigma_1^2 + 2w(1-w)\sigma_{12} + (1-w)^2\sigma_2^2 \end{aligned} \]Recall that \(\text{Var}(aX+bY) = a^2\text{Var}(X) + 2ab\text{Cov}(X,Y) + b^2\text{Var}(Y)\).
Feasible & Efficient Sets:
We seek the relationship between \(\sigma_P\) and \(\mu_P\) independent of \(w\).
This implies the points \((\sigma_P, \mu_P)\) lie on a hyperbola, provided \(\rho \neq \pm 1\). (If \(\rho = \pm 1\), the relation degenerates to straight lines).
The set \(F_P\) of feasible portfolios satisfies the hyperbola equation:
\[ \sigma_P^2 = \mathbb{A}\mu_P^2 + \mathbb{B}\mu_P + \mathbb{C} \]where the constants are:
\[ \begin{aligned} \mathbb{A} &= \frac{\sigma_1^2 - 2\rho \sigma_1 \sigma_2 + \sigma_2^2}{(\mu_1 - \mu_2)^2} \\ \mathbb{B} &= -2\frac{(\sigma_2^2 - \rho \sigma_1 \sigma_2)\mu_1 + (\sigma_1^2 - \rho \sigma_1 \sigma_2)\mu_2}{(\mu_1 - \mu_2)^2} \\ \mathbb{C} &= \frac{\sigma_1^2\mu_2^2 + \sigma_2^2\mu_1^2 - 2\rho \sigma_1 \sigma_2 \mu_1 \mu_2}{(\mu_1 - \mu_2)^2} \end{aligned} \]
Global Minimum Variance Portfolio (\(w_G\))
Since the variance \(\sigma_P^2(w) = aw^2 + bw + c\) is a parabola opening upward, the minimum occurs at the vertex \(w = -b/2a\).
Note: The Efficient Frontier \(M_E\) corresponds to the top portion of the hyperbola (all points above the global minimum variance portfolio).
The portfolio risk is:
\[ \sigma_P = \sqrt{w^2\sigma_1^2 + 2w(1-w)\rho \sigma_1 \sigma_2 + (1-w)^2 \sigma_2^2} \]Consider a portfolio of \(N\) securities with unit prices \(S_i\), for \(i= 1\) to \(N\). For each security, consider its return rate over \([t_0, t_f]\):
\[ R_i = \frac{S_i(t_f) - S_i(t_0)}{S_i(t_0)} + \frac{D_i}{S_i(t_0)} \quad (i= 1, \dots, N) \]As before, \(S_i, R_i, D_i\) are random variables.
What percentage of the money \(V_P(t_0)\) should you allocate today to each security to create an efficient portfolio?
\[ V_P(t_0) = \sum_{i=1}^n n_iS_i(t_0) = \underbrace{\frac{n_iS_i(t_0)}{V_P(t_0)}}_{w_i: \text{ weights}} V_P(t_0) \]Now, using vectors, we define:
\[ \vec{w} = \begin{bmatrix} w_1 \\ \vdots \\ w_N \end{bmatrix}, \vec{e} = \begin{bmatrix} 1 \\ \vdots \\ 1 \end{bmatrix} \]Note that \(\vec{w}^T\vec{e} = \vec{w} \cdot \vec{e} = w_1 + \dots + w_N = 1\).
\[ \begin{aligned} \text{Space of weights without short selling: } W_N^* &= \{\vec{w} \mid \vec{w}^T\vec{e} = 1, w_i \ge 0\} \\ \text{Space of weights with short selling: } W_N &= \{\vec{w} \mid \vec{w}^T\vec{e} = 1\} \end{aligned} \]We now have the following:
\[ \begin{aligned} \textbf{Portfolio Return: } R_P &= \sum_{i=1}^N w_iR_i = \vec{w}^T\vec{R} \\ \textbf{Expected Return: } \mu_P &= \sum_{i=1}^N w_i \mu_i = \vec{w}^T \vec{\mu} \\ \textbf{Portfolio Variance: } \sigma_P^2 &= \sum_{i=1}^N w_i^2 \sigma_i^2 + 2 \sum_{1 \leq i < j \le N} w_iw_j\underbrace{\rho \sigma_i \sigma_j}_{\sigma_{ij}} = \vec{w}^T V\vec{w} \end{aligned} \]Where:
\[ \vec{R} = \begin{bmatrix} R_1 \\ \vdots \\ R_N \end{bmatrix}, \vec{\mu} = \begin{bmatrix} \mu_1 \\ \vdots \\ \mu_N \end{bmatrix}, V = \begin{bmatrix} \sigma_1^2 & \dots & \sigma_{1N} \\ \vdots & \ddots & \vdots \\ \sigma_{N1} & \dots & \sigma_N^2 \end{bmatrix} = \begin{bmatrix} \sigma_{ij} \end{bmatrix} \]Notice:
- \(V\) is symmetric: \(V^T = V\) or \(\sigma_{ij} = \sigma_{ji}\).
- \(V\) is positive semi-definite: \(\vec{w}^TV\vec{w} \ge 0\).
- If \(V\) is positive definite: \(\vec{w}^T V \vec{w} > 0, \vec{w} \neq 0\), then \(V\) is invertible and \(V^{-1}\) is also symmetric.
To find the equation of the efficient frontier, we find the minimum variance portfolio for a given expected return \(\mu\).
Lagrange Multipliers: The minimum (or maximum) of a function \(f(\vec{x})\) under two constraints \(g(\vec{x}) = c_1\) and \(h(\vec{x}) = c_2\) occurs at point where
\[ \frac{\partial f}{\partial \vec{x}} = \lambda_1 \frac{\partial g}{\partial \vec{x}} + \lambda_2 \frac{\partial h}{\partial \vec{x}} \]Here, we minimize:
Derivation Steps:
Substituting the optimal weight vector \(\vec{w}\) back into the variance equation \(\sigma_P^2 = \vec{w}^T V \vec{w}\) yields the variance as a function of return \(\mu\):
\[ \begin{aligned} \sigma_P^2 &= \vec{w}^T (\lambda_1 \vec{e} + \lambda_2 \vec{\mu}) = \lambda_1(1) + \lambda_2(\mu) \\ &= \frac{C - B\mu}{AC - B^2} + \mu \frac{A\mu - B}{AC - B^2}\\ &= \frac{A\mu_P^2 - 2B\mu_P + C}{AC - B^2} \end{aligned} \]The Efficient Frontier (\(M_E\)) is the top part of this hyperbola.
To find the GMV portfolio, we minimize \(\sigma_P^2\) with respect to \(\mu_P\). Setting \(\frac{d\sigma_P^2}{d\mu_P} = 0\):
\[ \begin{aligned} 2A\mu_G - 2B &= 0 \implies \mu_G = \frac{B}{A} \\ \sigma_G^2 &= \frac{A(B/A)^2 - 2B(B/A) + C}{AC - B^2} = \frac{1}{A} \end{aligned} \]The weights of the GMV portfolio are found by plugging \(\mu = \frac{B}{A}\) into the weight formula. The term associated with \(V^{-1}\vec{\mu}\) vanishes (\(A(\frac{B}{A})-B = 0\)):
\[ \boxed{\vec{w}_G = \frac{1}{A}V^{-1}\vec{e}} \]The question of where to invest on \(M_E\) depends on the investor's risk tolerance, which can be measured using a utility function.
Definition: If \(x = \text{return (rate)}\) of a portfolio on \(M_E\), then a utility function is a function \(u(x)\) such that \(x_1 < x_2 \implies u(x_1) < u(x_2)\) (increasing), that measures the “degree of satisfaction”.
We want to find the optimal portfolio, which maximizes \(\mathbb{E}(u(R_p))\).
Example: Consider \(u(x) = ax-\frac{b}{2}x^2\), where \(a, b > 0, x < \frac{a}{b}\).
When \(u(x)=x^a\) or \(u(x) = x^{\frac{1}{3}}\), then we can use a Taylor approximation of order \(2\) about \(x = \mu_p\):
\[ u(x) \approx u(\mu_p) + u'(\mu_p)(x-\mu_p) + \frac{1}{2}u''(\mu_p)(x-\mu_p)^2 \]As \(x \to R_p\):
\[ \begin{aligned} u(R_p) &\approx u(\mu_p) + u'(\mu_p)(R_p-\mu_p) + \frac{1}{2}u''(\mu_p)(R_p-\mu_p)^2 \\ \mathbb{E}(u(R_p)) &\approx u(\mu_p) + u'(\mu_p) \cdot 0 + \frac{1}{2}u''(\mu_p)\sigma_p^2 \end{aligned} \]Hence, we must maximize \(\mathbb{E}(u(R_p)) = u(\mu_p) + \frac{1}{2}u''(\mu_p)\sigma_p^2\).
Example (2-Security Portfolio):
\[ \begin{aligned} \text{Security 1: }& \mu_1 = 13\%, \sigma_1 = 15\% \\ \text{Security 2: }& \mu_2 = 14\%, \sigma_2 = 20\% \quad (\rho_{12} = -.3) \end{aligned} \]An investor with utility function \(u(x) = x^{\frac{1}{3}}\) (risk averse) wants to invest \(1\) million dollars. Find their optimal portfolio.
We consider portfolios with \(N\) risky securities: \((\sigma_i, \mu_i)\) and a risk-free security with risk free rate \(r\): \((0, r)\). This is equivalent to portfolios with two securities:
Let \((\sigma_P, \mu_P)\) be the portfolio with weights \((w,1-w)\), with \(w\) in the risk-free security and \(1-w\) in a portfolio of the \(N\) securities.
Rearranging, we get the equation of a line:
\[ \boxed{\mu_P = \underbrace{\left( \frac{\mu_B - r}{\sigma_B} \right)}_{\text{Sharpe Ratio}} \cdot \sigma_P + r} \]Let \(M_{E, N}\) be the efficient frontier for only the \(N\) securities. The efficient frontier \(M_{E, N+1}\) for the \(N\) securities and the risk-free security is the highest CAL, the one that is tangent to \(M_{E, N}\).
Let \((\sigma_M, \mu_M)\) be the portfolio on \(M_E\) that makes the CAL through \((0, r)\) and \((\sigma_M, \mu_M)\) tangent to \(M_E\). The CAL is then called the Capital Market Line (CML) and \((\sigma_M, \mu_M)\) is called the market portfolio. \(M_{E, N+1} = \) CML.
Example: Suppose that you have \(\$2\,000\) to invest. Assume a risk-free rate of \(6\%\) and market expected return of \(12\%\).
Exercise: Find \(\mu_M\) in terms of \(A, B, C, V, \vec{\mu}, r\).
\[ \text{Line (CML):} \quad \mu_P = \frac{\mu_M - r}{\sigma_M}\sigma_P + r \tag{3} \] \[ \text{Frontier (Hyperbola):} \quad \sigma_P^2 = \frac{A\mu_P^2 - 2B\mu_P + C}{AC - B^2} \tag{4} \]At the tangency point \(M\), the slopes must be equal. We choose run over rise, so from (3), the inverse slope is:
\[ \frac{d\sigma_P}{d\mu_P}\bigg|_{(\sigma_M, \mu_M)} = \frac{\sigma_M}{\mu_M - r} \]Now, we differentiate (4) with respect to \(\mu_P\):
\[ \frac{d}{d\mu_P}(\sigma_P^2) = \frac{d}{d\mu_P}\left( \frac{A\mu_P^2 - 2B\mu_P + C}{AC - B^2} \right) \] \[ 2\sigma_P \cdot \frac{d\sigma_P}{d\mu_P} = \frac{2A\mu_P - 2B}{AC - B^2} \]Evaluating at point \(M\):
\[ \sigma_M \frac{d\sigma_P}{d\mu_P} = \frac{A\mu_M - B}{AC - B^2} \implies \frac{d\sigma_P}{d\mu_P} = \frac{1}{\sigma_M} \frac{A\mu_M - B}{AC - B^2} \]Now, set the geometric slope equal to the derivative slope:
\[ \begin{aligned} \frac{\sigma_M}{\mu_M - r} &= \frac{1}{\sigma_M} \frac{A\mu_M - B}{AC - B^2} \\ (AC - B^2)\sigma_M^2 &= (\mu_M - r)(A\mu_M - B) \end{aligned} \]Substitute (4) back in for the LHS and expand the RHS:
\[ \begin{aligned} A\mu_M^2 - 2B\mu_M + C &= A\mu_M^2 - B\mu_M - Ar\mu_M + Br \\ -2B\mu_M + C &= - B\mu_M - Ar\mu_M + Br \\ \mu_M(B - Ar) &= C - Br \end{aligned} \]Our final result is \(\boxed{\mu_M = \frac{C - Br}{B - Ar}}\).
Definition: For a security \((R_i, \mu_i, \sigma_i)\), we define its beta as
\[ \beta_i = \frac{\text{Cov}(R_i, R_M)}{\sigma_M^2} \]where \((R_M, \mu_M, \sigma_M)\) is the market portfolio.
Definition: If \(r\) is the risk free rate then
Theorem (CAPM): Assuming \(V\) is positive definite, \(\vec{\mu}\) and \(\vec{e}\) are linearly independent and \(0 \le r < \mu_G\), then
\[ \beta_i = \frac{\mu_i - r}{\mu_M - r} \implies \mu_i = r + \beta_i(\mu_M - r) \]A point under the Security Market Line is overpriced, while a point above is underpriced.
Proof:
\[ \beta_i = \frac{\text{Cov}(R_i, R_M)}{\text{Var}(R_M)} = \frac{\text{Cov}(R_i, R_M)}{\text{Cov}(R_M, R_M)} \]To compute these covariances, recall:
Substitute the matrix forms into the Beta fraction:
\[ \beta_i = \frac{\vec{e}_i^T V \vec{w}_M}{\vec{w}_M^T V \vec{w}_M} \]Now, substitute the definition of \(\vec{w}_M\) into the \(V \vec{w}_M\) term:
\[ V \vec{w}_M = V \left( V^{-1} \frac{\vec{\mu} - r\vec{e}}{B - Ar} \right) = \frac{\vec{\mu} - r\vec{e}}{B - Ar} \]Substitute this back into the numerator and denominator:
\[ \beta_i = \frac{\vec{e}_i^T \left( \frac{\vec{\mu} - r\vec{e}}{B - Ar} \right)}{\vec{w}_M^T \left( \frac{\vec{\mu} - r\vec{e}}{B - Ar} \right)} = \frac{\vec{e}_i^T (\vec{\mu} - r\vec{e})}{\vec{w}_M^T (\vec{\mu} - r\vec{e})} \]Distribute the transpose vectors:
We get the final result: \(\beta_i = \frac{\mu_i - r}{\mu_M - r}\).
Recall that the return of a security over the interval \([t_0, t_f]\) is:
\[ R_{t_0}(t_f) = \frac{S(t_f) - S(t_0) + D_{t_0}(t_f)}{S(t_0)} \]We apply the expectation operator \(\mathbb{E}\) to find the expected return \(\mu_i\):
\[ \mu_i = \mathbb{E}[R] = \frac{\mathbb{E}[S(t_f)] - S(t_0) + \mathbb{E}[D_{t_0}(t_f)]}{S(t_0)} \]We rearrange the equation, solving for the current price \(S(t_0)\):
\[ \begin{aligned} \mu_i S(t_0) &= \mathbb{E}[S(t_f)] - S(t_0) + \mathbb{E}[D_{t_0}(t_f)] \\ S(t_0) &= \frac{\mathbb{E}[S(t_f)] + \mathbb{E}[D_{t_0}(t_f)]}{1 + \mu_i} \end{aligned} \]From the CAPM Theorem, we know \(\mu_i = r + \beta_i(\mu_M - r)\). Plugging this into the denominator, we get the pricing formula:
\[ \boxed{S(t_0) = \frac{\mathbb{E}[S(t_f)] + \mathbb{E}[D_{t_0}(t_f)]}{1 + r + \beta_i(\mu_M - r)}} \]We fix a time interval \([t_0, t_f]\) and let \(\tau = t_f - t_0\). Next, we divide \([t_0, t_f]\) into \(n\) subintervals \([t_0, t_1], \dots, [t_{n-1}, t_n]\) of the same length \(h_n = \frac{\tau}{n}\).
Let \(S(t)\) be the price of the security at time \(t\), then we denote \(S_j = S(t_j)\). \(S_0\) is assumed known (current price), and \(S_1, \dots, S_n\) are random variables.
The gross returns are independent random variables:
\[ \frac{S_1}{S_0}, \frac{S_2}{S_1}, \dots, \frac{S_n}{S_{n-1}} \text{ (i.i.d)} \]where:
\[ \frac{S_j}{S_{j-1}} = \begin{cases} u_n \text{ with probability } p_n \\ d_n \text{ with probability } 1 - p_n \end{cases} \]If \(N_U\) is the random variable giving the number of upticks in the price, then \(N_U\) follows a \(\text{Binomial}(n, p_n)\):
\[ \mathbb{P}(N_U = k) = {n\choose k} p_n^k (1-p_n)^{n-k} = \mathbb{P}(S_n = S_0u_n^kd_n^{n-k}) \]The expected value of the future price \(S_n\) can be found in two ways:
Example: Suppose a stock is worth \(\$100\) today. Using a \(20\)-period binomial tree over \(1\) month, with \(u_{20} = 1.02, d_{20} = .98, p_{20} = .53\):
Solution:
The CRR tree model assumes that \(u_nd_n = 1\).
Proposition 1: \(\boxed{\mu_nh_n = p_n\ln u_n + (1-p_n)\ln d_n}\).
Proof: Recall the following:
\[ \mu_n = \frac{\mathbb{E}(\ln \frac{S_1}{S_0})}{h_n}, \frac{S_j}{S_{j-1}} = \begin{cases} u_n \text{ with probability } p \\ d_n \text{ with probability } 1-p \end{cases} \]We know:
\[ \ln \frac{S_1}{S_0} = \begin{cases} \ln u_n \text{ with probability } p \\ \ln d_n \text{ with probability } 1-p \end{cases} \]As such, \(\mu_nh_n = \mathbb{E}(\ln \frac{S_1}{S_0}) = p_n \ln u_n + (1-p_n)\ln d_n\).
Proposition 2: \(\boxed{\sigma_n^2 h_n = p_n(1-p_n)[\ln \frac{u_n}{d_n}]^2}\).
Proof: We know:
\[ (\ln \frac{S_1}{S_0})^2 = \begin{cases} (\ln u_n)^2 \text{ with probability } p \\ (\ln d_n)^2 \text{ with probability } 1-p \end{cases} \]As well, since \(\sigma_n^2 = \frac{\text{Var}(\ln \frac{S_1}{S_0})}{h_n}\):
\[ \begin{aligned} \sigma_n^2 h_n &= \text{Var}(\ln \frac{S_1}{S_0}) = \mathbb{E}[(\ln \frac{S_1}{S_0})^2] - [\mathbb{E}(\ln \frac{S_1}{S_0})]^2 \\ &= p_n(\ln u_n)^2 + (1-p_n)(\ln d_n)^2 - (p_n \ln u_n + (1-p_n)\ln d_n)^2 \\ &= p_n (1-p_n)[(\ln u_n)^2 + (\ln d_n)^2 - 2 \ln u_n \ln d_n] \\ &= p_n(1-p_n)[\ln u_n - \ln d_n]^2 \end{aligned} \]Proposition 3: \(\boxed{\mu = m - q - \frac{\sigma^2}{2}}\).
Proof: Let \(h_n = t_1 - t_0\) and \(q\) be the dividend rate such that \(D(t_0, t_1) = q S_0 h_n\). Define the return \(R_1 = \frac{S_1 - S_0}{S_0} = \frac{S_1}{S_0} - 1 \implies \frac{S_1}{S_0} = 1 + R_1\). Using the Taylor expansion \(\ln(1+x) \approx x - \frac{x^2}{2}\) (for \(x \approx 0\)):
\[ \begin{aligned} h_n \mu_n &= \mathbb{E}(\ln \frac{S_1}{S_0}) = \mathbb{E}(\ln(1+R_1)) \\ &\approx \mathbb{E}(R_1 - \frac{R_1^2}{2}) \\ &= \mathbb{E}(R_1) - \frac{1}{2}\mathbb{E}(R_1^2) \end{aligned} \]We evaluate the terms:
Substituting back:
\[ h_n \mu_n \approx (m-q)h_n - \frac{1}{2}\sigma^2 h_n \implies \mu \approx m - q - \frac{\sigma^2}{2} \]Theorem: Given a security with volatility \(\sigma\), drift \(\mu\), expected return \(m\), and dividend rate \(q\), the parameters \(u_n, d_n\) and \(p_n\) for a CRR tree are given by:
- \(u_n \approx e^{\sigma\sqrt{h_n}}\)
- \(d_n \approx e^{-\sigma\sqrt{h_n}}\)
- \(p_n \approx \frac{1}{2}\left(1 + \frac{\mu}{\sigma}\sqrt{h_n}\right) \approx \frac{e^{(m-q)h_n} - d_n}{u_n - d_n}\)
Proof of (a) and (b): We impose the symmetry constraint \(u_n d_n = 1 \implies \ln d_n = - \ln u_n\). From Proposition 2:
\[ \sigma^2 h_n = p_n(1-p_n)[\ln u_n - \ln d_n]^2 = p_n(1-p_n)[2 \ln u_n]^2 \]Assuming \(p_n \approx \frac{1}{2}\) (to first order):
\[ \begin{aligned} \sigma^2 h_n &\approx \frac{1}{2}\left(1-\frac{1}{2}\right) 4 (\ln u_n)^2 \\ \sigma^2 h_n &\approx (\ln u_n)^2 \implies \ln u_n \approx \sigma\sqrt{h_n} \end{aligned} \]Thus, \(u_n \approx e^{\sigma\sqrt{h_n}}\) and \(d_n = 1/u_n \approx e^{-\sigma\sqrt{h_n}}\).
Proof of (c) (Derivation of Exact Formula): We solve for \(p_n\) by matching the expected future price.
\[ \begin{aligned} \mathbb{E}\left(\frac{S_1}{S_0}\right) &= p_n u_n + (1-p_n)d_n \\ \mathbb{E}(1+R_1) &\approx 1 + (m-q)h_n \approx e^{(m-q)h_n} \end{aligned} \]Equating the two expressions:
\[ \begin{aligned} p_n u_n + d_n - p_n d_n &= e^{(m-q)h_n} \\ p_n(u_n - d_n) &= e^{(m-q)h_n} - d_n \\ p_n &= \frac{e^{(m-q)h_n} - d_n}{u_n - d_n} \end{aligned} \]Proof of (c) (Derivation of Approximation): Alternatively, using Proposition 1 and Parts A/B:
\[ \begin{aligned} \mu h_n &= p_n \ln u_n + (1-p_n) \ln d_n \\ \mu h_n &= p_n (\sigma\sqrt{h_n}) + (1-p_n)(-\sigma\sqrt{h_n}) \\ \mu h_n &= \sigma\sqrt{h_n} (2p_n - 1) \end{aligned} \]Solving for \(p_n\):
\[ 2p_n - 1 \approx \frac{\mu\sqrt{h_n}}{\sigma} \implies p_n \approx \frac{1}{2}\left(1 + \frac{\mu}{\sigma}\sqrt{h_n}\right) \]Question: Does the CRR tree model converge as \(n \to \infty\) to a continuous model?
Let the total time be \(t\). We divide the interval \([0, t]\) into \(n\) steps of length \(h_n = \frac{t}{n}\). The price at time \(t\), \(S_n(t)\), can be written as the product of independent increments:
\[ S_n(t) = S_0 \cdot \frac{S_1}{S_0} \cdot \frac{S_2}{S_1} \dots \frac{S_n}{S_{n-1}} \implies \ln \frac{S_n(t)}{S_0} = \sum_{j=1}^{n} \ln \frac{S_j}{S_{j-1}} \]Let \(Y_{n,j} = \ln \frac{S_j}{S_{j-1}}\) be the log returns, which are i.i.d for fixed \(n\). We define the standardized log return:
\[ X_{n,j} = \frac{Y_{n,j} - \mathbb{E}(Y_{n,j})}{\sqrt{\text{Var}(Y_{n,j})}} \]From previous propositions, we know \(\mathbb{E}(Y_{n,j}) \approx \mu h_n\) and \(\text{Var}(Y_{n,j}) \approx \sigma^2 h_n\). Substituting in:
\[ \ln \frac{S_n(t)}{S_0} \approx \sum_{j=1}^{n} (\mu h_n + \sigma\sqrt{h_n} X_{n,j}) = \mu (n h_n) + \sigma\sqrt{n h_n} \left( \frac{1}{\sqrt{n}} \sum_{j=1}^{n} X_{n,j} \right) \]Since \(n h_n = t\):
\[ S_n(t) = S_0 \exp \left[ \mu t + \sigma\sqrt{t} \left( \frac{1}{\sqrt{n}} \sum_{j=1}^{n} X_{n,j} \right) \right] \]We cannot use the classic Central Limit Theorem (CLT) because we have a triangular array of random variables (the distribution of \(X_{n,j}\) depends on \(n\)). Instead, we use the Lindeberg Central Limit Theorem (LCLT).
Theorem (LCLT): For a triangular array of independent random variables \(X_{n,j}\), if:
Then:
\[ \frac{1}{\sqrt{n}} \sum_{j=1}^{n} X_{n,j} \xrightarrow{d} Z \sim N(0,1) \]Checking the Condition: In the CRR model, \(X_{n,j}\) takes values relating to \(\ln u_n\) and \(\ln d_n\). As \(n \to \infty\), \(h_n \to 0\), so \(\ln u_n \approx \sigma\sqrt{h_n} \to 0\). Thus, for large \(n\), \(|X_{n,j}| < \epsilon\), satisfying the condition.
Conclusion: As \(n \to \infty\):
\[ S_n(t) \to S_t = S_0 \exp(X_t) \quad \text{where } X_t \sim N(\mu t, \sigma^2 t) \]\(S_t\) is a Lognormal random variable.
Definition: \(S_t\) is a continuous function of time satisfying:
Exercise: Let \(S_t = S_0 e^{X_t}\) where \(X_t \sim N(\mu t, \sigma^2 t)\). We know if \(X \sim N(a, b^2)\), then \(\mathbb{E}(e^X) = e^{a + \frac{b^2}{2}}\):
Definition: A collection of random variables \((X_n)_{n \ge 0}\) (discrete time) or \((X_t)_{t \ge 0}\) (continuous time) is called a stochastic process.
Definition: Given \(R_1, R_2, \dots\) i.i.d. random variables, the stochastic process
\[ W_k = \begin{cases} w_0 + R_1 + \dots + R_k & \text{if } k \ge 1 \\ w_0 & \text{if } k = 0 \end{cases} \]is called a random walk.
Consider the following game: flip a fair coin \(a\) times (\(n\) flips per unit time). The winnings for the \(k^{th}\) flip are:
\[ R_k = \begin{cases} +\frac{1}{\sqrt{n}} & \text{with probability } \frac{1}{2} \\ -\frac{1}{\sqrt{n}} & \text{with probability } \frac{1}{2} \\ \end{cases} \]Then, let \((W_k)_{k \ge 0}\) be the process defined by \(W_k = \sum_{i=1}^k R_i\) (with \(W_0=0\)), representing total winnings after \(k\) flips.
Let \((W^{(n)}(t))_{t \ge 0}\) be the continuous time extension of \((W_k)_{k \ge 0}\) such that if \(t_k = \frac{k}{n}\):
\[ \begin{cases} W^{(n)}(t_k) = W_k \\ W^{(n)}(t) \text{ is affine on } [t_k, t_{k+1}] \implies W^{(n)}(t) = W_k + \frac{t - t_k}{1/n}(W_{k+1} - W_k) \end{cases} \]Theorem: \((W^{(n)}(t))_{t \ge 0}\) converges (in distribution) to a stochastic process \((B(t))_{t \ge 0}\) called Standard Brownian Motion (SBM) or Wiener Process.
Definition: The stochastic process \((B(t))_{t \ge 0}\) is called Standard Brownian Motion if:
- \(B(0) = 0\).
- \(B(t)\) is continuous a.s.
- \(B(t) - B(s) \sim N(0, t-s)\), for all \(0 \le s < t\).
- \(B(t)\) has independent increments.
| SRW (\(W\)) | SBM (\(B\)) |
|---|---|
| \(t_k = \frac{k}{n}\) | \(t\) |
| \(t_{k+1} - t_k = \frac{1}{n}\) | \(dt\) |
| \(W_k = W(t_k)\) | \(B(t)\) |
| \(R_k = W_{k+1} - W_k\) | \(dB(t) = B(t+dt) - B(t)\) |
| \(R_i\) i.i.d. | \(dB(t_i)\) i.i.d. |
| \(\mathbb{E}(W_k) = 0, \quad \text{Var}(W_k) = t_k\) | \(\mathbb{E}(B(t)) = 0, \quad \text{Var}(B(t)) = t\) |
| Markov & Martingale | Markov & Martingale |
| QV: \(\sum (W(t_i) - W(t_{i-1}))^2 = t_k\) | QV: \(\sum (B(t_i) - B(t_{i-1}))^2 \xrightarrow[m.s.]{} t\) |
Definition: The stochastic process \((X(t))_{t \ge 0}\) is called a Brownian motion with drift and scaling if:
- \(X(0) = x_0\)
- \(X(t)\) is continuous a.s.
- \(X(t) - X(s) \sim N(\mu(t-s),\sigma^2(t-s))\), for all \(0 \leq s < t\).
- \(X(t)\) has independent increments.
The drift is \(\mu\). Note: \(X(t) \sim N(\mu t, \sigma^2 t) \implies \mathbb{E}(X(t)) = \mu t\).
If \(B\) is a Standard Brownian Motion, then:
\[ X(t) \stackrel{\text{d}}{=} x_0 + \mu t + \sigma B(t) \sim N(x_0 + \mu t, \sigma^2 t) \]This process satisfies the stochastic differential equation (SDE):
\[ \begin{aligned} dX(t) &= X(t+dt) - X(t) \\ &= \mu dt + \sigma (B(t+dt) - B(t)) \\ &= \boxed{\mu dt + \sigma dB(t)} \end{aligned} \]Note: If \(\mu = 0, \sigma = 1\), then \(dX(t) = dB(t) \implies X(t) = B(t)\).
Recall the continuous time model for security prices from Chapter 5 satisfies:
Let \(X(t) = \ln \frac{S_t}{S_0}\). Since \(X(0)=0\), \(X\) is continuous, and increments are normally distributed, \(X(t)\) is a Brownian motion with drift and scaling.
\[ \begin{aligned} X(t) &= \ln \frac{S_t}{S_0} \implies S_t = S_0e^{X(t)} = S_0e^{\mu t + \sigma B(t)} \end{aligned} \]Definition: The stochastic process \((S(t))_{t \ge 0}\) is called a Geometric Brownian Motion (GBM) if
\[ S(t) = S_0e^{X(t)} \]where \((X(t))_{t \ge 0}\) is a Brownian motion with drift and scaling.
We seek coefficients for \(dS(t) = (\dots) dt + (\dots) dB_t\).
1. Setup: \(\ln \frac{S_{t+dt}}{S_t} \sim N(\mu dt, \sigma^2 dt) \implies \frac{S_{t+dt}}{S_t} \stackrel{d}{=} e^{\mu dt + \sigma dB_t}\).
2. Expansion: Using Taylor series \(e^x \approx 1 + x + \frac{x^2}{2}\) with \(x = \mu dt + \sigma dB_t\):
\[ \begin{aligned} \frac{S_{t+dt}}{S_t} &\approx 1 + (\mu dt + \sigma dB_t) + \frac{1}{2}(\mu dt + \sigma dB_t)^2 \end{aligned} \]3. Apply Ito Rules: Recall \((dt)^2 \to 0\), \(dt \cdot dB_t \to 0\), and \((dB_t)^2 \to dt\).
\[ \begin{aligned} \frac{S_{t+dt}}{S_t} &= 1 + \mu dt + \sigma dB_t + \frac{1}{2}\sigma^2 (dB_t)^2 \\ &= 1 + \left(\mu + \frac{\sigma^2}{2}\right)dt + \sigma dB_t \end{aligned} \]4. Result: Rearranging for \(dS_t = S_{t+dt} - S_t\):
\[ \boxed{dS_t = S_t \underbrace{\left(\mu + \frac{\sigma^2}{2}\right)}_{\text{Drift Term}} dt + S_t \underbrace{\sigma}_{\text{Diffusion}} dB_t} \]Property: \((S(t))_{t \ge 0}\) is a GBM such that \(S(t) \stackrel{\text{d}}{=} S_0e^{\mu t + \sigma B(t)}\) iff
\[ dS(t) = \left( \mu + \frac{\sigma^2}{2}\right) S(t) dt + \sigma S(t)dB(t) \]Notation Note: Often written as \(dS_t = (m-q)S_t dt + \sigma S_t dB_t\), where \(m-q = \mu + \frac{\sigma^2}{2}\).
Example (Naive/Incorrect Approach): Solving \(dS_t = (m-q)S_t dt + \sigma S_t dB_t\).
Ito's Lemma: Let \(X\) be a process satisfying \(dX(t) = a(X,t) dt + b(X,t) dB(t)\). Let \(f(x,t)\) be continuously differentiable in \(t\) and twice in \(x\). Let \(Y(t)=f(X(t),t)\). Then:
\[ \begin{aligned} dY(t) &= \left( \frac{\partial f}{\partial t} + a \frac{\partial f}{\partial x} + \frac{1}{2}b^2 \frac{\partial^2f}{\partial x^2}\right)dt + b \frac{\partial f}{\partial x} dB(t) \end{aligned} \]In short: \(df = (f_t + af_x + \frac{1}{2}b^2f_{xx})dt + bf_xdB\).
Cases:
1. Find \(d(B^2(t))\):
\(f(x,t) = x^2\) and \(X=B\) (\(a=0, b=1\)).
2. Find \(d(t + e^{B(t)})\):
\(f(x,t) = t + e^x\) and \(X=B\) (\(a=0, b=1\)).
3. Find \(d(\ln S(t))\) for GBM:
\(f(S,t) = \ln S\). Process is \(dS = (m-q)S dt + \sigma S dB\) (so \(a=(m-q)S, b=\sigma S\)).
Applying Ito:
\[ \begin{aligned} d(\ln S) &= \left( 0 + (m-q)S \frac{1}{S} + \frac{1}{2}(\sigma S)^2 \left(-\frac{1}{S^2}\right) \right) dt + \sigma S \frac{1}{S} dB \\ &= \boxed{\left( m - q - \frac{\sigma^2}{2} \right) dt + \sigma dB} \end{aligned} \]To solve \(dS(t) = (m-q)S(t)dt + \sigma S(t)dB(t)\), we integrate the result from Example 3 above.
\[ \begin{aligned} \int_0^t d(\ln S) &= \int_0^t \left( m - q - \frac{\sigma^2}{2} \right) dt + \int_0^t \sigma dB \\ \ln S_t - \ln S_0 &= \left( m - q - \frac{\sigma^2}{2} \right)t + \sigma B_t \end{aligned} \]Exponentiating both sides:
\[ \boxed{S_t = S_0 e^{\left( m - q - \frac{\sigma^2}{2} \right)t + \sigma B_t}} \]Consider a partition \(0 = t_0 < \dots < t_n = t\). The Ito integral of a process \(f\) is:
\[ \int_{0}^{t}f(s)dB(s) = \lim_{n \to \infty}\sum_{i=1}^{n}f(t_{i-1})(B(t_i) - B(t_{i-1})) \]Key: The integrand is evaluated at the left endpoint \(t_{i-1}\) .
Example: Calculate \(\int_0^t B(s) dB(s)\). Using the identity \(x^2-y^2 = (x-y)^2 + 2y(x-y)\) on the sum \(\sum (B_{t_i}^2 - B_{t_{i-1}}^2)\):
\[ B(t)^2 - B(0)^2 = \underbrace{\sum (B_{t_i} - B_{t_{i-1}})^2}_{\to t \text{ (Quad. Var.)}} + 2 \underbrace{\sum B_{t_{i-1}}(B_{t_i} - B_{t_{i-1}})}_{\to \int B dB} \] \[ \implies B^2(t) = t + 2 \int_0^t B(s) dB(s) \implies \boxed{\int_0^t B(s) dB(s) = \frac{B^2(t)}{2} - \frac{t}{2}} \]SDE in Integral Form: \(dX(t) = a dt + b dB(t)\) is shorthand for:
\[ X(t) - X(0) = \int_{0}^{t}a(X(s),s) ds + \int_{0}^{t}b(X(s),s)dB(s) \]Properties of Ito's Integral (\(\mathcal{I}(t) = \int_{0}^{t} f(s)dB(s)\)):
European Options:
American Options: Can be exercised at any time prior to the exercise date.
Law of One Price: Suppose investment \(A\) costs \(C_A\) and investment \(B\) costs \(C_B\). If the payoff from investment \(A = \) payoff of investment \(B\), then either \(C_A = C_B\) or there is an arbitrage.
Assume no arbitrage in what follows!
The premium of an option contract is the amount that the buyer needs to pay and the seller receives at the time when both parties enter into the contract. A premium must be paid because otherwise there is an arbitrage since the payoff is \(\ge 0\).
Notations:
Call Option:
Put Option:
Example: On March 22, the call option strike price of Rolls-Royce was \(\$220\), with exercise date May 22. Suppose that the option sold for \(\$19.5\) and you bought it using a loan at \(5.23\%\). For what values of the stock price at expiration will you make a profit?
Jon and Paul both bought an American call option at time \(t_0 = 0\), with underlying asset price \(S(t)\), strike price \(K\), and expiration \(T\).
At time \(t\) (\(0 < t < T\)), assume \(S(t) > K\).
Conclusion: Paul's payoff is always \(\ge S(t) - K\). Therefore, it is never optimal to exercise an American call on a non-dividend paying stock early.
Theorem: Let \(C(t)\) and \(P(t)\) be the premium paid for a European call and put with strike price \(K\), underlying security price \(S(t)\), and exercise date \(T\). Let \(r\) be the risk-free rate. Then:
\[ \boxed{C(t) = S(t) + P(t) - K e^{-r(T-t)}} \]
| Component | \(\boldsymbol{S(T) \ge K}\) | \(\boldsymbol{S(T) < K}\) |
|---|---|---|
| Portfolio A (Call) | \(S(T) - K\) | \(0\) |
| Portfolio B | ||
| 1 Share (\(S(T)\)) | \(S(T)\) | \(S(T)\) |
| 1 Put | \(0\) | \(K - S(T)\) |
| Repay Loan | \(-K\) | \(-K\) |
| Total B | \(S(T) - K\) | \(0\) |
Conclusion (Law of One Price): Since the payoffs at time \(T\) are identical in all states of the world (\(V_A(T) = V_B(T)\)), the value of the portfolios at time \(t\) must be equal to prevent arbitrage.
\[ V_A(t) = V_B(t) \implies C(t) = S(t) + P(t) - Ke^{-r(T-t)} \]Question: What is the no-arbitrage price of a European call option with strike price \(\$100\)? (Assume the risk-free rate is \(5\%\)).
Idea: Create a replicating portfolio that has the same payoff as the call option. Then, by the Law of One Price, the price of the call at time \(t_0\) is equal to the value of the portfolio at time \(t_0\).
Consider the portfolio: long \(\Delta(t_0)\) units of the security and short (borrow) \(b_0\) at risk-free rate \(r\).
The value of the portfolio at \(t_0\) is then:
\[ V(t_0) = \Delta(t_0)S(t_0) - b_0 \]We need to find \(\Delta(t_0)\) and \(b_0\) so that \(V(t_1) = C(t_1)\).
Let \(h = t_1 - t_0\). The value of the portfolio at time \(t_1\) must match the option payoff \(C(t_1)\) in both the “up” and “down” states:
\[ V(t_1) = \underbrace{\Delta(t_1)S(t_1)}_{e^{qh}\Delta(t_0)S(t_1)} - b_0 e^{rh} = C(t_1) \]This yields a system of 2 equations with 2 unknowns (\(\Delta(t_0)\) and \(b_0\)):
\[ \text{Up state: } e^{qh}\Delta(t_0)S(t_0)u - b_0 e^{rh} = C_u \tag{5} \] \[ \text{Down state: } e^{qh}\Delta(t_0)S(t_0)d - b_0 e^{rh} = C_d \tag{6} \]Solving the system:
Pricing the Call Option: Using the Law of One Price (LOP), \(C(t_0) = V(t_0)\):
\[ C(t_0) = \Delta(t_0)S(t_0) - b_0 \]Substituting the expressions for \(\Delta(t_0)\) and \(b_0\):
\[ C(t_0) = e^{-qh}\frac{C_u - C_d}{u-d} - e^{-rh} \frac{dC_u - uC_d}{u-d} \]Simplifying this expression yields the 1-Step Binomial Pricing Formula:
\[ \boxed{C(t_0) = e^{-rh} \left[ \frac{e^{(r-q)h} - d}{u-d}C_u + \frac{u - e^{(r-q)h}}{u-d}C_d \right]} \]Example: A stock price is \(\$20\). After 3 months, it is either \(\$22\) or \(\$18\). How much should a call option to buy the stock for \(\$21\) in 3 months cost? Assume a risk-free rate of \(5\%\) and no dividends (\(q=0\)).
Parameters:
Step 1: Calculate Payoffs at \(t_1\)
\[ \begin{aligned} C_u &= \max(S_u - K, 0) = \max(22 - 21, 0) = 1 \\ C_d &= \max(S_d - K, 0) = \max(18 - 21, 0) = 0 \end{aligned} \]Step 2: Apply Formula
\[ \begin{aligned} C(t_0) &= e^{-0.05(0.25)} \left[ \frac{e^{(0.05)(0.25)} \cdot 1 - 0.9 \cdot 1}{1.1 - 0.9}(1) + 0 \right] \\ &= e^{-0.0125} \left[ \frac{1.0126 - 0.9}{0.2} \right] \\ &\approx 0.9875 \left[ \frac{0.1126}{0.2} \right] \\ &\approx \boxed{\$0.56} \end{aligned} \]Observation: In the previous 1-step pricing formula, the result was independent of the real-world probability \(p\) of the stock moving up or down.
The Risk Neutral Hypothesis: We assume a theoretical world where investors are indifferent to risk. In this world, the expected return on the stock is exactly the risk-free rate (adjusted for dividends \(q\)).
\[ \mathbb{E}[S(t_1)] = S(t_0)e^{(r-q)h} \]Let \(p^*\) be the risk neutral probability of an up move. We solve for \(p^*\) such that the expected future stock price matches the forward price:
\[ \begin{aligned} p^* S(t_0) u + (1-p^*) S(t_0) d &= S(t_0) e^{(r-q)h} \\ p^* u + (1-p^*) d &= e^{(r-q)h} \\ p^*(u-d) + d &= e^{(r-q)h} \end{aligned} \]Solving for \(p^*\):
\[ \boxed{p^* = \frac{e^{(r-q)h} - d}{u - d}} \]Note: For no arbitrage to exist, we must have \(d < e^{(r-q)h} < u\), which implies \(p^* \in (0,1)\).
Risk Neutral Pricing Formula: Substituting \(p^*\) into the 1-step pricing formula derived previously, the price becomes the discounted expected payoff under the risk-neutral measure:
\[ \begin{aligned} C(t_0) &= e^{-rh} [ p^* C_u(t_1) + (1-p^*) C_d(t_1) ] \\ &= e^{-rh} \mathbb{E}^* [ C(t_1) ] \end{aligned} \]We extend to \(t_2 = 2h\). The tree now branches from \(t_0 \to t_1 \to t_2\).
Using the recursive property of risk-neutral valuation:
Substituting the expressions for \(t_1\) into the equation for \(t_0\) gives the 2-step formula:
\[ C(t_0) = e^{-r(2h)} \left[ (p^*)^2 C_{u^2} + 2p^*(1-p^*) C_{ud} + (1-p^*)^2 C_{d^2} \right] \]By mathematical induction (see TW#9, Problem 6), we can generalize this to an \(n\)-step tree where the total time is \(T = nh\).
Theorem: The price of a European Option at time \(t_0\) for an \(n\)-step binomial tree is:
\[ \boxed{C(t_0; n) = e^{-(nh)r} \sum_{i=0}^{n} \binom{n}{i} (p^*)^i (1-p^*)^{n-i} C_{u^i d^{n-i}}(t_n)} \]Where:
Goal: We want to find the limit of the Binomial Option Pricing formula as the number of steps \(n \to \infty\) (and consequently the time step \(h \to 0\)), converging to a continuous-time model.
Consider the \(n\)-step binomial formula over the interval \([t, T]\). Let \(\tau = T-t\) and \(n h = \tau\).
\[ C(t) = e^{-r\tau} \sum_{i=0}^{n} \binom{n}{i} (p^*)^i (1-p^*)^{n-i} C_{u^i d^{n-i}}(T) \]where the payoff is \(C_{u^i d^{n-i}}(T) = \max \{ S_t u^i d^{n-i} - K, 0 \}\).
1. Separating In-the-Money Paths: Let \(k_*\) be the threshold number of “up” moves such that the option expires in the money:
We can split the summation into two parts starting from \(k_*\):
\[ \begin{aligned} C(t) &= e^{-r\tau} \sum_{i=k_*}^{n} \binom{n}{i} (p^*)^i (1-p^*)^{n-i} (S_t u^i d^{n-i} - K) \\ &= \underbrace{S_t e^{-r\tau} \sum_{i=k_*}^{n} \binom{n}{i} (p^*)^i (1-p^*)^{n-i} u^i d^{n-i}}_{\text{Asset Term}} - \underbrace{K e^{-r\tau} \sum_{i=k_*}^{n} \binom{n}{i} (p^*)^i (1-p^*)^{n-i}}_{\text{Strike Term}} \end{aligned} \]2. Interpreting Probabilities:
Thus:
\[ C(t) = S_t e^{-q\tau} P(Y_n \ge k_*) - K e^{-r\tau} P(X_n \ge k_*) \]3. Applying the Central Limit Theorem (CLT): As \(n \to \infty\), the binomial distributions converge to the Normal distribution.
\[ \begin{aligned} P(Y_n \ge k_*) &\xrightarrow{n \to \infty} N(d_1) \quad (\text{or } N(d_+)) \\ P(X_n \ge k_*) &\xrightarrow{n \to \infty} N(d_2) \quad (\text{or } N(d_-)) \end{aligned} \]Theorem (BSM Formula): The price of a European call option at time \(t\) with expiry \(T\) is:
\[ \boxed{C(t) = e^{-q(T-t)} S(t) N(d_1) - e^{-r(T-t)} K N(d_2)} \]where:
Generalization (Time-Dependent Parameters): If \(r, q, \sigma\) are functions of time, the formula holds with the replacements:
\[ r(T-t) \to \int_t^T r(s) ds, \quad q(T-t) \to \int_t^T q(s) ds, \quad \sigma\sqrt{T-t} \to \sqrt{\int_t^T \sigma^2(s) ds} \]Problem: A security with volatility \(20\%\) sells for \(\$30\). Risk-free rate is \(8\%\). Find the no-arbitrage cost of a call option with strike \(\$34\) expiring in 3 months. Dividend rate \(q=0\).
Parameters: \(\sigma = 0.2, S_t = 30, r = 0.08, T-t = 0.25, K = 34, q = 0\).
We assume that the price of the underlying security \(S(t)\) follows a Geometric Brownian Motion (GBM):
\[ dS = (m-q)S dt + \sigma SdB \]Let \(f(S(t),t)\) be the price of the derivative at time \(t\) (call option, put option, \(\dots\)).
Consider the portfolio:
So at time \(t\) its value is \(V(t) = S(t)\Delta(t) - f(S(t),t)\). We must find \(\Delta\) such that the portfolio is risk-free.
We calculate the differential change in the portfolio value \(dV\) over a small time step \(dt\).
\[ dV = d(S\Delta) - df \]Using the product rule for differentials \(d(XY) = XdY + YdX + dXdY\):
\[ dV = (\Delta dS + S d\Delta + dS d\Delta) - df \]Assumption (Dividend Reinvestment): The change in the number of units held (\(S d\Delta + dS d\Delta\)) comes from reinvesting the dividends paid by the security.
\[ S d\Delta + dS d\Delta = (q S \Delta) dt \]Thus, the equation becomes:
\[ dV = \Delta dS + q S \Delta dt - df \tag{7} \]Expansions using Itô's Lemma
We substitute the dynamics of \(S\) and \(f\):
Substituting these into Eq (7) and grouping terms by \(dt\) and \(dB\):
\[ \begin{aligned} dV &= \Delta [(m-q)S dt + \sigma S dB] + qS\Delta dt \\ &\quad - \left[ \left( \frac{\partial f}{\partial t} + (m-q)S \frac{\partial f}{\partial S} + \frac{1}{2}\sigma^2 S^2 \frac{\partial^2 f}{\partial S^2} \right) dt + \sigma S \frac{\partial f}{\partial S} dB \right] \end{aligned} \]Rearranging to isolate the stochastic part (\(dB\)):
\[ dV = \left[ (m-q)S\left(\Delta - \frac{\partial f}{\partial S}\right) + Sq\Delta - \frac{\partial f}{\partial t} - \frac{1}{2}\sigma^2 S^2 \frac{\partial^2 f}{\partial S^2} \right] dt + \sigma S \left( \Delta - \frac{\partial f}{\partial S} \right) dB \]Delta Hedging and No Arbitrage
To make the portfolio risk-free, we eliminate the \(dB\) term by choosing:
\[ \Delta = \frac{\partial f}{\partial S} \quad \text{(“Delta Hedging”)} \]Substituting this back into the equation for \(dV\), the stochastic term vanishes, and the \((m-q)S(\dots)\) term becomes zero:
\[ dV = \left[ Sq\Delta - \frac{\partial f}{\partial t} - \frac{1}{2}\sigma^2 S^2 \frac{\partial^2 f}{\partial S^2} \right] dt \]Since the portfolio is now risk-free, by the No Arbitrage principle, it must earn risk-free rate \(r\):
\[ dV = r V dt = r(S\Delta - f) dt \]Equating the two expressions for \(dV\):
\[ Sq\Delta - \frac{\partial f}{\partial t} - \frac{1}{2}\sigma^2 S^2 \frac{\partial^2 f}{\partial S^2} = r(S\Delta - f) \]Substituting \(\Delta = \frac{\partial f}{\partial S}\) and rearranging terms yields the Black-Scholes-Merton PDE:
\[ \boxed{\frac{\partial f}{\partial t} + (r-q)S \frac{\partial f}{\partial S} + \frac{1}{2}\sigma^2 S^2 \frac{\partial^2 f}{\partial S^2} = rf} \]Boundary conditions for a European call option \(f = C\):
We must solve:
\[ \frac{\partial C}{\partial t} + \frac{1}{2}\sigma^2 S^2 \frac{\partial^2C}{\partial s^2} + (r-q)S\frac{\partial C}{\partial s} - rC = 0 \]We turn this into a constant coefficient equation by doing a change of variables:
The BSM PDE becomes (exercise):
\[ \frac{\partial v}{\partial \tau} = \frac{\partial^2v}{\partial x^2} + (a-1)\frac{\partial v}{\partial x} - av \]We can turn this into the heat equation with the following substitution:
\[ v(x,\tau) = e^{-\frac{1}{2}(a-1)x - \frac{1}{4}(a+1)^2\tau}u(x, \tau) \]This gives us:
\[ \frac{\partial u}{\partial \tau } = \frac{\partial ^2 u }{\partial x^2} \]However, by changing variables, we changed the boundary and final conditions. In particular, the final condition at \(t = T\), i.e., \(\tau = 0\) is:
\[ v(x,0) = \frac{\max{\{S(T)-K, 0}\}}{K} = \max\{e^x - 1, 0\} \]and for the heat equation, the condition is then:
\[ u(x,0) = e^{\frac{1}{2}(a-1)x}v(x,0) = \max\{e^{\frac{1}{2}(a+1)x}-e^{\frac{1}{2}(a-1)x}, 0\} := u_0(x) \]The heat equation has solution:
\[ u(x,\tau) = \frac{1}{2\sqrt{\pi \tau}}\int_{-\infty}^{\infty} u_0(s)e^{-\frac{(x-s)^2}{4\tau}}ds \]Hence, we get:
\[ \begin{aligned} u(x,\tau) &= \frac{1}{2\sqrt{\pi \tau}}\int_{-\infty}^{\infty} \max\{e^{\frac{1}{2}(a+1)s}-e^{\frac{1}{2}(a-1)s}, 0\}e^{-\frac{(x-s)^2}{4\tau}}ds \\ &= e^{\frac{1}{2}(a+1)x+\frac{1}{4}(a+1)^2\tau}\frac{1}{\sqrt{2\pi}}\int_{-\frac{x}{\sqrt{2\tau}}-\frac{1}{2}(a+1)\sqrt{2\tau}}^{\infty} e^{-\frac{u^2}{2}}du \\ &- e^{\frac{1}{2}(a-1)x+\frac{1}{4}(a-1)^2\tau}\frac{1}{\sqrt{2\pi}}\int_{-\frac{x}{\sqrt{2\tau}}-\frac{1}{2}(a-1)\sqrt{2\tau}}^{\infty} e^{-\frac{u^2}{2}}du \end{aligned} \]And finally, since \(C(S(t),t) = Kv(x,\tau) = Ke^{-\frac{1}{2}(a-1)x-\frac{1}{4}(a+1)^2\tau}u(x,\tau)\), we have:
\[ C(S(t),t) = S(t)N(d_+) - Ke^{-r(T-t)}N(d_{-}) \]To get the pricing formula for the put option, you can redo the same thing with different boundary condition:
Fin!