Stochastic Calculus for Finance II : Continuous-Time Models

Steven E. Shreve

Chapter 4 Stochastic Calculus - all with Video Answers

Educators

Chapter Questions

Problem 1

Suppose $M(t), 0 \leq t \leq T$, is a martingale with respect to some filtration $\mathcal{F}(t), 0 \leq t \leq T$. Let $\Delta(t), 0 \leq t \leq T$, be a simple process adapted to $\mathcal{F}(t)$ (i.e., there is a partition $\Pi=\left\{t_0, t_1, \ldots, t_n\right\}$ of $[0, T]$ such that, for every $j, \Delta\left(t_j\right)$ is $\mathcal{F}\left(t_j\right)$-measurable and $\Delta(t)$ is constant in $t$ on each subinterval $\left[t_j, t_{j+1}\right)$ ). For $t \in\left[t_k, t_{k+1}\right)$, define the stochastic integral
$$
I(t)=\sum_{j=0}^{k-1} \Delta\left(t_j\right)\left[M\left(t_{j+1}\right)-M\left(t_j\right)\right]+\Delta\left(t_k\right)\left[M(t)-M\left(t_k\right)\right] .
$$

We think of $M(t)$ as the price of an asset at time $t$ and $\Delta\left(t_j\right)$ as the number of shares of the asset held by an investor between times $t_j$ and $t_{j+1}$. Then $I(t)$ is the capital gains that accrue to the investor between times 0 and $t$. Show that $I(t), 0 \leq t \leq T$, is a martingale.

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Problem 2

. Let $W(t), 0 \leq t \leq T$, be a Brownian motion, and let $\mathcal{F}(t)$, $0 \leq t \leq T$, be an associated filtration. Let $\Delta(t), 0 \leq t \leq T$, be a nonrandom simple process (i.e., there is a partition $\Pi=\left\{t_0, t_1, \ldots, t_n\right\}$ of $[0, T]$ such that
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4 Stochastic Calculus
for every $j, \Delta\left(t_j\right)$ is a nonrandom quantity and $\Delta(t)=\Delta\left(t_j\right)$ is constant in $t$ on the subinterval $\left[t_j, t_{j+1}\right)$ ). For $t \in\left[t_k, t_{k+1}\right]$, define the stochastic integral
$$
I(t)=\sum_{j=0}^{k-1} \Delta\left(t_j\right)\left[W\left(t_{j+1}\right)-W\left(t_j\right)\right]+\Delta\left(t_k\right)\left[W(t)-W\left(t_k\right)\right] .
$$
(i) Show that whenever $0 \leq s<t \leq T$, the increment $I(t)-I(s)$ is independent of $\mathcal{F}(s)$. (Simplification: If $s$ is between two partition points, we can always insert $s$ as an extra partition point. Then we can relabel the partition points so that they are still called $t_0, t_1, \ldots, t_n$, but with a larger value of $n$ and now with $s=t_k$ for some value of $k$. Of course, we must set $\Delta(s)=\Delta\left(t_{k-1}\right)$ so that $\Delta$ takes the same value on the interval $\left[s, t_{k+1}\right)$ as on the interval $\left[t_{k-1}, s\right)$. Similarly, we can insert $t$ as an extra partition point if it is not already one. Consequently, to show that $I(t)-I(s)$ is independent of $\mathcal{F}(s)$ for all $0 \leq s<t \leq T$, it suffices to show that $I\left(t_k\right)-I\left(t_{\ell}\right)$ is independent of $\mathcal{F}\left(t_{\ell}\right)$ whenever $t_k$ and $t_{\ell}$ are two partition points with $t_{\ell}<t_k$. This is all you need to do.)
(ii) Show that whenever $0 \leq s<t \leq T$, the increment $I(t)-I(s)$ is a normally distributed random variable with mean zero and variance $\int_s^t \Delta^2(u) d u$.
(iii) Use (i) and (ii) to show that $I(t), 0 \leq t \leq T$, is a martingale.
(iv) Show that $I^2(t)-\int_0^t \Delta^2(u) d u, 0 \leq t \leq T$, is a martingale.

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01:34

Problem 3

We now consider a case in which $\Delta(t)$ in Exercise 4.2 is simple but random. In particular, let $t_0=0, t_1=s$, and $t_2=t$, and let $\Delta(0)$ be nonrandom and $\Delta(s)=W(s)$. Which of the following assertions is true? Justify your answers.
(i) $I(t)-I(s)$ is independent of $\mathcal{F}(s)$.
(ii) $I(t)-I(s)$ is normally distributed. (Hint: Check if the fourth moment is three times the square of the variance; see Exercise 3.3 of Chapter 3.)
(iii) $\mathbb{E}[I(t) \mid \mathcal{F}(s)]=I(s)$.
(iv) $\mathbb{E}\left[I^2(t)-\int_0^t \Delta^2(u) d u \mid \mathcal{F}(s)\right]=I^2(s)-\int_0^s \Delta^2(u) d u$.

Manik Pulyani

Numerade Educator

Problem 4

(Stratonovich integral). Let $W(t), t \geq 0$, be a Brownian motion. Let $T$ be a fixed positive number and let $\Pi=\left\{t_0, t_1, \ldots, t_n\right\}$ be a partition of $[0, T]$ (i.e., $0=t_0<t_1<\cdots<t_n=T$ ). For each $j$, define $t_j^*=\frac{t_j+t_{2+1}}{2}$ to be the midpoint of the interval $\left[t_j, t_{j+1}\right]$.
(i) Define the half-sample quadratic variation corresponding to $\Pi$ to be
$$
Q_{n / 2}=\sum_{j=0}^{n-1}\left(W\left(t_j^*\right)-W\left(t_j\right)\right)^2
$$

Show that $Q_{\Pi / 2}$ has limit $\frac{1}{2} T$ as $\|\Pi\| \rightarrow 0$. (Hint: It suffices to show that $\mathbf{E} Q_{\Pi / 2}=\frac{1}{2} T$ and $\lim _{\|\Pi\| \rightarrow 0} \operatorname{Var}\left(Q_{\Pi / 2}\right)=0$.)
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(ii) Define the Stratonovich integral of $W(t)$ with respect to $W(t)$ to be
$$
\int_0^T W(t) \circ d W(t)=\lim _{\|\Pi\| \rightarrow 0} \sum_{j=0}^{n-1} W\left(t_j^*\right)\left(W\left(t_{j+1}\right)-W\left(t_j\right)\right) .
$$

In contrast to the Itô integral $\int_0^T W(t) d W(t)=\frac{1}{2} W^2(T)-\frac{1}{2} T$ of (4.3.4), which evaluates the integrand at the left endpoint of each subinterval $\left[t_j, t_{j+1}\right]$, here we evaluate the integrand at the midpoint $t_j^*$. Show that
$$
\int_0^T W(t) \circ d W(t)=\frac{1}{2} W^2(T) .
$$
(Hint: Write the approximating sum in (4.10.1) as the sum of an approximating sum for the Itô integral $\int_0^T W(t) d W(t)$ and $Q_{\Pi / 2}$. The approximating sum for the Itô integral is the one corresponding to the partition $0=t_0<t_0^*<t_1<t_1^*<\cdots<t_{n-1}^*<t_n=T$, not the partition $\Pi$. .)

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Problem 5

(Solving the generalized geometric Brownian motion equation). Let $S(t)$ be a positive stochastic process that satisfies the generalized geometric Brownian motion differential equation (see Example 4.4.8)
$$
d S(t)=\alpha(t) S(t) d t+\sigma(t) S(t) d W(t),
$$
where $\alpha(t)$ and $\sigma(t)$ are processes adapted to the filtration $\mathcal{F}(t), t \geq 0$, associated with the Brownian motion $W(t), t \geq 0$. In this exercise, we show that $S(t)$ must be given by formula (4.4.26) (i.e., that formula provides the only solution to the stochastic differential equation (4.10.2)). In the process, we provide a method for solving this equation.
(i) Using (4.10.2) and the Itô-Doeblin formula, compute $d \log S(t)$. Simplify so that you have a formula for $d \log S(t)$ that does not involve $S(t)$.
(ii) Integrate the formula you obtained in (i), and then exponentiate the answer to obtain (4.4.26).

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Problem 6

Let $S(t)=S(0) \exp \left\{\sigma W(t)+\left(\alpha-\frac{1}{2} \sigma^2\right) t\right\}$ be a geometric Brownian motion. Let $p$ be a positive constant. Compute $d\left(S^p(t)\right)$, the differential of $S(t)$ raised to the power $p$.

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Problem 7

(i) Compute $d W^4(t)$ and then write $W^4(T)$ as the sum of an ordinary (Lebesgue) integral with respect to time and an Itô integral.
(ii) Take expectations on both sides of the formula you obtained in (i), use the fact that $\mathbb{E} W^2(t)=t$, and derive the formula $\mathbb{E} W^4(T)=3 T^2$.
(iii) Use the method of (i) and (ii) to derive a formula for $\mathbb{E} W^6(T)$.

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09:47

Problem 8

Solving the Vasicek equation). The Vasicek intercst rate stochastic differential equation (4.4.32) is
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4 Stochastic Calculus
$$
d R(t)=(\alpha-\beta R(t)) d t+\sigma d W(t),
$$
where $\alpha, \beta$, and $\sigma$ are positive constants. The solution to this equation is given in Example 4.4.10. This exercise shows how to derive this solution.
(i) Use (4.4.32) and the Itô-Doeblin formula to compute $d\left(e^{\beta t} R(t)\right)$. Simplify it so that you have a formula for $d\left(e^{\beta t} R(t)\right)$ that does not involve $R(t)$.
(ii) Integrate the equation you obtained in (i) and solve for $R(t)$ to obtain (4.4.33).

Nick Johnson

Numerade Educator

Problem 9

. For a European call expiring at time $T$ with strike price $K$, the Black-Scholes-Merton price at time $t$, if the time- $t$ stock price is $x$, is
$$
c(t, x)=x N\left(d_{+}(T-t, x)\right)-K e^{-r(T-t)} N\left(d_{-}(T-t, x)\right),
$$
where
$$
\begin{aligned}
& d_{+}(\tau, x)=\frac{1}{\sigma \sqrt{\tau}}\left[\log \frac{x}{K}+\left(r+\frac{1}{2} \sigma^2\right) \tau\right], \\
& d_{-}(\tau, x)=d_{+}(\tau, x)-\sigma \sqrt{\tau},
\end{aligned}
$$

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Problem 10

(Self-financing trading). The fundamental idea behind noarbitrage pricing is to reproduce the payoff of a derivative security by trading in the underlying asset (which we call a stock) and the money market account. In discrete time, we let $X_k$ denote the value of the hedging portfolio at time $k$ and let $\Delta_k$ denote the number of shares of stock held between times $k$ and $k+1$. Then, at time $k$, after rebalancing (i.e., moving from a position of $\Delta_{k-1}$ to a position $\Delta_k$ in the stock), the amount in the money market account is $X_k-S_k \Delta_k$. The value of the portfolio at time $k+1$ is
$$
X_{k+1}=\Delta_k S_{k+1}+(1+r)\left(X_k-\Delta_k S_k\right) .
$$
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4 Stochastic Calculus
This formula can be rearranged to become
$$
X_{k+1}-X_k=\Delta_k\left(S_{k+1}-S_k\right)+r\left(X_k-\Delta_k S_k\right),
$$
which says that the gain between time $k$ and time $k+1$ is the sum of the capital gain on the stock holdings, $\Delta_k\left(S_{k+1}-S_k\right)$, and the interest earnings on the money market account, $r\left(X_k-\Delta_k S_k\right)$. The continuous-time analogue of $(4.10 .8)$ is
$$
d X(t)=\Delta(t) d S(t)+r(X(t)-\Delta(t) S(t)) d t .
$$

Alternatively, one could define the value of a share of the money market account at time $k$ to be
$$
M_k=(1+r)^k
$$
and formulate the discrete-time model with two processes, $\Delta_k$ as before and $\Gamma_k$ denoting the number of shares of the money market account held at time $k$ after rebalancing. Then
$$
X_k=\Delta_k S_k+\Gamma_k M_k
$$
so that (4.10.7) becomes
$$
X_{k+1}=\Delta_k S_{k+1}+(1+r) \Gamma_k M_k=\Delta_k S_{k+1}+\Gamma_k M_{k+1} .
$$

Subtracting (4.10.10) from (4.10.11), we obtain in place of (4.10.8) the equation
$$
X_{k+1}-X_k=\Delta_k\left(S_{k+1}-S_k\right)+\Gamma_k\left(M_{k+1}-M_k\right),
$$
which says that the gain between time $k$ and time $k+1$ is the sum of the capital gain on stock holdings, $\Delta_k\left(S_{k+1}-S_k\right)$, and the earnings from the money market investment, $\Gamma_k\left(M_{k+1}-M_k\right)$.

But $\Delta_k$ and $\Gamma_k$ cannot be chosen arbitrarily. The agent arrives at time $k+1$ with some portfolio of $\Delta_k$ shares of stock and $\Gamma_k$ shares of the money market account and then rebalances. In terms of $\Delta_k$ and $\Gamma_k$, the value of the portfolio upon arrival at time $k+1$ is given by (4.10.11). After rebalancing, it is
$$
X_{k+1}=\Delta_{k+1} S_{k+1}+\Gamma_{k+1} M_{k+1} .
$$

Setting these two values equal, we obtain the discrete-time self-financing con-

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Problem 11

Exercise 4.11. Let
$$
c(t, x)=x N\left(d_{+}(T-t, x)\right)-K e^{-r(T-t)} N\left(d_{-}(T-t, x)\right)
$$
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197
denote the price for a European call, expiring at time $T$ with strike price $K$, where
$$
d_{ \pm}(T-t, x)=\frac{1}{\sigma_1 \sqrt{T-t}}\left[\log \frac{x}{K}+\left(r \pm \frac{\sigma_1^2}{2}\right)(T-t)\right] .
$$

This option price assumes the underlying stock is a geometric Brownian motion with volatility $\sigma_1>0$. For this problem, we take this to be the market price of the option.

Suppose, however, that the underlying asset is really a geometric Brownian motion with volatility $\sigma_2>\sigma_1$, i.e.,
$$
d S(t)=\alpha S(t) d t+\sigma_2 S(t) d W(t) .
$$

Consequently, the market price of the call is incorrect.
We set up a portfolio whose value at each time $t$ we denote by $X(t)$. We begin with $X(0)=0$. At each time $t$, the portfolio is long one European call, is short $c_x(t, S(t))$ shares of stock, and thus has a cash position
$$
X(t)-c(t, S(t))+S(t) c_x(t, S(t))
$$
which is invested at the constant interest rate $r$. We also remove cash from this portfolio at a rate $\frac{1}{2}\left(\sigma_2^2-\sigma_1^2\right) S^2(t) c_{x x}(t, S(t))$. Therefore, the differential of the portfolio value is
$$
\begin{aligned}
d X(t)=d c( & t, S(t))-c_x(t, S(t)) d S(t) \\
& +r\left[X(t)-c(t, S(t))+S(t) c_x(t, S(t))\right] d t \\
& -\frac{1}{2}\left(\sigma_2^2-\sigma_1^2\right) S^2(t) c_{x x}(t, S(t)) d t, 0 \leq t \leq T .
\end{aligned}
$$

Show that $X(t)=0$ for all $t \in[0, T]$. In particular, because $c_{x x}(t, S(t))>0$ and $\sigma_2>\sigma_1$, we have an arbitrage opportunity; we can start with zero initial capital, remove cash at a positive rate between times 0 and $T$, and at time $T$ have zero liability. (Hint: Compute $d\left(e^{-r t} X(t)\right)$.)

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Problem 12

(i) Use formulas (4.5.23)-(4.5.25), (4.5.26), and (4.5.29) to determine the delta $p_x(t, x)$, the gamma $p_{x x}(t, x)$, and the theta $p_t(t, x)$ of a European put.
(ii) Show that an agent hedging a short position in the put should have a short position in the underlying stock and a long position in the money market account.
(iii) Show that $f(t, x)$ of (4.5.26) and $p(t, x)$ satisfy the same Black-ScholesMerton partial differential equation (4.5.14) satisfied by $c(t, x)$.

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Problem 13

(Decomposition of correlated Brownian motions into independent Brownian motions). Suppose $B_1(t)$ and $B_2(t)$ are Brownian motions and
$$
d B_1(t) d B_2(t)=\rho(t) d t
$$
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where $\rho$ is a stochastic process taking values strictly between -1 and 1 . Define processes $W_1(t)$ and $W_2(t)$ such that
$$
\begin{aligned}
& B_1(t)=W_1(t), \\
& B_2(t)=\int_0^t \rho(s) d W_1(s)+\int_0^t \sqrt{1-\rho^2(s)} d W_2(s),
\end{aligned}
$$
and show that $W_1(t)$ and $W_2(t)$ are independent Brownian motions.

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Problem 14

In the derivation of the Itô-Doeblin formula, Theorem 4.4.1, we considered only the case of the function $f(x)=\frac{1}{2} x^2$, for which $f^{\prime \prime}(x)=1$. This made it easy to determine the limit of the last term,
$$
\frac{1}{2} \sum_{j=0}^{n-1} f^{\prime \prime}\left(W\left(t_j\right)\right)\left[W\left(t_{j+1}\right)-W\left(t_j\right)\right]^2,
$$
appearing in (4.4.5). Indeed,
$$
\begin{aligned}
\lim _{\|\Pi\| \rightarrow 0} \sum_{j=0}^{n-1} f^{\prime \prime}\left(W\left(t_j\right)\right)\left[W\left(t_{j+1}\right)-W\left(t_j\right)\right]^2 & =\lim _{\|\Pi\| \rightarrow 0} \sum_{j=0}^{n-1}\left[W\left(t_{j+1}\right)-W\left(t_j\right)\right]^2 \\
& =[W, W](T)=T \\
& =\int_0^T f^{\prime \prime}(W(t)) d t .
\end{aligned}
$$

If we had been working with an arbitrary function $f(x)$, we could not replace $f^{\prime \prime}\left(W\left(t_j\right)\right)$ by 1 in the argument above. It is tempting in this case to just argue that $\left[W\left(t_{j+1}\right)-W\left(t_j\right)\right]^2$ is approximately equal to $\left(t_{j+1}-t_j\right)$, so that
$$
\sum_{j=0}^{n-1} f^{\prime \prime}\left(W\left(t_j\right)\right)\left[W\left(t_{j+1}\right)-W\left(t_j\right)\right]^2
$$
is approximately equal to
$$
\sum_{j=0}^{n-1} f^{\prime \prime}\left(W\left(t_j\right)\right)\left(t_{j+1}-t_j\right)
$$

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Problem 15

(Creating correlated Brownian motions from independent ones). Let $\left(W_1(t), \ldots, W_d(t)\right)$ be a $d$-dimensional Brownian motion. In particular, these Brownian motions are independent of one another. Let $\left(\sigma_{i j}(t)\right)_{i=1, \ldots, m ; j=1, \ldots, d}$ be an $m \times d$ matrix-valued process adapted to the filtration associated with the $d$-dimensional Brownian motion. For $i=1, \ldots, m$, define
$$
\sigma_i(t)=\left[\sum_{j=1}^d \sigma_{i j}^2(t)\right]^{\frac{1}{2}},
$$
and assume this is never zero. Define also
$$
B_i(t)=\sum_{j=1}^d \int_0^t \frac{\sigma_{i j}(u)}{\sigma_i(u)} d W_j(u) .
$$
(i) Show that, for each $i, B_i$ is a Brownian motion.
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(ii) Show that $d B_i(t) d B_k(t)=\rho_{i k}(t)$, where
$$
\rho_{i k}(t)=\frac{1}{\sigma_i(t) \sigma_k(t)} \sum_{j=1}^d \sigma_{i j}(t) \sigma_{k j}(t) .
$$

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Problem 16

(Creating independent Brownian motions to represent correlated ones). Let $B_1(t), \ldots, B_m(t)$ be $m$ one-dimensional Brownian motions with
$$
d B_i(t) d B_k(t)=\rho_{i k}(t) d t \text { for all } i, k=1, \ldots, m,
$$
where $\rho_{i k}(t)$ are adapted processes taking values in $(-1,1)$ for $i \neq k$ and $\rho_{i k}(t)=1$ for $i=k$. Assume that the symmetric matrix
$$
C(t)=\left[\begin{array}{cccc}
\rho_{11}(t) & \rho_{12}(t) & \cdots & \rho_{1 m}(t) \\
\rho_{21}(t) & \rho_{22}(t) & \cdots & \rho_{2 m}(t) \\
\vdots & \vdots & & \vdots \\
\rho_{m 1}(t) & \rho_{m 2}(t) & \cdots & \rho_{m m}(t)
\end{array}\right]
$$
is positive definite for all $t$ almost surely. Because the matrix $C(t)$ is symmetric and positive definite, it has a matrix square root. In other words, there is a matrix
$$
A(t)=\left[\begin{array}{cccc}
a_{11}(t) & a_{12}(t) & \cdots & a_{1 m}(t) \\
a_{21}(t) & a_{22}(t) & \cdots & a_{2 m}(t) \\
\vdots & \vdots & & \vdots \\
a_{m 1}(t) & a_{m 2}(t) & \cdots & a_{m m}(t)
\end{array}\right]
$$
such that $C(t)=A(t) A^{\operatorname{tr}}(t)$, which when written componentwise is

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Problem 17

Exercise 4.17 (Instantaneous correlation). Let
$$
\begin{aligned}
& X_1(t)=X_1(0)+\int_0^t \Theta_1(u) d u+\int_0^t \sigma_1(u) d B_1(u), \\
& X_2(t)=X_2(0)+\int_0^t \Theta_2(u) d u+\int_0^t \sigma_2(u) d B_2(u),
\end{aligned}
$$
where $B_1(t)$ and $B_2(t)$ are Brownian motions satisfying $d B_1(t) d B_2(t)=\rho(t)$ and $\rho(t), \Theta_1(t), \Theta_2(t), \sigma_1(t)$, and $\sigma_2(t)$ are adapted processes. Then
$$
d X_1(t) d X_2(t)=\sigma_1(t) \sigma_2(t) d B_1(t) d B_2(t)=\rho(t) \sigma_1(t) \sigma_2(t) d t .
$$

We call $\rho(t)$ the instantaneous correlation between $X_1(t)$ and $X_2(t)$ for the reason explained by this exercise.
We first consider the case when $\rho, \Theta_1, \Theta_2, \sigma_1$, and $\sigma_2$ are constant. Then
$$
\begin{aligned}
& X_1(t)=X_1(0)+\Theta_1 t+\sigma_1 B_1(t), \\
& X_2(t)=X_2(0)+\Theta_2 t+\sigma_2 B_2(t) .
\end{aligned}
$$

Fix $t_0>0$, and let $\epsilon>0$ be given.
(i) Use Itô's product rule to show that
$$
\mathbf{E}\left[\left(B_1\left(t_0+\epsilon\right)-B_1\left(t_0\right)\right)\left(B_2\left(t_0+\epsilon\right)-B_2\left(t_0\right)\right) \mid \mathcal{F}\left(t_0\right)\right]=\rho \epsilon .
$$
(ii) Show that, conditioned on $\mathcal{F}\left(t_0\right)$, the pair of random variables
$$
\left(X_1\left(t_0+\epsilon\right)-X_1\left(t_0\right), X_2\left(t_0+\epsilon\right)-X_2\left(t_0\right)\right)
$$
has means, variances, and covariance
$$
\begin{aligned}
M_i(\epsilon)= & \mathbf{E}\left[X_i\left(t_0+\epsilon\right)-X_i\left(t_0\right) \mid \mathcal{F}\left(t_0\right)\right]=\Theta_i \epsilon \text { for } i=1,2 \\
V_i(\epsilon)= & \mathbf{E}\left[\left(X_i\left(t_0+\epsilon\right)-X_i\left(t_0\right)\right)^2 \mid \mathcal{F}\left(t_0\right)\right]-M_i^2(\epsilon) \\
= & \sigma_i^2 \epsilon \text { for } i=1,2 \\
C(\epsilon)= & \mathbf{E}\left[\left(X_1\left(t_0+\epsilon\right)-X_1\left(t_0\right)\right)\left(X_2\left(t_0+\epsilon\right)-X_2\left(t_0\right)\right) \mid \mathcal{F}\left(t_0\right)\right] \\
& \quad-M_1(\epsilon) M_2(\epsilon)=\rho \sigma_1 \sigma_2 \epsilon
\end{aligned}
$$
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In particular, conditioned on $\mathcal{F}\left(t_0\right)$, the correlation between the increments $X_1\left(t_0+\epsilon\right)-X_1\left(t_0\right)$ and $X_2\left(t_0+\epsilon\right)-X_2\left(t_0\right)$ is
$$
\frac{C(\epsilon)}{\sqrt{V_1(\epsilon) V_2(\epsilon)}}=\rho .
$$

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Problem 18

Let a stock price be a geometric Brownian motion
$$
d S(t)=\alpha S(t) d t+\sigma S(t) d W(t),
$$
and let $r$ denote the interest rate. We define the market price of risk to be
$$
\theta=\frac{\alpha-r}{\sigma}
$$
and the state price density process to be
$$
\zeta(t)=\exp \left\{-\theta W(t)-\left(r+\frac{1}{2} \theta^2\right) t\right\} .
$$
(i) Show that
$$
d \zeta(t)=-\theta \zeta(t) d W(t)-r \zeta(t) d t .
$$
(ii) Let $X$ denote the value of an investor's portfolio when he uses a portfolio process $\Delta(t)$. From (4.5.2), we have
$$
d X(t)=r X(t) d t+\Delta(t)(\alpha-r) S(t) d t+\Delta(t) \sigma S(t) d W(t) .
$$

Show that $\zeta(t) X(t)$ is a martingale. (Hint: Show that the differential $d(\zeta(t) X(t))$ has no $d t$ term.)
(iii) Let $T>0$ be a fixed terminal time. Show that if an investor wants to begin with some initial capital $X(0)$ and invest in order to have portfolio value $V(T)$ at time $T$, where $V(T)$ is a given $\mathcal{F}(T)$-measurable random variable, then he must begin with initial capital
$$
X(0)=\mathbb{E}[\zeta(T) V(T)] .
$$

In other words, the present value at time zero of the random payment $V(T)$ at time $T$ is $\mathbb{E}[\zeta(T) V(T)]$. This justifies calling $\zeta(t)$ the state price density process.

ons to conclude that

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Problem 19

. Let $W(t)$ be a Brownian motion, and define
$$
B(t)=\int_0^t \operatorname{sign}(W(s)) d W(s),
$$
where
$$
\operatorname{sign}(x)= \begin{cases}1 & \text { if } x \geq 0, \\ -1 & \text { if } x<0\end{cases}
$$
4.10 Exercises
205
(i) Show that $B(t)$ is a Brownian motion.
(ii) Use Itô's product rule to compute $d[B(t) W(t)]$. Integrate both sides of the resulting equation and take expectations. Show that $\mathbb{E}[B(t) W(t)]=0$ (i.e., $B(t)$ and $W(t)$ are uncorrelated).
(iii) Verify that
$$
d W^2(t)=2 W(t) d W(t)+d t .
$$
(iv) Use Itô's product rule to compute $d\left[B(t) W^2(t)\right]$. Integrate both sides of the resulting equation and take expectati

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Problem 20

(Local time). Let $W(t)$ be a Brownian motion. The ItôDoeblin formula in differential form says that
$$
d f(W(t))=f^{\prime}(W(t)) d W(t)+\frac{1}{2} f^{\prime \prime}(W(t)) d t .
$$

In integrated form, this formula is
$$
f(W(T))=f(W(0))+\int_0^T f^{\prime}(W(t)) d W(t)+\frac{1}{2} \int_0^T f^{\prime \prime}(W(t)) d t .
$$

The usual statement of this formula assumes that the function $f^{\prime \prime}(x)$ is defined for every $x \in \mathbb{R}$ and is a continuous function of $x$. In fact, the formula still holds if there are finitely many points $x$ where $f^{\prime \prime}(x)$ is undefined, provided that $f^{\prime}(x)$ is defined for every $x \in \mathbb{R}$ and is a continuous function of $x$ (and provided that $\left|f^{\prime \prime}(x)\right|$ is bounded so that the integral $\int_0^T f^{\prime \prime}(W(t)) d t$ is defined). However, if $f^{\prime}(x)$ is not defined at some point, naive application of the Itô-Doeblin formula can give wrong answers, as this problem demonstrates.
(i) Let $K$ be a positive constant, and define $f(x)=(x-K)^{+}$. Compute $f^{\prime}(x)$ and $f^{\prime \prime}(x)$. Note that there is a point $x$ where $f^{\prime}(x)$ is not defined, and note also that $f^{\prime \prime}(x)$ is zero everywhere except at this point, where $f^{\prime \prime}(x)$ is also undefined.
(ii) Substitute the function $f(x)=(x-K)^{+}$into (4.10.42), replacing the term $\frac{1}{2} \int_0^T f^{\prime \prime}(W(t)) d t$ by zero since $f^{\prime \prime}$ is zero everywhere except at one point, where it is not defined. Show that the two sides of this equation cannot be equal by computing their expected values.
(iii) To get some idea of what is going on here, define a sequence of functions $\left\{f_n\right\}_{n=1}^{\infty}$ by the formula
$$
f_n(x)= \begin{cases}0 & \text { if } x \leq K-\frac{1}{2 n}, \\ \frac{n}{2}(x-K)^2+\frac{1}{2}(x-K)+\frac{1}{8 n} & \text { if } K-\frac{1}{2 n} \leq x \leq K+\frac{1}{2 n}, \\ x-K & \text { if } x \geq K+\frac{1}{2 n}\end{cases}
$$

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