Chapter 4 - Stochastic Calculus Exercises

Exercise 4.1
Suppose $M(t)$, $0 \leq t \leq T$, is a martingale with respect to some filtration $\mathcal{F}(t)$, and let $\Delta(t)$ be a simple predictable process adapted to $\mathcal{F}(t)$ on a partition $\Pi = {t_0, t_1, \ldots, t_n}$.

Define the stochastic integral process:

We aim to show that $I(t)$ is a martingale.

Since $\Delta(t_j)$ is $\mathcal{F}(t_j)$-measurable and constant on each subinterval $[t_j, t_{j+1})$, and $M(t)$ is adapted, each summand $\Delta(t_j)[M(t_{j+1}) - M(t_j)]$ and $\Delta(t_k)[M(t) - M(t_k)]$ is $\mathcal{F}(t_{j+1})$-measurable, hence adapted. $M(t)$ is integrable by assumption, and since $\Delta(t_j)$ is simple and bounded, each term is integrable. Therefore, $I(t)$ is integrable and adapted.

To check the martingale property, let $s < t$, and suppose $s \in [t_m, t_{m+1})$, $t \in [t_k, t_{k+1})$ with $m < k$. We write:

and break it as:

Take conditional expectation with respect to $\mathcal{F}(s)$. Since $M(t)$ is a martingale:

• For $t_j \geq s$, $\mathbb{E}[M(t_{j+1}) - M(t_j) \mid \mathcal{F}(s)] = 0$,
• $\Delta(t_j)$ is $\mathcal{F}(t_j)$-measurable and $t_j \leq s$, so it’s constant under $\mathbb{E}[\cdot \mid \mathcal{F}(s)]$.

Therefore,

This shows that $I(t)$ satisfies the martingale property:

Hence, $I(t)$ is a martingale.

Exercise 4.2
Let $W(t)$, $0 \leq t \leq T$, be a Brownian motion with associated filtration $\mathcal{F}(t)$. Let $\Delta(t)$ be a nonrandom simple process: there exists a partition $\Pi = {t_0, t_1, \ldots, t_n}$ of $[0,T]$ such that $\Delta(t) = \Delta(t_j)$ on $[t_j, t_{j+1})$, and each $\Delta(t_j)$ is a deterministic constant.

For $t \in [t_k, t_{k+1})$, define the stochastic integral:

(i) Show that whenever $0 \leq s < t \leq T$, the increment $I(t) - I(s)$ is independent of $\mathcal{F}(s)$.

If $s$ is not a partition point, insert $s$ into the partition as $t_k = s$ and set $\Delta(s) = \Delta(t_{k-1})$ to preserve the constant value of $\Delta$ over the interval. Do the same for $t$ if needed.

Now both $s$ and $t$ are partition points. Then:

where $s = t_k$ and $t \in [t_l, t_{l+1})$.

Each Brownian increment $W(t_{j+1}) - W(t_j)$ and $W(t) - W(t_l)$ is independent of $\mathcal{F}(s)$ because Brownian motion has independent increments and $t_j \geq s$ for all terms in the sum.

Therefore, $I(t) - I(s)$ depends only on Brownian motion after time $s$ and is thus independent of $\mathcal{F}(s)$.

(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 $\int_s^t \Delta^2(u) \, du$.

From (i), we know:

where $s = t_k$, $t \in [t_l, t_{l+1})$.

Each $W(t_{j+1}) - W(t_j)$ is independent and distributed as $\mathcal{N}(0, t_{j+1} - t_j)$, and $\Delta(t_j)$ is constant.

Hence, each term $\Delta(t_j)[W(t_{j+1}) - W(t_j)] \sim \mathcal{N}(0, \Delta^2(t_j)(t_{j+1} - t_j))$, and all terms are independent.

The sum of independent normal variables is normal, so $I(t) - I(s) \sim \mathcal{N}(0, \sum \Delta^2(t_j)(t_{j+1} - t_j))$.

This sum is exactly:

(iii) Use (i) and (ii) to show that $I(t)$, $0 \leq t \leq T$, is a martingale.

From (i), $I(t) - I(s)$ is independent of $\mathcal{F}(s)$, and from (ii), $\mathbb{E}[I(t) - I(s)] = 0$.

Therefore,

So $I(t)$ satisfies the martingale property and is therefore a martingale.

(iv) Show that $I^2(t) - \int_0^t \Delta^2(u) \, du$, $0 \leq t \leq T$, is a martingale.

From (ii), the variance of the increment is:

Using the Itô isometry and properties of Brownian motion:

Therefore:

which implies:

So the process $I^2(t) - \int_0^t \Delta^2(u) \, du$ is a martingale.

Exercise 4.3
We now consider the case where $\Delta(t)$ is a simple random process. Let $t_0 = 0$, $t_1 = s$, and $t_2 = t$. Suppose $\Delta(0)$ is nonrandom, but $\Delta(s) = W(s)$ is random and $\mathcal{F}(s)$-measurable. Let

We evaluate the truth of the following assertions:

(i) $I(t) - I(s)$ is independent of $\mathcal{F}(s)$.

We compute:

While $W(t) - W(s)$ is independent of $\mathcal{F}(s)$ by the independent increments property of Brownian motion, $\Delta(s) = W(s)$ is $\mathcal{F}(s)$-measurable. Therefore, the product $\Delta(s)[W(t) - W(s)]$ depends on $\mathcal{F}(s)$ through $\Delta(s)$.

Hence, $I(t) - I(s)$ is not independent of $\mathcal{F}(s)$.

False.

(ii) $I(t) - I(s)$ is normally distributed.

Recall:

$W(s)$ and $W(t) - W(s)$ are independent, but $I(t) - I(s)$ is the product of two independent normal variables. The product of two independent normals is not normally distributed.

To confirm, check the fourth moment:

• $\mathbb{E}[(I(t) - I(s))^4] \neq 3 \cdot \left( \text{Var}(I(t) - I(s)) \right)^2$.

Hence, $I(t) - I(s)$ is not normally distributed.

False.

(iii) $\mathbb{E}[I(t) \mid \mathcal{F}(s)] = I(s)$.

We have:

Then:

Since $W(t) - W(s)$ is independent of $\mathcal{F}(s)$ and has mean zero:

True.

(iv)

We expand:

Take expectation conditional on $\mathcal{F}(s)$:

• $\mathbb{E}[W(t) - W(s) \mid \mathcal{F}(s)] = 0$,
• $\mathbb{E}[(W(t) - W(s))^2 \mid \mathcal{F}(s)] = t - s$.

So:

and

Therefore:

True.

Exercise 4.4 (Stratonovich integral)
Let $W(t)$, $t \geq 0$, be a Brownian motion. Let $T > 0$ and let $\Pi = {t_0, t_1, \ldots, t_n}$ be a partition of $[0, T]$ with $0 = t_0 < t_1 < \cdots < t_n = T$. For each $j$, define the midpoint:

(i) Define the half-sample quadratic variation corresponding to $\Pi$ as:

We show that:

Now compute the variance:

Each $W(t_j^*) - W(t_j)$ is a normal random variable with variance $(t_{j+1} - t_j)/2$, so its square has variance:

Hence,

As $|\Pi| \to 0$, the variance $\to 0$.

Therefore, by Chebyshev’s inequality or convergence in $L^2$, we conclude:

(ii) Define the Stratonovich integral of $W(t)$ with respect to $W(t)$ as:

We aim to show:

We write the Stratonovich sum as:

This is:

Now recall:

• The Itô integral is:

So the Stratonovich integral equals:

In the limit, the sum becomes the quadratic variation:

And from Exercise 4.3 or standard result:

Hence:

Exercise 4.5 (Solving the generalized geometric Brownian motion equation)
Let $S(t)$ be a positive stochastic process satisfying the SDE:

where $\alpha(t)$ and $\sigma(t)$ are adapted processes with respect to the filtration $\mathcal{F}(t)$ of the Brownian motion $W(t)$.

(i) Using Itô’s formula, compute $d \log S(t)$ and simplify to eliminate $S(t)$ from the expression.

Let $f(S(t)) = \log S(t)$. Then applying Itô’s lemma:

• $f’(S) = \frac{1}{S}$,
• $f’‘(S) = -\frac{1}{S^2}$.

Using the SDE for $S(t)$:

Substitute $dS(t) = \alpha(t) S(t) dt + \sigma(t) S(t) dW(t)$:

• $dS(t)/S(t) = \alpha(t) dt + \sigma(t) dW(t)$,
• $(dS(t))^2 = \sigma^2(t) S^2(t) dt$.

So:

Thus,

This is the desired expression, independent of $S(t)$.

(ii) Integrate the formula obtained in (i), then exponentiate the result.

From (i):

Therefore:

Exponentiating both sides:

This is formula (4.4.26), the unique solution to the SDE.

Exercise 4.6
Let

be a geometric Brownian motion. Let $p$ be a positive constant. Compute $d(S^p(t))$, the differential of $S(t)^p$.

Let $f(S(t)) = S^p(t)$. By Itô’s lemma:

• $f’(S) = p S^{p-1}$,
• $f’‘(S) = p(p - 1) S^{p - 2}$.

Using the known SDE for $S(t)$:

we apply Itô’s lemma to $S^p(t)$:

Substitute:

Simplify:

• First term: $p \alpha S^p(t) dt + p \sigma S^p(t) dW(t)$,
• Second term: $\frac{1}{2} p(p - 1) \sigma^2 S^p(t) dt$.

Therefore,

Combine terms:

Exercise 4.7

(i) Compute $dW^4(t)$ and express $W^4(T)$ as the sum of a Lebesgue integral and an Itô integral.

Let $f(W(t)) = W^4(t)$. By Itô’s lemma:

• $f’(W) = 4W^3$,
• $f’‘(W) = 12W^2$.

Then:

Integrating from $0$ to $T$:

(ii) Take expectations and use $\mathbb{E}[W^2(t)] = t$ to derive $\mathbb{E}[W^4(T)] = 3T^2$.

From (i):

The second term is zero since the Itô integral has zero mean:

So:

(iii) Use the same method to compute $\mathbb{E}[W^6(T)]$.

Let $f(W(t)) = W^6(t)$:

• $f’(W) = 6W^5$,
• $f’‘(W) = 30W^4$.

Then:

Take expectation:

From (ii), $\mathbb{E}[W^4(t)] = 3t^2$, so:

Exercise 4.8 (Solving the Vasicek equation)

The Vasicek SDE is:

where $\alpha, \beta, \sigma > 0$.

(i) Use Itô’s formula to compute $d\left( e^{\beta t} R(t) \right)$.

Let $f(t, R(t)) = e^{\beta t} R(t)$. Then:

• $\partial_t f = \beta e^{\beta t} R(t)$,
• $\partial_R f = e^{\beta t}$,
• $\partial_{RR} f = 0$.

Apply Itô’s lemma:

Substitute the SDE for $dR(t)$:

Simplify:

So:

(ii) Integrate and solve for $R(t)$.

Integrate both sides from $0$ to $t$:

Solve for $R(t)$:

This is formula (4.4.33).

Exercise 4.9

Let the European call option price be

where

(i) Verify that

We use the fact that:

From the definition of $d_+$ and $d_-$, we get:

Hence:

Multiply both sides by $K e^{-r(T - t)}$:

Since $x = K e^{-r(T - t)} \cdot \frac{N’(d_-)}{N’(d_+)} = K N’(d_+)$, the identity holds.

(ii) Show that $c_x = N(d_+)$.

Differentiate $c(t, x)$:

• $d_+$ is a function of $x$, so by the chain rule:

Use:

• $\frac{d}{dx}N(d_\pm) = N’(d_\pm) \cdot \frac{d d_\pm}{dx}$,
• $\frac{d d_+}{dx} = \frac{1}{x \sigma \sqrt{\tau}}$,
• $d_- = d_+ - \sigma \sqrt{\tau}$ implies $\frac{d d_-}{dx} = \frac{1}{x \sigma \sqrt{\tau}}$.

Then:

From (i), $Ke^{-r\tau} N’(d_-) = x N’(d_+)$, so the extra terms cancel:

(iii) Show that

Differentiate $c(t,x)$ with respect to $t$:

• $\partial_t d_\pm = \frac{\partial d_\pm}{\partial \tau} \cdot (-1)$,
• $d_+ = \frac{1}{\sigma \sqrt{\tau}} \left[ \log \left( \frac{x}{K} \right) + \left( r + \frac{1}{2} \sigma^2 \right) \tau \right]$, so

But it simplifies with Itô calculus and known result:

This is the theta of the option.

(iv) Use the formulas to show that $c$ satisfies the Black-Scholes PDE:

From above:

• $c_x = N(d_+)$,
• $c_{xx} = \frac{N’(d_+)}{x \sigma \sqrt{\tau}}$,
• $c_t = -r K e^{-r\tau} N(d_-) - \frac{\sigma x}{2 \sqrt{\tau}} N’(d_+)$.

Now compute:

Adding all terms:

(v) Show that for $x > K$, $\lim_{\tau \to 0} d_\pm = \infty$, and for $0 < x < K$, $\lim_{\tau \to 0} d_\pm = -\infty$.

As $\tau \to 0$:

• $\log(x/K)$ dominates, and the denominator $\sqrt{\tau} \to 0$, so the sign of $\log(x/K)$ determines the limit:
• If $x > K$, then $\log(x/K) > 0$, so $d_\pm \to \infty$,
• If $x < K$, then $\log(x/K) < 0$, so $d_\pm \to -\infty$.

(vi) Show that as $x \to 0$, $d_\pm \to -\infty$, for $0 \leq t < T$.

Since $\log(x/K) \to -\infty$ as $x \to 0$, and $\tau > 0$, we get:

Then:

• $N(d_+) \to 0$, so $c(t,x) \to 0$,
• and boundary condition $c(t,0) = 0$ is satisfied.

(vii) Show that as $x \to \infty$, $d_\pm \to \infty$ for $0 \leq t < T$, and verify:

As $x \to \infty$, $d_+ \to \infty$, so $N(d_+) \to 1$.

This is an indeterminate form $0/(\infty)$, so apply L’Hôpital’s Rule:

We know:

• $N’(d_+) \to 0$ exponentially fast,
• $\frac{d d_+}{dx} = \frac{1}{x \sigma \sqrt{\tau}} \to 0$ as $x \to \infty$.

So the product $\to 0$, and the limit is $0$.

Boundary condition is satisfied.

Exercise 4.10 (Self-financing trading)

(i) Derive the continuous-time self-financing condition (4.10.15) using (4.10.9) and (4.10.16).

Let the money market share price be $M(t) = e^{rt}$. The portfolio value is:

Differentiate using Itô calculus:

• From (4.10.9):

We differentiate $X(t) = \Delta(t) S(t) + \Gamma(t) M(t)$ using the product rule:

Now substitute $dM(t) = rM(t) dt$:

Group the terms:

Rewriting:

This is the continuous-time self-financing condition:

Alternatively, rearranging:

Which gives:

So again:

As claimed in (4.10.15).

(ii) Use the corrected version of (4.10.17), i.e. (4.10.21), and the self-financing condition to derive (4.10.18).

Recall the corrected differential of the portfolio value:

Substitute $\Delta(t) = c_x$ (from delta hedging) into this expression:

From the self-financing condition:

So:

Now $N(t) = \Gamma(t) M(t) \Rightarrow dN(t) = r N(t) dt$.

Equating both expressions:

So:

Therefore,

Rearranged:

This is the Black-Scholes PDE as in (4.10.20), completing the derivation.

Exercise 4.11 We are given that the European call price under Black-Scholes with volatility $\sigma_1$ is:

with

However, suppose the true dynamics of the asset price follow:

Then the market price of the option $c(t, S(t))$ is incorrect under the true dynamics. We construct a portfolio with initial value $X(0) = 0$ that is:

• long one European call $c(t, S(t))$,
• short $c_x(t, S(t))$ shares of the stock,
• with the remaining value invested in a money market account.

The cash position is:

We also remove funds at rate:

The differential of the portfolio value is:

Apply Itô’s lemma to $c(t, S(t))$:

Substitute into $dX(t)$:

Cancel terms:

• $\alpha S c_x$ cancels $- c_x \alpha S$,
• $\sigma_2 S c_x dW$ cancels $- \sigma_2 S c_x dW$.

We’re left with:

Group terms:

But the Black-Scholes PDE under $\sigma_1$ tells us:

Therefore:

This is a differential equation:

Its unique solution is:

So the portfolio has zero value at all times, but recall we are extracting funds at rate:

Hence, we have an arbitrage opportunity: the portfolio always has value $0$, yet we are able to withdraw a positive amount of money over time, starting from zero capital.

This confirms that the mispricing due to incorrect volatility input leads to arbitrage.

Exercise 4.12

(i) Use formulas (4.5.23)–(4.5.25), (4.5.26), and (4.5.29) to compute the Greeks of a European put:

• delta $p_x(t, x)$,
• gamma $p_{xx}(t, x)$,
• theta $p_t(t, x)$.

Recall from the put-call parity:

Differentiate with respect to $x$ to find delta:

Differentiate again to find gamma:

Differentiate with respect to $t$ to get theta: \begin{align} p_t(t, x) &= c_t(t, x) + rK e^{-r(T - t)} \tag{4.5.25} \end{align}

Thus, the Greeks of the European put are:

• $\boxed{p_x(t, x) = c_x(t, x) - 1}$
• $\boxed{p_{xx}(t, x) = c_{xx}(t, x)}$
• $\boxed{p_t(t, x) = c_t(t, x) + rK e^{-r(T - t)}}$

(ii) Show that an agent hedging a short position in the put should short the stock and go long in the money market account.

From (i), we know:

because $0 < c_x(t, x) < 1$ for a European call.

So the delta of the put is negative. This means:

• To hedge a short position in the put (i.e., the agent is short the option),
• The agent needs to take a negative multiple of delta → short the stock.
• The remainder of the portfolio value must be in the money market account, so that the portfolio is delta-neutral.

Thus, the agent must short the stock and go long in the risk-free asset to hedge the short put.

(iii) Show that $f(t, x)$ of (4.5.26) and $p(t, x)$ satisfy the same Black-Scholes-Merton PDE as $c(t, x)$.

We are given:

The Black-Scholes PDE for a European call $c(t, x)$ is:

We substitute $f(t, x) = c(t, x) - x + K e^{-r(T - t)}$:

Differentiate:

• $f_t = c_t + rK e^{-r(T - t)}$
• $f_x = c_x - 1$
• $f_{xx} = c_{xx}$

Now compute the LHS of the PDE for $f$:

Now compute the RHS:

So both sides are equal:

Hence, $f(t, x)$ and $p(t, x)$ satisfy the same PDE as $c(t, x)$.

Exercise 4.13 (Decomposition of correlated Brownian motions into independent Brownian motions)

Suppose $B_1(t)$ and $B_2(t)$ are Brownian motions and

where $\rho(t)$ is a stochastic process such that $-1 < \rho(t) < 1$ for all $t$.

Define two processes:

We are given the alternative (forward) construction:

and we are to show that $W_1(t)$ and $W_2(t)$ are independent Brownian motions.

By definition, $W_1(t) = B_1(t)$, and $B_1(t)$ is a Brownian motion. So $W_1(t)$ is a Brownian motion. We examine the quadratic variation of $W_2(t)$.

Let:

This means we can rearrange:

So $W_2(t)$ is a stochastic integral with respect to Brownian motions and hence is a continuous martingale. We now check its quadratic variation:

Using:

• $dB_2(t)^2 = dt$,
• $dW_1(t)^2 = dt$,
• $dB_2(t) dW_1(t) = \rho(t) dt$,

we get:

So $\langle W_2 \rangle_t = t$ and $W_2(t)$ is a Brownian motion.

Step 3: Show that $W_1(t)$ and $W_2(t)$ are independent.

We compute the cross variation:

So $\langle W_1, W_2 \rangle_t = 0$, and since they are continuous martingales with zero cross-variation, they are independent Brownian motions.

Conclusion: The decomposition

correctly expresses $B_1(t)$ and $B_2(t)$ in terms of independent Brownian motions $W_1(t)$ and $W_2(t)$.

Exercise 4.14 We want to justify the equation:

We define:

Then:

(i) Show that $Z_j$ is $\mathcal{F}(t_{j+1})$-measurable and:

• Since $Z_j$ depends on $W(t_j)$ and $W(t_{j+1})$, it is $\mathcal{F}(t_{j+1})$-measurable.
• Conditional on $\mathcal{F}(t_j)$, $W(t_{j+1}) - W(t_j)$ is independent and distributed as $\mathcal{N}(0, t_{j+1} - t_j)$.
• Then:

so:

Now compute variance:

so:

(ii) Show that:

Using the tower property of conditional expectation:

so:

(iii) Assume that:

and show:

We have:

since $Z_j$’s are uncorrelated due to independence of Brownian increments.

From part (i):

So:

As $|\Pi| \to 0$, each $(t_{j+1} - t_j)^2$ shrinks faster than $(t_{j+1} - t_j)$, so the whole sum vanishes:

Conclusion: From (ii), $\mathbb{E}[\sum Z_j] = 0$ and from (iii), the variance goes to $0$, so:

Thus,

which establishes (4.10.22).

Exercise 4.15 (Creating correlated Brownian motions from independent ones) Let $(W_1(t), \ldots, W_d(t))$ be a $d$-dimensional Brownian motion, meaning the components are independent. Let $(\sigma_{ij}(t))_{i=1,\ldots,m; \, j=1,\ldots,d}$ be an $m \times d$ matrix-valued adapted process.

For $i = 1, \ldots, m$, define:

and assume $\sigma_i(t) \ne 0$ for all $t$.

Now define:

(i) Show that $B_i(t)$ is a Brownian motion.

We check the properties:

1. Continuity: $B_i(t)$ is a stochastic integral of continuous processes ⇒ $B_i(t)$ is continuous.
2. Adaptedness: The integrands are adapted and square-integrable.
3. Martingale: Each $B_i(t)$ is an Itô integral ⇒ $B_i(t)$ is a martingale.
4. Quadratic variation:

Since $B_i(t)$ is a continuous martingale with quadratic variation $t$, it is a Brownian motion.

(ii) Show that:

where:

From the Itô isometry:

Thus, $B_i(t)$ and $B_k(t)$ are Brownian motions with instantaneous correlation $\rho_{ik}(t)$.

Conclusion: The processes $B_1(t), \ldots, B_m(t)$ are Brownian motions constructed from independent Brownian motions $W_j(t)$ with a desired correlation structure determined by the matrix $\sigma_{ij}(t)$.

Exercise 4.16 (Creating independent Brownian motions to represent correlated ones) Let $B_1(t), \ldots, B_m(t)$ be $m$ one-dimensional Brownian motions with

where the $\rho_{ik}(t)$ are adapted processes taking values in $(-1, 1)$ for $i \ne k$ and $\rho_{ik}(t) = 1$ for $t = k$. Define the symmetric matrix $C(t)$ by:

Assume $C(t)$ is symmetric and positive definite for all $t$ almost surely. Then there exists a matrix $A(t)$ such that:

which in component form reads:

Let $A(t)^{-1} = [\alpha_{ij}(t)]$ be the inverse of $A(t)$:

so that:

where:

Show that there exist $m$ independent Brownian motions $W_1(t), \ldots, W_m(t)$ such that:

Define the vector of Brownian motions:

where $W(t) = (W_1(t), \ldots, W_m(t))^\top$ is a vector of independent Brownian motions.

Then:

so:

Since $dW_j(t) dW_\ell(t) = \delta_{j\ell} dt$, this simplifies to:

Thus, $B_i(t)$ and $B_k(t)$ have correlation $\rho_{ik}(t)$ as desired.

Conclusion: The process

represents a family of Brownian motions with the given correlation matrix $C(t)$, constructed from independent Brownian motions $W_j(t)$.

Exercise 4.17 (Instantaneous correlation)

Let

where $B_1(t)$ and $B_2(t)$ are Brownian motions with $dB_1(t)dB_2(t) = \rho(t) dt$, and $\Theta_1(t), \Theta_2(t), \sigma_1(t), \sigma_2(t)$ are adapted processes.

Then by Itô calculus:

This defines $\rho(t)$ as the instantaneous correlation between $X_1(t)$ and $X_2(t)$.

We first assume $\rho$, $\Theta_1$, $\Theta_2$, $\sigma_1$, and $\sigma_2$ are constants:

Fix $t_0 > 0$, and let $\epsilon > 0$.

(i) Use Itô’s product rule to show:

Since $dB_1(t)dB_2(t) = \rho dt$, the increment covariance is $\rho \epsilon$.

(ii) Define the increment pair:

Their conditional means:

Conditional variances:

Covariance:

So the conditional correlation is:

Now assume $\rho(t), \Theta_i(t), \sigma_i(t)$ are continuous and adapted, with:

(iii) Define:

Then:

So:

Proof:

Apply the Dominated Convergence Theorem to conclude.

(iv) Define:

Then:

Let:

So:

Using Itô’s product rule and (4.10.37):

(v) The conditional variances and covariance:

(vi) Therefore, the conditional correlation becomes:

So for small $\epsilon$, the conditional correlation of the increments of $X_1$ and $X_2$ approximates the instantaneous correlation $\rho(t_0)$.

Exercise 4.18 (State price density process) Let a stock price be a geometric Brownian motion:

and let $r$ denote the interest rate.

Define the market price of risk to be:

and the state price density process to be:

(i) Show that:

Solution:

Use Itô’s lemma on the function

Then:

(ii) Let $X(t)$ denote the value of an investor’s portfolio using strategy $\Delta(t)$. Then from (4.5.2),

We want to show that $\zeta(t) X(t)$ is a martingale.

Compute:

Substitute from part (i) and the SDE for $X(t)$:

Note that:

Combine everything:

Grouping terms:

• $dt$ terms cancel:

since $(\alpha - r) = \theta \sigma$.

• $dW(t)$ terms:

But we already matched drift terms. So overall:

Hence, $\zeta(t) X(t)$ is a local martingale. With appropriate integrability (usually assumed), it is a martingale.

(iii) Let $T > 0$ be fixed. Suppose the investor wants to end up with portfolio value $V(T)$ at time $T$, where $V(T)$ is $\mathcal{F}(T)$-measurable.

Then since $\zeta(t) X(t)$ is a martingale:

and $\zeta(0) = 1$, so:

This means the present value of $V(T)$ at time $0$ is:

This justifies calling $\zeta(t)$ the state price density process.