Chapter 6, Connections with Partial Differential Equations Video Solutions, Stochastic Calculus for Finance II : Continuous-Time Models

Problem 1

Consider the stochastic differential equation
$$
d X(u)=(a(u)+b(u) X(u)) d u+(\gamma(u)+\sigma(u) X(u)) d W(u),
$$
where $W(u)$ is a Brownian motion relative to a filtration $\mathcal{F}(u), u \geq 0$, and we allow $a(u), b(u), \gamma(u)$, and $\sigma(u)$ to be processes adapted to this filtration. Fix an initial time $t \geq 0$ and an initial position $x \in \mathbb{R}$. Define
$$
\begin{aligned}
& Z(u)=\exp \left\{\int_t^u \sigma(v) d W(v)+\int_t^u\left(b(v)-\frac{1}{2} \sigma^2(v)\right) d v\right\}, \\
& Y(u)=x+\int_t^u \frac{a(v)-\sigma(v) \gamma(v)}{Z(v)} d v+\int_t^u \frac{\gamma(v)}{Z(v)} d W(v) .
\end{aligned}
$$
(i) Show that $Z(t)=1$ and
$$
d Z(u)=b(u) Z(u) d u+\sigma(u) Z(u) d W(u), u \geq t .
$$
(ii) By its very definition, $Y(u)$ satisfies $Y(t)=x$ and
$$
d Y(u)=\frac{a(u)-\sigma(u) \gamma(u)}{Z(u)} d u+\frac{\gamma(u)}{Z(u)} d W(u), u \geq t .
$$

Show that $X(u)=Y(u) Z(u)$ solves the stochastic differential equation (6.2.4) and satisfies the initial condition $X(t)=x$.

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Problem 2

(No-arbitrage derivation of bond-pricing equation). In Section 6.5, we began with the stochastic differential equation (6.5.1) for the interest rate under the risk-neutral measure $\widetilde{\mathbb{P}}$, used the risk-neutral pricing formula (6.5.3) to consider a zero-coupon bond maturing at time $T$ whose price $B(t, T)$ at time $t$ before maturity is a function $f(t, R(t))$ of the time and the interest rate, and derived the partial differential equation (6.5.4) for the function $f(t, r)$. In this exercise, we show how to derive this partial differential equation from no-arbitrage considerations rather than by using the risk-neutral pricing formula.

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08:31

Problem 3

(Solution of Hull-White model). This exercise solves the ordinary differential equations (6.5.8) and (6.5.9) to produce the solutions $C(t, T)$ and $A(t, T)$ given in (6.5.10) and (6.5.11).
(i) Use equation (6.5.8) with $s$ replacing $t$ to show that
$$
\frac{d}{d s}\left[e^{-\int_0^s b(v) d v} C(s, T)\right]=-e^{-\int_0^* b(v) d v} .
$$
(ii) Integrate the equation in (i) from $s=t$ to $s=T$, and use the terminal condition $C(T, T)$ to obtain (6.5.10).
(iii) Replace $t$ by $s$ in (6.5.9), integrate the resulting equation from $s=t$ to $s=T$, use the terminal condition $A(T, T)=0$, and obtain (6.5.11).

Stanley Enemuo

Numerade Educator

Problem 4

(Solution of Cox-Ingersoll-Ross model). This exercise solves the ordinary differential equations $(6.5 .14)$ and (6.5.15) to produce the solutions $C(t, T)$ and $A(t, T)$ given in (6.5.16) and (6.5.17).
(i) Define the function
$$
\varphi(t)=\exp \left\{\frac{1}{2} \sigma^2 \int_t^T C(u, T) d u\right\} .
$$

Show that
$$
\begin{aligned}
C(t, T) & =-\frac{2 \varphi^{\prime}(t)}{\sigma^2 \varphi(t)}, \\
C^{\prime}(t, T) & =-\frac{2 \varphi^{\prime \prime}(t)}{\sigma^2 \varphi(t)}+\frac{1}{2} \sigma^2 C^2(t, T) .
\end{aligned}
$$
(ii) Use the equation (6.5.14) to show that
$$
\varphi^{\prime \prime}(t)-b \varphi^{\prime}(t)-\frac{1}{2} \sigma^2 \varphi(t)=0 .
$$

This is a constant-coefficient linear ordinary differential equation. All solutions are of the form
$$
\varphi(t)=a_1 e^{\lambda_1 t}+a_2 e^{\lambda_2 t},
$$
where $\lambda_1$ and $\lambda_2$ are solutions of the so-called characteristic equation $\lambda^2-$ $b \lambda-\frac{1}{2} \sigma^2=0$, and $a_1$ and $a_2$ are constants.
(iii) Show that $\varphi(t)$ must be of the form
$$
\varphi(t)=\frac{c_1}{\frac{1}{2} b+\gamma} e^{-\left(\frac{1}{2} b+\gamma\right)(T-t)}-\frac{c_2}{\frac{1}{2} b-\gamma} e^{-\left(\frac{1}{2} b-\gamma\right)(T-t)}
$$
for some constants $c_1$ and $c_2$, where $\gamma=\frac{1}{2} \sqrt{b^2+2 \sigma^2}$.

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Problem 5

(Two-dimensional Feynman-Kac).
(i) With $g\left(t, x_1, x_2\right)$ and $f\left(t, x_1, x_2\right)$ defined by (6.6.1) and (6.6.2), show that $g\left(t, X_1(t), X_2(t)\right)$ and $e^{-r t} f\left(t, X_1(t), X_2(t)\right)$ are martingales.
(ii) Assuming that $W_1$ and $W_2$ are independent Brownian motions, use the Itô-Doeblin formula to compute the differentials of $g\left(t, X_1(t), X_2(t)\right)$ and $e^{-r t} f\left(t, X_1(t), X_2(t)\right)$, set the $d t$ term to zero, and thereby obtain the partial differential equations (6.6.3) and (6.6.4).
(iii) Now consider the case that $d W_1(t) d W_2(t)=\rho d t$, where $\rho$ is a constant. Compute the differentials of $g\left(t, X_1(t), X_2(t)\right)$ and $e^{-r t} f\left(t, X_1(t), X_2(t)\right)$, set the $d t$ term to zero, and obtain the partial differential equations
$$
\begin{aligned}
& g_t+\beta_1 g_{x_1}+\beta_2 g_{x_2}+\left(\frac{1}{2} \gamma_{11}^2+\rho \gamma_{11} \gamma_{12}+\frac{1}{2} \gamma_{12}^2\right) g_{x_1 x_1} \\
&+\left(\gamma_{11} \gamma_{21}+\rho \gamma_{11} \gamma_{22}+\rho \gamma_{12} \gamma_{21}+\gamma_{12} \gamma_{22}\right) g_{x_1 x_2}+\left(\frac{1}{2} \gamma_{21}^2+\rho \gamma_{21} \gamma_{22}+\frac{1}{2} \gamma_{22}^2\right) g_{x_2 x_2}=0 \\
& f_t+\beta_1 f_{x_1}+\beta_2 f_{x_2}+\left(\frac{1}{2} \gamma_{11}^2+\rho \gamma_{11} \gamma_{12}+\frac{1}{2} \gamma_{12}^2\right) f_{x_1 x_1} \\
&+\left(\gamma_{11} \gamma_{21}+\rho \gamma_{11} \gamma_{22}+\rho \gamma_{12} \gamma_{21}+\gamma_{12} \gamma_{22}\right) f_{x_1 x_2} \\
&+\left(\frac{1}{2} \gamma_{21}^2+\rho \gamma_{21} \gamma_{22}+\frac{1}{2} \gamma_{22}^2\right) f_{x_2 x_2}=r f
\end{aligned}
$$

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Problem 6

(Moment-generating function for Cox-Ingersoll-Ross process).
(i) Let $W_1, \ldots, W_d$ be independent Brownian motions and let $a$ and $\sigma$ be positive constants. For $j=1, \ldots, d$, let $X_j(t)$ be the solution of the OrnsteinUhlenbeck stochastic differential equation
6.9 Exercises
287
$$
d X_j(t)=-\frac{b}{2} X_j(t) d t+\frac{1}{2} \sigma d W_j(t) .
$$

Show that
$$
X_j(t)=e^{-\frac{1}{2} b t}\left[X_j(0)+\frac{\sigma}{2} \int_0^t e^{\frac{1}{2} b u} d W_j(u)\right] .
$$

Show further that for fixed $t$, the random variable $X_j(t)$ is normal with
$$
\mathbf{E} X_j(t)=e^{-\frac{1}{2} b t} X_j(0), \quad \operatorname{Var}\left(X_j(t)\right)=\frac{\sigma^2}{4 b}\left[1-e^{-b t}\right]
$$

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Problem 7

Exercise 6.7 (Heston stochastic volatility model). Suppose that under a risk-neutral measure $\widetilde{\mathbb{P}}$ a stock price is governed by
$$
d S(t)=r S(t) d t+\sqrt{V(t)} S(t) d \widetilde{W}_1(t),
$$
where the interest rate $r$ is constant and the volatility $\sqrt{V(t)}$ is itself a stochastic process governed by the equation
$$
d V(t)=(a-b V(t)) d t+\sigma \sqrt{V(t)} d \widetilde{W}_2(t) .
$$
6.9 Exercises
289
The parameters $a, b$, and $\sigma$ are positive constants, and $\widetilde{W}_1(t)$ and $\widetilde{W}_2(t)$ are correlated Brownian motions under $\widetilde{\mathbb{P}}$ with
$$
d \widetilde{W}_1(t) d \widetilde{W}_2(t)=\rho d t
$$
for some $\rho \in(-1,1)$. Because the two-dimensional process $(S(t), V(t))$ is governed by the pair of stochastic differential equations (6.9.23) and (6.9.24), it is a two-dimensional Markov process.

So long as trading takes place only in the stock and money market account, this model is incomplete. One can create a one-parameter family of risk-neutral measures by changing the $d t$ term in (6.9.24) without affecting (6.9.23).

At time $t$, the risk-neutral price of a call expiring at time $T \geq t$ in this stochastic volatility model is $\mathbb{E}\left[e^{-r(T-t)}(S(T)-K)^{+} \mid \mathcal{F}(t)\right]$. Because of the Markov property, there is a function $c(t, s, v)$ such that
$$
c(t, S(t), V(t))=\tilde{\mathbb{E}}\left[e^{-r(T-t)}(S(T)-K)^{+} \mid \mathcal{F}(t)\right], \quad 0 \leq t \leq T .
$$

This problem shows that the function $c(t, s, v)$ satisfies the partial differential equation
$$
c_t+r s c_s+(a-b v) c_v+\frac{1}{2} s^2 v c_{s s}+\rho \sigma s v c_{s v}+\frac{1}{2} \sigma^2 v c_{v v}=r c
$$
in the region $0 \leq t<T, s \geq 0$, and $v \geq 0$. The function $c(t, s, v)$ also satisfies the boundary conditions
$$
\begin{aligned}
c(T, s, v) & =(s-K)^{+} \text {for all } s \geq 0, v \geq 0, \\
c(t, 0, v) & =0 \text { for all } 0 \leq t \leq T, v \geq 0, \\
c(t, s, 0) & =\left(s-e^{-r(T-t)} K\right)^{+} \text {for all } 0 \leq t \leq T, s \geq 0, \\
\lim _{s \rightarrow \infty} \frac{c(t, s, v)}{s-K} & =1 \text { for all } 0 \leq t \leq T, v \geq 0, \\
\lim _{v \rightarrow \infty} c(t, s, v) & =s \text { for all } 0 \leq t \leq T, s \geq 0 .
\end{aligned}
$$

In this problem, we shall be concerned only with (6.9.27).
(i) Show that $e^{-r t} c(t, S(t), V(t))$ is a martingale under $\widetilde{\mathbb{P}}$, and use this fact to obtain (6.9.26).
(ii) Suppose there are functions $f(t, x, v)$ and $g(t, x, v)$ satisfying

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Problem 8

Exercise 6.8 (Kolmogorov backward equation). Consider the stochastic differential equation
$$
d X(u)=\beta(u, X(u)) d u+\gamma(u, X(u)) d W(u) .
$$

We assume that, just as with a geometric Brownian motion, if we begin a process at an arbitrary initial positive value $X(t)=x$ at an arbitrary initial time $t$ and evolve it forward using this equation, its value at each time $T>t$ could be any positive number but cannot be less than or equal to zero. For $0 \leq t<T$, let $p(t, T, x, y)$ be the transition density for the solution to this equation (i.e., if we solve the equation with the initial condition $X(t)=x$, then the random variable $X(T)$ has density $p(t, T, x, y)$ in the $y$ variable). We are assuming that $p(t, T, x, y)=0$ for $0 \leq t<T$ and $y \leq 0$.
Show that $p(t, T ; x, y)$ satisfies the Kolmogorov backward equation
$$
-p_t(t, T, x, y)=\beta(t, x) p_x(t, T, x, y)+\frac{1}{2} \gamma^2(t, x) p_{x x}(t, T, x, y) .
$$
(Hint: We know from the Feynman-Kac Theorem, Theorem 6.4.1, that, for any function $h(y)$, the function
$$
g(t, x)=\mathbb{E}^{t, x} h(X(T))=\int_0^{\infty} h(y) p(t, T, x, y) d y
$$
satisfies the partial differential equation
$$
g_t(t, x)+\beta(t, x) g_x(t, x)+\frac{1}{2} \gamma^2(t, x) g_{x x}(t, x)=0 .
$$

Use (6.9.44) to compute $g_t, g_x$, and $g_{x x}$, and then argue that the only way (6.9.45) can hold regardless of the choice of the function $h(y)$ is for $p(t, T, x, y)$ to satisfy the Kolmogorov backward equation.)

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Problem 9

Exercise 6.8 (Kolmogorov backward equation). Consider the stochastic differential equation
$$
d X(u)=\beta(u, X(u)) d u+\gamma(u, X(u)) d W(u) .
$$

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