Chapter 6 - Partial Differential Equations Exercises

Exercise 6.1

Consider the stochastic differential equation

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

(i) Show that $Z(t) = 1$ and

Let

so that $Z(u) = \exp(f(u))$. Applying Itô’s formula:

where

• $df(u) = \sigma(u)\,dW(u) + \left(b(u) - \frac{1}{2}\sigma^2(u)\right)\,du$,
• $df = \sigma^2(u)\,du$.

Then,

Also, since the integrals in the exponential defining $Z(u)$ start at $t$, we have:

(ii) By its very definition, $Y(u)$ satisfies $Y(t) = x$, and

Now define $X(u) = Y(u)Z(u)$ and apply the product rule for Itô calculus:

Substitute:

• $dZ(u) = b(u)Z(u)\,du + \sigma(u)Z(u)\,dW(u)$,
• $dY(u) = \frac{a(u) - \sigma(u)\gamma(u)}{Z(u)}\,du + \frac{\gamma(u)}{Z(u)}\,dW(u)$,
• The cross term is:

Putting everything together:

since $X(u) = Y(u)Z(u)$.

Also, $X(t) = Y(t)Z(t) = x \cdot 1 = x$, so the initial condition is satisfied.

Hence, $X(u) = Y(u)Z(u)$ solves the stochastic differential equation and satisfies the initial condition.

Exercise 6.2 (No-arbitrage derivation of bond-pricing equation)

Suppose the interest rate is given by the stochastic differential equation

where $W(t)$ is a Brownian motion under a probability measure $\mathbb{P}$ (not necessarily risk-neutral). The price of a $T$-maturity zero-coupon bond at time $t$ is given by a function $f(t, R(t), T)$.

Assume that $f_r(t, r, T) \neq 0$ and define

Then the PDE for $f$ is

(i) Consider two maturities $0 \leq T_1 \leq T_2$, and a portfolio that holds at time $t \leq T_1$:

• $\Delta_1(t)$ units of the bond maturing at $T_1$,
• $\Delta_2(t)$ units of the bond maturing at $T_2$,

and finances this position by investing or borrowing at the interest rate $R(t)$. Let $X(t)$ denote the value of the portfolio. Then:

Grouping terms and substituting the definition of $\beta(t, r, T)$:

This is equation (6.9.4) from the textbook.

(ii) Define the sign function:

and let

Set the portfolio weights as:

• $\Delta_1(t) = S(t) f_r(t, R(t), T_2)$,
• $\Delta_2(t) = -S(t) f_r(t, R(t), T_1)$.

Substitute into the drift term of $d(D(t)X(t))$:

This becomes:

Since $S(t)$ has the same sign as the expression in the brackets, this term is strictly positive unless $\beta(t, R(t), T_1) = \beta(t, R(t), T_2)$. Hence, there is an arbitrage unless $\beta(t, r, T)$ is independent of $T$. Since $T_1$ and $T_2$ are arbitrary, we conclude:

(iii) Now let $T > 0$ be given and consider a portfolio that only invests in the $T$-maturity bond. Then:

where all functions are evaluated at $(t, R(t), T)$. This is equation (6.9.5).

If $f_r(t, r, T) = 0$ but the drift term above is non-zero, then there is an arbitrage opportunity. Therefore, in the case $f_r = 0$, we must have:

as in equation (6.9.6).

Hence, even if $f_r = 0$, the bond pricing PDE still holds.

Conclusion: If zero-coupon bond markets are arbitrage-free, then for all $t, r, T$ such that $f_r(t, r, T) \neq 0$, the function $\beta(t, r, T)$ defined in (6.9.2) must not depend on $T$. Therefore, the bond price $f(t, r, T)$ satisfies:

which is equation (6.5.4).

Under the measure $\widetilde{\mathbb{P}}$ defined by:

$\widetilde{W}(t)$ is a Brownian motion, and the SDE for $R(t)$ becomes the risk-neutral one, so $\widetilde{\mathbb{P}}$ is risk-neutral.

Exercise 6.3 (Solution of Hull-White model)

This exercise solves the ODEs (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 $t$. The equation is:

Multiply both sides by $\exp\left(-\int_0^s b(v)\,dv\right)$:

This is the same as:

So we have shown:

(ii) Integrate both sides from $s = t$ to $s = T$:

The left-hand side becomes:

Using $C(T, T) = 0$, we have:

Multiplying both sides by $-1$:

This is equation (6.5.10).

(iii) Replace $t$ by $s$ in (6.5.9):

Integrate from $s = t$ to $s = T$ and use $A(T, T) = 0$:

Then:

This is equation (6.5.11).

Exercise 6.4 (Solution of Cox-Ingersoll-Ross model)

This exercise solves the ODEs (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

Differentiate:

• First, show

• Then, differentiate $C(t, T)$:

This gives equations (6.9.8) and (6.9.9).

(ii) Use equation (6.5.14):

Substitute $C(t, T) = -\frac{2\varphi’(t)}{\sigma^2 \varphi(t)}$ and plug into the equation above. After simplification, the condition for equality is:

This is equation (6.9.10).

(iii) This is a second-order linear ODE with constant coefficients. The general solution is:

where $\lambda_{1,2}$ are the roots of the characteristic equation:

(iv) Solving the characteristic equation:

So the general solution is:

This is equation (6.9.11).

(v) Differentiate $\varphi(t)$:

Use $C(T, T) = 0$ in the formula for $C(t, T)$:

(vi) Plug $c_1 = c_2$ into $\varphi(t)$:

Simplify the expression:

Then:

which matches equation (6.5.16).

(vii) From (6.5.15) and (6.9.8):

Replace $t$ by $s$ and integrate from $s = t$ to $s = T$:

This matches equation (6.5.17).

Exercise 6.5 (Two-dimensional Feynman-Kac)

(i) With $g(t, x_1, x_2)$ and $f(t, x_1, x_2)$ defined by (6.6.1) and (6.6.2), we want to show that:

• $g(t, X_1(t), X_2(t))$ and
• $e^{-rt}f(t, X_1(t), X_2(t))$

are martingales.

This follows from the fact that both $g$ and $f$ solve their respective PDEs and that the SDEs for $X_1(t)$ and $X_2(t)$ are Markovian. If the discounted process $e^{-rt}f(t, X_1(t), X_2(t))$ has zero drift (i.e., the $dt$ term vanishes), then it is a martingale under the risk-neutral measure.

(ii) Assume $W_1(t)$ and $W_2(t)$ are independent Brownian motions. Apply Itô’s formula to $g(t, X_1(t), X_2(t))$:

Let the dynamics be:

Then the Itô formula gives:

Using independence, $dW_1 dW_2 = 0$, and computing $(dX_i)^2$:

• $(dX_1)^2 = \gamma_{11}^2\,dt + \gamma_{12}^2\,dt$,
• $(dX_2)^2 = \gamma_{21}^2\,dt + \gamma_{22}^2\,dt$,
• $dX_1 dX_2 = \gamma_{11}\gamma_{21}\,dt + \gamma_{12}\gamma_{22}\,dt$.

Setting the drift term to zero leads to:

which is PDE (6.6.3).

Now apply Itô’s formula to $e^{-rt}f(t, X_1(t), X_2(t))$ and set $dt$ term to zero:

which gives the Feynman-Kac PDE (6.6.4).

(iii) Now assume:

so the Brownian motions are correlated. Then:

• $dX_1 dX_2$ contains cross terms:

Thus the mixed second-order term becomes:

Putting everything together, the PDE for $g$ becomes:

which is equation (6.9.13).

Similarly, for $f$, include the $-rf$ term:

which is equation (6.9.14).

Exercise 6.6 (Moment-generating function for Cox-Ingersoll-Ross process)

(i) Let $X_j(t)$ solve the Ornstein-Uhlenbeck SDE:

To solve this, use integrating factor $e^{\frac{b}{2}t}$:

Then:

which is equation (6.9.16).

Now show $X_j(t)$ is normal with:

• Mean:

• Variance:

which gives equation (6.9.17).

(ii) Define:

Apply Itô’s lemma to $X_j^2(t)$:

Then:

Let $a = \frac{d\sigma^2}{4}$ and define:

Then:

which is the CIR process in equation (6.9.19). Since $B(t)$ is constructed as a stochastic integral of orthonormal components, by Lévy’s theorem, $B(t)$ is a Brownian motion — this is equation (6.9.20).

(iii) Suppose $R(0) > 0$ is given, and define:

Then all $X_j(t)$ are independent and identically distributed normal random variables with:

• Mean:

• Variance:

so $X_j(t) \sim \mathcal{N}(\mu(t), v(t))$.

(iv) Since $R(t) = \sum_{j=1}^d X_j^2(t)$ is a sum of squares of i.i.d. normal variables with nonzero mean, it follows a noncentral chi-squared distribution. To compute the MGF of $R(t)$, first compute the MGF of $X_j^2(t)$:

which is equation (6.9.21), using completion of the square and integration of the normal density.

(v) Since $R(t) = \sum_{j=1}^d X_j^2(t)$, and the $X_j^2(t)$ are i.i.d., the moment-generating function of $R(t)$ is:

Using $d = \frac{4a}{\sigma^2}$, this becomes:

valid for all $u < \frac{1}{2v(t)}$, which is equation (6.9.22).

Exercise 6.7 (Heston stochastic volatility model)

(i) Show that $e^{-rt}c(t, S(t), V(t))$ is a martingale under $\widetilde{\mathbb{P}}$.

Using Itô’s formula on $c(t, S(t), V(t))$ where $S(t)$ and $V(t)$ follow (6.9.23) and (6.9.24), and applying the product rule to $e^{-rt}c(t, S(t), V(t))$, we require the drift of this process to be zero. Setting the $dt$ term to zero gives the PDE:

which is equation (6.9.26).

(ii) Suppose $f(t, x, v)$ and $g(t, x, v)$ satisfy the PDEs:

Then, define

Show that $c(t, s, v)$ satisfies (6.9.26).

To verify this, apply the chain rule:

• $c_s = f + s f_x$,
• $c_{ss} = \frac{1}{s}f_x + f_{xx} + s f_{xx}$,
• $c_t = s f_t - re^{-r(T - t)}K g - e^{-r(T - t)}K g_t$,
• $c_v = s f_v - e^{-r(T - t)}K g_v$,
• $c_{sv}, c_{vv}$ likewise follow from applying product/chain rule.

Then substitute into the LHS of (6.9.26) and simplify using (6.9.32) and (6.9.33).

(iii) Let $(X(t), V(t))$ solve:

Define:

Using Feynman-Kac, $f$ solves (6.9.32) with terminal condition:

(iv) Now let $(X(t), V(t))$ follow:

Define:

Then $g$ satisfies (6.9.33) and boundary condition:

(v) Since $f$ and $g$ satisfy their PDEs and terminal conditions as shown, the function

satisfies the PDE (6.9.26) and boundary condition (6.9.27):

Exercise 6.8 (Kolmogorov backward equation)

We are given the SDE:

Let $p(t, T; x, y)$ denote the transition density of $X(T)$ given $X(t) = x$. We want to show that $p$ satisfies the Kolmogorov backward equation:

We use the Feynman-Kac Theorem, which says for any $h(y)$,

satisfies the PDE:

Now differentiate under the integral in (6.9.44):

• $g_t = \int h(y) p_t\,dy$,
• $g_x = \int h(y) p_x\,dy$,
• $g_{xx} = \int h(y) p_{xx}\,dy$.

Substitute into (6.9.45):

Since this holds for all $h(y)$, we conclude:

which is equation (6.9.43).

Exercise 6.9 (Kolmogorov forward equation)

Again, consider:

Let $p(t, T; x, y)$ be the transition density. We aim to show that $p$ satisfies the Kolmogorov forward (Fokker-Planck) equation:

(i) Let $h_b(y)$ be a test function with compact support on $(0, b)$, satisfying $h_b(0) = h_b’(0) = h_b(b) = h_b’(b) = 0$.

Apply Itô’s lemma to $h_b(X(u))$:

Take expectation and integrate from $t$ to $T$:

(ii) Replace $\mathbb{E}[h_b(X(T))]$ using the transition density:

(iii) Apply integration by parts in the $y$ variable to the RHS:

(iv) Differentiate (6.9.49) w.r.t. $T$:

(v) Since $h_b(y)$ is arbitrary, the integrand must be zero:

which is equation (6.9.47), the Kolmogorov forward equation.

Exercise 6.10 (Implying the volatility surface)

We are given the SDE:

and define the transition density $\widetilde{p}(t, T; x, y)$ of $S(T)$ given $S(t) = x$.

According to Exercise 6.9, this satisfies the Kolmogorov forward equation:

Define the time-zero price of a European call as:

Differentiate this in $T$:

(i) Use integration by parts to show:

Assuming the boundary term vanishes:

(ii) Now compute:

Integrate by parts twice, applying (6.9.57) and (6.9.58) which ensure boundary terms vanish:

(iii) Plug everything back into equation (6.9.53):

Start with:

and use (6.9.51), (6.9.54), (6.9.56) to get:

Note:

• $\int_K^\infty \widetilde{p} = \frac{\partial}{\partial K} c(0, T, x, K)$,
• $\widetilde{p}(0, T; x, K) = \frac{\partial^2}{\partial K^2} c(0, T, x, K)$.

So:

This is the final equation. Under the assumption $c_{KK}(0, T, x, K) \neq 0$, we can solve (6.9.59) for $\sigma^2(T, K)$ in terms of:

• $c_T(0, T, x, K)$,
• $c_K(0, T, x, K)$,
• $c_{KK}(0, T, x, K)$,

which are quantities that can be inferred from observable option prices.