Problem 18.13 Medium Free 1 variation

Mean of the Vasicek Rate

Let $W_t$ be a standard Brownian motion and let $R_t$ satisfy the Vasicek SDE

dR_t = (a - bR_t)\,dt + \sigma\,dW_t, \qquad R_0 = r,

for constants $a,b,\sigma > 0$ and $r>0$ . Write $\mathbb{E}[R_T]$ as an explicit function of $a,b,r,T$ (note that it does not depend on $\sigma$ ). Evaluate this function for $T=10$ , $a=0.2$ , $b=0.1$ , $r=1$ : the result has the form $A + Be^{C}$ for integers $A,B,C$ . Find $A+B+C$ .

Work it out, then compare with the answer.

SolutionShow the worked solution

Key idea. The drift is linear in $R_t$ , so the integrating factor $e^{bt}$ removes $R_t$ from the drift. The noise term then has deterministic integrand and mean zero.

By Itô’s product rule (the factor $e^{bt}$ has finite variation, so there is no cross-variation term),

d\big(e^{bt}R_t\big) = be^{bt}R_t\,dt + e^{bt}\,dR_t = ae^{bt}\,dt + \sigma e^{bt}\,dW_t .

Integrating from $0$ to $T$ gives

R_T = re^{-bT} + \frac{a}{b}\big(1-e^{-bT}\big) + \sigma\int_0^T e^{-b(T-s)}\,dW_s .

The last integral has a deterministic, square-integrable integrand. It is therefore a centred Gaussian random variable, with mean $0$ . Hence

\mathbb{E}[R_T] = re^{-bT} + \frac{a}{b}\big(1-e^{-bT}\big),

which does not involve $\sigma$ . Equivalently, $m(t)=\mathbb{E}[R_t]$ solves $m'=a-bm$ with $m(0)=r$ .

With $T=10$ , $a=0.2$ , $b=0.1$ and $r=1$ , we have $bT=1$ and $a/b=2$ . So

\mathbb{E}[R_{10}] = e^{-1} + 2\big(1-e^{-1}\big) = 2 - e^{-1}.

This gives $A=2$ , $B=-1$ , $C=-1$ . The representation is unique because $e$ is transcendental. Therefore

\boxed{\mathbb{E}[R_T] = re^{-bT} + \tfrac{a}{b}\big(1-e^{-bT}\big),\qquad A+B+C = 0}

Variations

The same idea with a twist. Each solution gives only the changed step.

Variation 18.13a

Let $W_t$ be a standard Brownian motion and let $X_t$ satisfy the CIR-type SDE

dX_t = \kappa(\theta - X_t)\,dt - \sigma\sqrt{X_t}\,dW_t, \qquad X_0 = x > 0,

for constants $\kappa,\theta,\sigma>0$ . Assume $2\kappa\theta > \sigma^2$ , which guarantees $X_t>0$ almost surely, so $\sqrt{X_t}$ is well defined. Write $\mathbb{E}[X_T]$ as an explicit function of $x,\kappa,\theta,T$ (it does not depend on $\sigma$ ). Evaluate this function for $T=10$ , $\kappa=0.2$ , $\theta=2$ , $x=5$ .

Work it out, then compare with the answer.

Solution

The same integrating factor gives

d\big(e^{\kappa t}X_t\big) = \kappa\theta e^{\kappa t}\,dt - \sigma e^{\kappa t}\sqrt{X_t}\,dW_t .

Now the stochastic integral has a random integrand, so we must check that it has mean zero.

Step 1: $\mathbb{E}[X_t]$ is finite. Localise with stopping times $\tau_n\uparrow\infty$ at which the stochastic integral is stopped. This gives

\mathbb{E}\big[e^{\kappa(t\wedge\tau_n)}X_{t\wedge\tau_n}\big] \le x+\theta(e^{\kappa t}-1).

Since $X\ge 0$ , Fatou’s lemma yields $\sup_{t\le T}\mathbb{E}[X_t]<\infty$ .

Step 2: the integral is a true martingale. By Step 1, $\mathbb{E}\int_0^T e^{2\kappa s}X_s\,ds<\infty$ , so the integral is a true martingale with mean zero.

Taking expectations, $m(t)=\mathbb{E}[X_t]$ solves $m'=\kappa(\theta-m)$ with $m(0)=x$ . Hence

\mathbb{E}[X_T]=\theta+(x-\theta)e^{-\kappa T}.

With $\kappa T=2$ , $\theta=2$ and $x=5$ , this is $2+3e^{-2}\approx 2.406$ .

\boxed{\mathbb{E}[X_T]=\theta+(x-\theta)e^{-\kappa T},\qquad \mathbb{E}[X_{10}]=2+3e^{-2}}

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