I need help to understand a couple of calculations in this Girsanov theorem related SDE problem. I have five questions as stated below.

Let $X_t$ solve the Ornstein-Uhlenbeck equation $$dX_t = X_t\, dt + dB_t, \quad X_0 = x$$ and show: $$E[f(X_t)] = E\left[\exp\left\{\frac{1}{2}(W_t^2-t)+xW_t - \int_0^t(W_s + x)^2ds\right\}f(W_t+x)\right]$$

The solution specifies that we set:

$dX_t = dW_t, \quad X_0 = x$

Question 1: How do we know what to set $dX_t$ to?

We use Girsanov transformation

$dQ = L(T)dP$


$dL_t = X_t L_t dW_t$ $L_0 = 1$

The Girsanov theorem then tells us that $$dW_t = X_tdt + dB_t$$ where $B$ is a $Q$-Brownian motion. The SDE for X thus becomes $$dX_t = X_tdt + dB_t,$$ which means that $$X_t = x + \int_0^t X_s ds + B_t.$$

Question 2: Is this not the same as given in the problem statement? In this case why did we need to use Girsanov to draw this circle conclusion?

Now using that $Z \in F_t$ then:

$$E^Q[Z] = E^P[L_tZ]$$

Question 3: How do we know this?

we obtain $$E^Q[f(X_t)] = E^P[L_tf(X_t)]. $$ The solution of $dL_t$ with $L_0 = 1$ is given by $$L_t = \exp \left\{\int_0^t X_s dW_s - \frac{1}{2} \int_0^t X_s^2 ds\right\}.$$

Question 4: how does the term $-\frac{1}{2}\int_0^t X_s^2 ds$ get there?

From calculations I only get: $$dL_t = X_tL_tdW_t, \quad L_0 = 1\\ \ln(L_t) = \int_0^tX_s dW_s + L_0 = \int_0^tX_s dW_s + 1\\ L_t = \exp\left\{\int_0^tX_s dw_s + 1\right\},$$ which is not the same as stated in the problem solution.

Now lastly using $X_t = x + W_t$ under $P$ we obtain $$\int_0^t X_s dW_s = \int_0^t (x+W_t) dW_s = \int_0^t x dW_s + \int_0^t W_tdW_s = xW_t + \frac{1}{2}(W_t^2 - t).$$

Question 5: How do we find that: $$\int_0^t W_t dW_s = \frac{1}{2}(W_t^2 -t ).$$

Now inserting this gives the expectation as asked for.


Question 1 (Girsanov's theorem) Let $W_t$ be a Brownian motion under the physical measure $\mathbb{P}$. Define $$L_t := \exp \left\{-\int_{0}^{t} X_s dW_s - \frac{1}{2} \int_{0}^{t} X_s^2ds \right\},$$ and define an equivalent martingale measure $\mathbb{Q}$ by setting $d\mathbb{Q}/d\mathbb{P} = L_t$, then $B_t = W_t + \int_{0}^{t} X_s ds$ is a standard Brownian motion under $\mathbb{Q}$.

Question 2+3 It follows from Girsanov's theorem: for any measurable subset $A$ $$\mathbb{E}_{\mathbb{Q}}[Z] = \int_{A} Z d\mathbb{Q} = \int_{A} Z L_t d\mathbb{P}.$$

Question 4 It's how $L_t$ is defined, show instead by applying Ito's lemma that $dL_t:= -L_t X_t dW_t$.

Question 5 This follows by Ito's formula. If $f$ is a twice differentiable real valued function then $$f(W_t) = f(W_0) + \int_{0}^{t} f'(W_s)dW_s + \frac{1}{2} \int_{0}^{t} f''(W_s)ds.$$ Apply Ito's formula to $f(x) = x^2$ and the result follows.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.