# change of measure expectation How to find expectation of this stochastic process? Also, to show that the expectation of a stochastic process expression [Xt - St] in one measure is equal to expectation of another expression (of the mentioned stochastic process) in another measure?

Given: $$S_t=S_0e^{\sigma W_t+(r-\sigma^2/2)t}$$

$$dS_t = rS_tdt + \sigma S_tdW_t$$

• What is your dynamic of $S_t$? – starovoitovs Mar 27 '19 at 3:23
• $S_t=S_0e^{\sigma W_t+(r-\sigma^2/2)t}$ It took me forever to type this up using latex. – happyGiraffe Mar 27 '19 at 4:09
• Hint: take the log of $S_t$, write out the integral in the exponent and you'll have one non-random term and one term which is $\sigma \int_0^t W_u \mathrm{d}u$. Then see quant.stackexchange.com/questions/29504. This integral is normally distributed which should help you compute the expectation you are looking for. – LocalVolatility Mar 27 '19 at 8:34
• Please show your work. What have you tried, and how far have you gotten? Also, if this is a homework assignment, it should be marked as such. – AdB Mar 28 '19 at 9:57

Your process for $$(S_t)$$ is a geoemtric Brownian motion and since $$S_t=S_0 e^{\left(r-\frac{1}{2}\sigma^2\right)t+\sigma W_t}$$, we have \begin{align*} \ln(S_t) &= \ln(S_0)+\left(r-\frac{1}{2}\sigma^2\right)t+\sigma W_t \\ &\sim N\left(\ln(S_0)+\left(r-\frac{1}{2}\sigma^2\right)t,\sigma^2 t\right). \end{align*} Thus, \begin{align*} X_t &= e^{\frac{1}{t}\int_0^t \ln(S_u)\mathrm{d}u} \\ &= e^{\frac{1}{t}\int_0^t \left(\ln(S_0)+\left(r-\frac{1}{2}\sigma^2\right)u\right)\mathrm{d}u}\cdot e^{\frac{1}{t}\sigma\int_0^t W_u\mathrm{d}u} \\ &= S_0\cdot e^{\frac{1}{2}\left(r-\frac{1}{2}\sigma^2\right)t}\cdot e^{\frac{1}{t}\sigma\int_0^t W_u\mathrm{d}u}. \end{align*} Fortunately, the time integral of a Brownian motion is well-known to be normally distributed with mean zero and variance $$\frac{1}{3}t^3$$, see here. Remember that if $$Z\sim N(0,1)$$, then $$\mathbb{E}\left[e^Z\right]=e^{\frac{1}{2}}$$ and thus $$\mathbb{E}\left[e^{m+s Z}\right]=e^{m+\frac{1}{2}s^2}$$. As a consequence, $$X_t$$ is log-normally distributed with \begin{align*} \mathbb{E}[X_t] &= S_0\cdot e^{\frac{1}{2}\left(r-\frac{1}{2}\sigma^2\right)t}\cdot \mathbb{E}\left[e^{\frac{1}{t}\sigma\int_0^t W_u\mathrm{d}u}\right] \\ &= S_0\cdot e^{\frac{1}{2}\left(r-\frac{1}{2}\sigma^2\right)t}\cdot \mathbb{E}\left[e^{\frac{1}{t}\sigma \sqrt{\frac{1}{3}t^3}Z} \right] \\ &= S_0\cdot e^{\frac{1}{2}\left(r-\frac{1}{2}\sigma^2\right)t}\cdot \mathbb{E}\left[e^{ \sqrt{\frac{1}{3}t\sigma^2}Z} \right] \\ &= S_0\cdot e^{\frac{1}{2}\left(r-\frac{1}{2}\sigma^2\right)t}\cdot e^{ \frac{1}{2}\frac{1}{3}\sigma^2t} \\ &= S_0\cdot e^{\frac{1}{2}\left(r-\frac{1}{6}\sigma^2\right)t}. \end{align*}