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

$dS_t = rS_tdt + \sigma S_tdW_t$

  • $\begingroup$ What is your dynamic of $S_t$? $\endgroup$ – starovoitovs Mar 27 '19 at 3:23
  • $\begingroup$ $S_t=S_0e^{\sigma W_t+(r-\sigma^2/2)t}$ It took me forever to type this up using latex. $\endgroup$ – happyGiraffe Mar 27 '19 at 4:09
  • 1
    $\begingroup$ 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. $\endgroup$ – LocalVolatility Mar 27 '19 at 8:34
  • $\begingroup$ 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. $\endgroup$ – 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*}

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