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Suppose we have a stochastic process

$$dX_t = X_t\mu dt+X_t\sigma dW_t$$ where $W_t$ is a Wiener process. How does one evaluate the following expected value:

$$E\left[{\frac{1}{\sqrt{X_t}}}\right]$$

or in more detalized fashion,

$$E\left[X_0^\frac{-1}{2}e^{-\frac{1}{2}\left((\mu-\frac{1}{2}\sigma^{2})t+\sigma W_t\right)} \right]$$

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To evaluate the expectation, we can rearrange the expression you have given us as $$ E\left[X_0^{-\frac{1}{2}} e^{-\frac{1}{2}\left(\left(\mu - \frac{1}{2}\sigma^2\right)t + \sigma W_t\right)}\right] = E\left[X_0^{-\frac{1}{2}} e^{-\sigma\frac{1}{2}W_t - \frac{1}{4}\sigma^2t}e^{\frac{1}{2}\left(\sigma^2 - \mu\right)t}\right] = e^{\frac{1}{2}\left(\sigma^2 - \mu \right)t}E\left[X_0^{-\frac{1}{2}}\right]E\left[e^{-\frac{1}{2}\sigma W_t - \frac{1}{4}\sigma^2 t}\right]$$ The right most expectation is 1 as $\exp\left(-\frac{1}{2}\sigma W_t - \frac{1}{4}\sigma^2 t\right)$ is a martingale. This is an example of an exponential martingale. It is easy to check that for any $\theta \in \mathbb{R}$, the process $\left(\exp{\left(\theta W_t - \frac{1}{2}\theta^2 t\right)}\right)_{t\geq 0}$ is a martingale. Thus $$E\left[\frac{1}{\sqrt{X_t}}\right] = e^{\frac{1}{2}\left(\sigma^2 - \mu\right)t}E\left[X_0^{-\frac{1}{2}}\right]$$

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