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changed bracket notation around the expectation
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lkdo
lkdo
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I am wondering if the following expression can be processed a bit analytically, $$ E \left< e^{aX} \int_0^X e^{bu}dW(u)\right>, $$$$ E \left[ e^{aX} \int_0^X e^{bu}dW(u)\right], $$ where $W_u$ is the normal Brownian motion (1D Wiener process), and $X$ is a random variable.

I know that $E \left< \int_0^T e^{bu}dW(u)\right>$$E \left[ \int_0^T e^{bu}dW(u)\right]$ is zero, so $E \left< \int_0^X e^{bu}dW(u)\right>$$E \left[ \int_0^X e^{bu}dW(u)\right]$ should also be zero. However $e^{aX}$ and $\int_0^X e^{bu}dW(u)$ are clearly correlated, so the expectation of their product is not trivial.

I am wondering if the following expression can be processed a bit analytically, $$ E \left< e^{aX} \int_0^X e^{bu}dW(u)\right>, $$ where $W_u$ is the normal Brownian motion (1D Wiener process), and $X$ is a random variable.

I know that $E \left< \int_0^T e^{bu}dW(u)\right>$ is zero, so $E \left< \int_0^X e^{bu}dW(u)\right>$ should also be zero. However $e^{aX}$ and $\int_0^X e^{bu}dW(u)$ are clearly correlated, so the expectation of their product is not trivial.

I am wondering if the following expression can be processed a bit analytically, $$ E \left[ e^{aX} \int_0^X e^{bu}dW(u)\right], $$ where $W_u$ is the normal Brownian motion (1D Wiener process), and $X$ is a random variable.

I know that $E \left[ \int_0^T e^{bu}dW(u)\right]$ is zero, so $E \left[ \int_0^X e^{bu}dW(u)\right]$ should also be zero. However $e^{aX}$ and $\int_0^X e^{bu}dW(u)$ are clearly correlated, so the expectation of their product is not trivial.

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lkdo
lkdo
  • 41
  • 4

Expected value of a stochastic integral expression

I am wondering if the following expression can be processed a bit analytically, $$ E \left< e^{aX} \int_0^X e^{bu}dW(u)\right>, $$ where $W_u$ is the normal Brownian motion (1D Wiener process), and $X$ is a random variable.

I know that $E \left< \int_0^T e^{bu}dW(u)\right>$ is zero, so $E \left< \int_0^X e^{bu}dW(u)\right>$ should also be zero. However $e^{aX}$ and $\int_0^X e^{bu}dW(u)$ are clearly correlated, so the expectation of their product is not trivial.