# Expected value of the stochastic integral $\int_0^t e^{as} dW_s$

I am trying to calculate a stochastic integral $\mathbb{E}[\int_0^t e^{as} dW_s]$. I tried breaking it up into a Riemann sum $\mathbb{E}[\sum e^{as_{t_i}}(W_{t_i}-W_{t_{i-1}})]$, but I get expected value of $0$, since $\mathbb{E}(W_{t_i}-W_{t_{i-1}}) =0$. But I think it's wrong. Thanks!

And I want to calculate $\mathbb{E}[W_t \int_0^t e^{as} dW_s]$ as well, I write $W_t=\int_0^t dW_s$ and get $\mathbb{E}[W_t \int_0^t e^{as} dW_s]=\mathbb{E}[\int_0^t e^{as} dW_s]$.

Is that ok?

($W_t$ is brownian motion.)

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@Srivatsan: $W_s$ is very common notation for standard Brownian motion or Wiener process (en.wikipedia.org/wiki/Wiener_process) – George Lowther Oct 26 '11 at 2:49
And the expected value is zero, for the reason mentioned in the question (although to be rigorous you should be careful taking the limit). – George Lowther Oct 26 '11 at 2:50
I think our [stochastic integral] tread is quite poor, so maybe we can re-tag appropriate question with this tag. – Ilya Oct 26 '11 at 8:53

The expectation of the Ito integral $\mathbb{E}( \int_0^t \mathrm{e}^{a s} \mathrm{d} W_s )$ is zero as George already said.
To compute $\mathbb{E}( W_t \int_0^t \mathrm{e}^{a s} \mathrm{d} W_s )$, write $W_t = \int_0^t \mathrm{d} W_s$. Then use Ito isometry:
$$\mathbb{E}( W_t \int_0^t \mathrm{e}^{a s} \mathrm{d} W_s ) = \mathbb{E}\left( \int_0^t \mathrm{d} W_s \cdot \int_0^t \mathrm{e}^{a s} \mathrm{d} W_s \right) = \int_0^t (1 \cdot \mathrm{e}^{a s}) \mathrm{d} t = \frac{\mathrm{e}^{a t} - 1}{a}$$