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Suppose we have the following dynamics: $$dX_t = a_t X_t dt + b_t X_t dW_t; \;\;\; X_0 = x.$$

Would the expectation of $X_t$ at $t=T$ be $$\mathbf{E} [X_T] = x + \int_0^T a_t X_t dt$$?

I know that it would be correct if the $X_t$ wasn't there in the SDE, but with $X_t$, it doesn't quite seem right. Would anyone be able to help me on this? Thanks!

spideyonthego
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  • The expectation needs to be applied on both sides. Then you could ask if the time integral and the expectation can be interchanged. That is usually true. – Lutz Lehmann Oct 25 '21 at 12:20
  • Thanks @LutzLehmann, I get $E[X_T] = x + \int_0^T E[a_t X_t] dt$. I don't see how I could simplify that further; do you know if there is a way? – spideyonthego Oct 25 '21 at 12:34
  • $a_t$ is (usually) a constant at any fixed time $t$, compared to $X_t(\omega)$, so $E(a_tX_t)=a_tE(X_t)$. This then gives a separable ODE for $E(X_t)$. – Lutz Lehmann Oct 25 '21 at 12:55
  • Right; I did the maths and I get $E[X_T] = xe^{\int_0^T a_t dt}$. Is this correct? Thanks very much – spideyonthego Oct 25 '21 at 13:03
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    Yes, that is the solution of the resulting ODE. – Lutz Lehmann Oct 25 '21 at 13:43

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From here Solution to General Linear SDE, we have the solution

\begin{align*} X_t =& X_0\exp\left( \int_0^t\left( a(s) - \frac{1}{2}b^2(s) \right) \mathrm{d}s + \int_0^t b(s)\mathrm{d}B_s\right). \end{align*}

Then here we use that the Itô integral $\int_0^t b(s)\mathrm{d}B_s$ is a Gaussian with second moment $ \int_0^t\frac{1}{2}b^2(s)ds$, so indeed the mean is

$$E[X_t]= X_0\exp\left( \int_0^t a(s) \mathrm{d}s \right).$$

Thomas Kojar
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