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I have the following SDE

$$ dx = -at^2 xdt + gdW $$

that I apply the following transformation to, $y=e^{at^3/3}$, which gives me the e^{at^3/3}for $y$

$$ dy = e^{at^3/3}gdW $$

Now, solving this is easy and I get (after transforming back to $x$)

$$ x(t) = x_0e^{-at^3/3} + ae^{-at^3/3}\int_0^t{e^{as^3/3}dW(s)} $$

Now, I am asked to find the mean and variance of $x$. For the mean I get

$$ \langle x \rangle = \langle x_0e^{-at^3/3} \rangle + \langle ae^{-at^3/3}\int_0^t{e^{as^3/3}dW(s)} \rangle \\ = x_0e^{-at^3/3}+ ae^{-at^3/3}\int_0^t{e^{as^3/3}\langle dW(s)} \rangle \\ = x_0e^{-at^3/3} $$

How would I go about finding the variance? Can I use the solution $x$ I found, or do I have to solve $d(x^2)$?

Tyler D
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    You can use the solution you found as long as you know how to evaluate $$\left\langle \left(\int_0^t g(s),dW(s)\right)^2\right\rangle$$ which amounts to knowing the quadratic variation of the brownian motion. – Brian Moehring Aug 30 '19 at 07:25
  • @BrianMoehring My idea is that $\langle ( \int_0^t g(s) dW(s) )^2\rangle = Var(\int_0^t g(s) dW(s)) = \int_0^t g(s)^2 ds$. Is this correct? – Tyler D Aug 30 '19 at 15:25
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    Yes, that's what you need to use. Just use $$\text{Var}(x) = \left\langle\left(x - \langle x\rangle\right)^2\right\rangle$$ along with that identity. – Brian Moehring Aug 31 '19 at 02:29
  • So I find that the variance of $x$ is given by $V(x) = g^2 e^{-2at^3/3}\int_0^t e^{2as^3/3}ds$. However, that integral doesn't look right to me. Do you agree with me? – Tyler D Aug 31 '19 at 08:09
  • So after solving the SDE, we have $m_t = E[x_t] = x_0e^{-at^3/3} + ae^{-at^3/3}E[\int_0^t{e^{as^3/3}dW(s)}] = x_0e^{-at^3/3}$ as the stochastic integral is a Wiener process. The variance: we have $v_t = E[x_t^2] - m_t^2$. But $E[x_t^2]= (x_0e^{-at^3/3})^2 + E[(ae^{-at^3/3}\int_0^t{e^{as^3/3}dW(s))^2}] + E[x_0ae^{-2at^3/3} \int_0^t{e^{as^3/3}dW(s)}] = (x_0e^{-at^3/3})^2+a^2e^{-2at^3/3}\int_0^t{e^{2as^3/3}ds}$ – Sesame Sep 10 '19 at 17:25
  • I used the ito isometry for the 2nd term and the mean of the stochastic integral is $0$ because it is a wiener process. Finally, $v_t = a^2e^{-2at^3/3}\int_0^t{e^{2as^3/3}ds}$ – Sesame Sep 10 '19 at 17:27

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Indeed using the integration factor Solution to General Linear SDE we obtain

$$x(t) = x_0e^{-at^3/3} + ae^{-at^3/3}\int_0^t{e^{as^3/3}dW(s)}.$$

The Itô-integral has mean zero. For the variance, we can use Itô isometry to get

$$Ex^{2}(t) = E\left[\left(ae^{-at^3/3}\int_0^t{e^{as^3/3}dW(s)}\right)^{2}\right]=a^{2} e^{-a2t^3/3}\int_0^t e^{a2s^3/3}ds.$$

Thomas Kojar
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