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What is the mean and the variance of $y_t$, given the following SDE:

$dy_t = -x_t y_t dt + \sigma_1 dW^1_t$

$dx_t = -\sigma_2 y_t dW^2_t$

$W^1$ and $W^2$ are (possibly correlated) Wiener processes.

Especially, is it true that $Var[dy_t] = \sigma_1^2 dt$, or is that not necessarily correct due to the process $x_t$ in the drift of $y_t$?

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closed as off-topic by zhoraster, Did, E. Joseph, tilper, Watson Dec 22 '16 at 18:58

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    $\begingroup$ The equation essentially is giving you everything expect a direct answer. $\sigma_1$ and $\sigma_2$ are the standard deviations for the equations. If you reduce the noise, equivalently if you lower the standard deviation what happens to $dx_t$. Propagate that effect into the equation for $dy_t$. $\endgroup$ – Zach466920 Dec 22 '16 at 6:49
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Start from the second equation, by direct integrating, we have $$ X_t=X_0-\sigma_2\int_0^t Y_sdW_s^2 $$ Plug into the first equation, we have $$ dY_t=-(X_0-\sigma_2\int_0^t Y_sdW_s^2)Y_tdt+\sigma_1dW_t^1\\ Y_t-Y_0=-X_0\int_0^t Y_sds-\sigma_2\int_0^t\int_0^sY_\tau dW_\tau^2Y_sds+\sigma_1 W_t^1\cdots(*) $$ Assuming $E(Y_t)=m_t$ which is a function of time. Take expectation on both sides of $(*)$, we have $$ m_t+X_0\int_0^t m_s ds=Y_0 $$ Here I am using zero expection property of Ito's integral. This is an integral equation and it is not hard to find the solution is $$ m_t=Y_0\exp(-X_0 t) $$ Variance can be found by squaring $(*)$ and follow the similar procedure to find $E(Y_t^2)$.

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