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I need to find the autocovariance $C_{YY}(t,s)$ of the stochastic process $Y(t) = t^2 X(t) -2X'(t)$ where $C_{XX}(t,s) = e^{-t^2 -s^2}$ is given.

Using known properties I can calculate the autocovariance of both $Y_1(t)=t^2 X(t)$ and $Y_2(t)=-2 X'(t)$, but how about the sum of these?

To restate the problem: it is noticeable (by definition of autocovariance) that $C_{YY}(t,s) \neq C_{Y_1 Y_1}(t,s) + C_{Y_2 Y_2}(t,s)$, but what is missing?

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  • $\begingroup$ Obviously one needs the covariance of $X(t)$ and $X'(s)$ (but you might be interested to know that the autocovariance of $X$ you indicate implies that $X(t)=\mathrm e^{-t^2/2}X(0)$ for every $t$). $\endgroup$ – Did Sep 8 '13 at 10:45
  • $\begingroup$ I am sorry, my handwritten notes are very poor and it has been a while. I am not sure how you came up with $X(t)$. Furthermore I doubt I can proceed in a straightforward way without knowing $\operatorname{E}[X(0)]$. $\endgroup$ – Pranasas Sep 8 '13 at 11:52
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By assuming regularity conditions that allow interchange between limit and expectation you can calculate the crosscovariance of X and its derivative: $C_{X,X'}(t,s)=E[(X(t)-E[X(t)])(\lim_{h\rightarrow0}{X(s+h)-E[X(s+h)] -(X(s)-E[X(s)])})=\lim_{h\rightarrow0}\frac{C_{X,X'}(t,s+h)-C_{X,X'}(t,s)}{h}=\frac{\partial C_{X,X}(t,s)}{\partial s}$ The resulting autocovariance is: $C_{Y,Y}(t,s)=C_{Y_1,Y_1}(t,s)+C_{Y_2,Y_2}(t,s)+C_{Y_1,Y_2}(t,s)+C_{Y_1,Y_2}(s,t)$ where, $C_{Y_1,Y_1}(t,s)=(st)^2C_{X,X}(t,s) \\ C_{Y_2,Y_2}(t,s)=4C_{X,X}\frac{\partial^2 C_{X,X}(t,s)}{\partial t\partial s} \\ C_{Y_1,Y_2}(t,s)=-2t^2\frac{\partial C_{X,X}(t,s)}{\partial s}$

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  • $\begingroup$ This appears to be a part of the autocovariance in question. Hopefully you could express it in terms of already mentioned characteristics (including this crosscovariance). $\endgroup$ – Pranasas Sep 8 '13 at 13:26
  • $\begingroup$ See the edited answer. $\endgroup$ – SBM Sep 8 '13 at 13:53

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