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$X$ and $W$ are independent random variables. $$ Z=X+W $$ $$ W \sim \mathcal{N}(0,\sigma) $$ $$ E[X]=\bar{x} $$ I want to calculate $E[Z]$ with respect to the joint pdf $p(z,x)$ $$ E[Z]=\int\int (x+w)p(x,z)dxdz = \bar{x} + \int\int wp(x,z)dxdz $$ How to calculate $$ \int\int wp(x,z)dxdz $$ ?

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  • $\begingroup$ It's not clear what you mean by "$E[Z]$ with respect to the joint pdf $p(z,x)$" as distinct from just plain old $\ E[Z]\ $ by itself, unqualified by any further restriction. *By definition* $$ E[Z] = \int\int z p(z,x)dxdz = \int z p(z) dz\ .$$ Once you have the joint pdf $\ p(z, x)\ $, you can calculate $\ E[Z]\ $ from the above expression without any reference to $\ W\ $ at all. Also, without even knowing $\ p(z,x)\ $, you still know that $$ E[Z] = E[X]+E[W] = \overline{x} + 0\ . $$ $\endgroup$ – lonza leggiera Apr 20 at 14:08
  • $\begingroup$ My question is related to proving that maximum a posterior estimation is unbiased. By definition from the book: If $x$ is a random variable with prior $p(x)$, then the unbiasedness property is written as $$E[\hat{x}(Z)]=E[X] $$ where the expectation on the left-hand is with respect to the joint pdf $p(z,x)$ and the one on the right hand side is with respect to $p(x)$. Now the problem is as I stated before but with additional assumption that $p(x)$ is Gaussian. $$\hat{x}(Z) = \alpha + \beta Z$$ $$E[\alpha + \beta Z] = \alpha + \beta [\bar{x} -E[W]]$$ Why $E_{p(z,x)}[W]=E_{p(w)}[W]$ ? $\endgroup$ – Valjean Apr 20 at 19:07

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