Change of variable to calculate expected value

$$X$$ 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 $$\int\int wp(x,z)dxdz$$ ?

• 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 $$E[Z] = E[X]+E[W] = \overline{x} + 0\ .$$ – lonza leggiera Apr 20 at 14:08
• 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 $E_{p(z,x)}[W]=E_{p(w)}[W]$ ? – Valjean Apr 20 at 19:07