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Let $A$ and $B$ be a random variables with continues PDF $f_A$ e $f_B$. Let $Y=A+B$ and let $\hat{A}(Y)=Y$ be an estimation of $A$. I have to evaluate expectation of the square of the estimation errorr $\tilde{A}(A,Y)=A-\hat{A}(Y)$, i.e. the MSE $$\mathbb{E}_{A,Y}[\tilde{A}^2]=\iint_{\mathbb{R}^2} (a-y)^2 f_{A,Y} (a,y) \text{ d}a\text{d}y$$ where $f_{A,Y}$ is the joint PDF of $A$ and $Y$.

Since $\tilde{A}(A,Y)=A-\hat{A}(Y)=-B$, I think $$\mathbb{E}_{A,Y}[\tilde{A}^2]=\mathbb{E}_B[B^2]$$ But is it correct my intuition? I'm not sure because in explicit terms $$\iint_{\mathbb{R}^2} (a-y)^2 f_{A,Y} (a,y) \text{ d}a\text{d}y=\int_\mathbb{R} b^2 f_B (b) \text{ d}b$$ is not clear if these two integrals gets the same value.

Gost91
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  • Express the left hand side in terms of $(a,b)$ and integrate out $a:$ done. – whuber Jul 04 '18 at 16:44
  • How can I do that? $$\iint_{\mathbb{R}^2} (a-y)^2 f_{A,Y} (a,y) \text{ d}a\text{d}y=\iint_{\mathbb{R}^2} b^2 f_{A,A+B} (a,a+b) \text{ d}a\text{d}(a+b)$$, but now how can I go on? – Gost91 Jul 04 '18 at 19:56
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    See any thread that discusses *Jacobians.* An alternative approach, [using forms and wedge products,](https://stats.stackexchange.com/a/154298/919) makes this very easy--it goes from the right hand side of your comment to the right hand side in your question in a single step--but is not well known among statisticians. – whuber Jul 04 '18 at 21:16

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