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A random variable $Z$ is the sum of two independent random variables $X$ and $Y$, with known probability densities $f_X$ and $f_Y$, respectively.

Now suppose you sample $Z_1=X_1+Y_1$ but you don't know the values of $X_1$ and $Y_1$. How can we determine the probabilities for the values that $X_1$ may have taken, knowing that $Z=Z_1$ ?

My intuition tells me that the "probability density" for $X_1$ is $f_X(x) f_Y(y-Z_1)$ but i can't seem to prove it.

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  • $\begingroup$ Your intuition is leading you astray and your alleged formula for the probability density of $X_1$ has three arguments $x$, $y$, and $Z_1$ instead of the usual single argument. It is fortunate that you have not found a proof of what your intuition is telling you. $\endgroup$ – Dilip Sarwate Oct 31 '14 at 13:29
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A similar computation process can be referred in statistics textbooks.

Set u = x and z = x + y,
=> $x = u, y = z - u $
=> $f(u, z) = f_{x,y}(u, z-u)*|J| = f_{x,y}(u, z-u) = f_{x}(u)*f_{y}(z-u) $
=> $f(z) = ∫ f_{x}(u)*f_{y}(z-u)du $

The probability density of $X$1 given $Z=Z$1 equals to
$f(X=X_{1}|Z = Z_{1}) = \frac{f(x, z)} {f(z)} = \frac {f_{x}(x)*f_{y}(z-x)} { ∫ f_{x}(x)*f_{y}(z-x)dx} $ for $X, Z∈ τ$

And if there are errors welcome commenters to point them out.

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  • $\begingroup$ The question indicates the random variables are continuous. So your expression is really for a density function, not probabilities. $\endgroup$ – soakley Nov 1 '14 at 23:22

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