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This answered question shows how to solve the problem but I still do not understand how to get the conditional density function, i.e.

${"}$Let $Z=X+Y$, then the density $f_{X,Z}$ of $(X,Z)$ is defined by $f_{X,Z}(x,z)=f_X(x)f_Y(z-x)$ because $X$ and $Y$ are independent hence the conditional distribution of $X$ conditionally on $Z=z$ is proportional to $f_X(x)f_Y(z-x)$, that is, $$ f_{X\mid Z}(x\mid z)=\frac1{c(z)}f_X(x)f_Y(z-x),\qquad c(z)=\displaystyle\int f_X(t)f_Y(z-t)\mathrm dt." $$ I keep getting that $$ f_{X\mid Z}(x\mid z)=\frac1{c(z)}f_X(x)f_Z(z)=\frac1{c(z)}f_X(x)\int f_X(x)f_Y(z-x),\qquad c(z)=\displaystyle\int f_X(x)f_Y(z-x)\mathrm dx. $$ which just ends up being $$ f_{X\mid Z}(x\mid z)=f_X(x). $$ I know that I'm making a fundamental error, could someone please explain in more detail what it is?

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The general formula for conditional density is $$ f_{X\mid Z}(x\mid z) = {f_{X,Z}(x,z)\over f_Z(x)}.\tag1 $$ Your error is in replacing the numerator in (1) with $$f_{X,Z}(x,z)=f_X(x)f_Z(z), $$ which is true only if $X$ and $Z$ are independent. What you should do is replace the numerator in (1) with $$ f_{X,Z}(x,z)=f_X(x)f_Y(z-x) $$ and the denominator in (1) with $$ f_Z(z)=\int f_{X,Z}(x,z)\,\mathrm dx = \int f_X(x)f_Y(z-x)\,\mathrm dx=:c(z), $$ and you'll get the desired conditional density function.

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  • $\begingroup$ Thank you. I'm guessing that the error is that Z and X are not indpt. since Z=X+Y, and X is not indpt. of X. $\endgroup$
    – litmus
    Dec 10, 2015 at 11:42

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