Let $X$ and $Y$ be independent random variables, both of exponential distibution, such that $EX=\frac{1}{2},EY=1$. Calculate the density of conditional distribution $f_{X+Y|X}(z|x)$. In fact I am interested in $s=3$.
My idea: I know that I should apply the formula $f_{Z|X}(z,x)=\frac{f_{Z,X}(z,x)}{f_X(x)}$. However I do not know how to obtain $f_{Z,X}(z,x)$. Could you please help?
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Let $Z = X+Y \mid X=x$. The complementary cumulative distribution function of $Z$ is easy to work out: $$ S_Z(z) = \Pr(Z > z) = \Pr\left(X+Y > z \mid X=x\right) = \Pr\left(Y > z-x \mid X=x\right) = \Pr\left(Y > z-x \right) $$ Hence $$ S_Z(z) = \begin{cases} \mathrm{e}^{-(z-x)} & z \geqslant x \cr 0 & z <x \end{cases} $$ The probability density function then is obtained by differentiation: $$ f_{X+Y\mid X}\left(z\mid x\right) = -S_Z^\prime(z) = \begin{cases} \mathrm{e}^{-(z-x)} & z \geqslant x \cr 0 & z < x \end{cases} $$
Hint: Given that $X = x$, the conditional distribution of $Y$ is the same as the unconditional distribution of $Y$ because $X$ and $Y$ are given to be independent random variables, and so knowing the value of $X$ tells us nothing that we did not already know about $Y$. Thus, the conditional distribution of $X+Y$, given that $X$ has taken on the value $x$ is the same as the unconditional distribution of $x+Y$. Can you work this out? Remember that $x$ is just a constant. Replace $x$ by $3$ if you need to.