Given that $X$ and $Y$ are independent exponential random variables with parameters $\lambda_1$ and $\lambda_2$. If $Z=min(X,Y)$, find the conditional distribution of $Z$ given $Z=X$
My try:
I found the pdf of the random variable $Z$ using the fact that the CDF of $Z$ is: $$F_{z}(z)=P(Z \leq z)=P(\min (X, Y) \leq z)=1-P(\min (X, Y)>z)$$ So we have: $$F_Z(z)=1-P(X>z)P(Y>z)=1-\left(1-F_{X}(z)\right)\left(1-F_{Y}(z)\right)$$
using the CDF's of $X$ and $Y$ we get: $$F_{z}(z)=\left\{\begin{array}{cc} 0 & z \leq 0 \\ 1-e^{-\left(\lambda_{1}+\lambda_{2}\right) z} & z>0 \end{array}\right\}$$
So the PDF of $Z$ is: $$f_{Z}(z)=\left\{\begin{array}{cc} \left(\lambda_{1}+\lambda_{2}\right) e^{-\left(\lambda_{1}+\lambda_{2}\right) z} & z>0 \\ 0 & \text { else } \end{array}\right\}$$
Now can i get any hint to find conditional distribution of $Z$ given $Z=X$?
Just use Bayes' Rule. Note: that $\{Z=z, X=Z\}=\{X=z, X<Y\}$
$$f_Z(z\mid Z=X)~{=f_{X}(z\mid X<Y)\\~\vdots}$$