Given that $X $~$ $exp$(\lambda_1)$, $Y $~$ $exp$(\lambda_2)$, and $X$,$Y$ are independt.
$Z$=$max\{X,Y\}$
Find the conditional density function of $Z$ given $Z$=$X$.
I get pdf of $Z$ is $\lambda_1 e^{-\lambda_1z} (1 - e^{-\lambda_2z})+\lambda_2 e^{-\lambda_2z} (1 - e^{-\lambda_1z})$
$F_Z(z)=Pr(Z\le z)=Pr(X \le z) \cap Pr(Y \le z) = Pr(X \le z)Pr(Y \le z) = (1 - e^{-\lambda_1z})(1 - e^{-\lambda_2z}), \forall z \ge 0$
Density function of Z is $f_Z(z) = \frac{\partial F_Z(z)}{\partial z} = \lambda_1 e^{-\lambda_1z} (1 - e^{-\lambda_2z})+\lambda_2 e^{-\lambda_2z} (1 - e^{-\lambda_1z})$
But how to write the conditional density function of $Z|Z=X$?
From the definition of Conditional continuous distributions,
$f_Z(z|X=x) = \frac{f_{Z,X}(z,x)}{f_X(x)}$
To fit into this problem, I rewrite the equation above as $f_Z(z|X=z) = \frac{f_{Z,X}(z,z)}{f_X(z)}$. But not sure how to process the next steps.