Uniform joint PDF's Let $X$ and $Y$ be two random variables. Their joint PDF is uniform in the region $0$ to $1$ (inclusive). Let $Z$ be a random variable defined as $Z = \min\{X,Y\}$. Determine $f_Z (z)$, $f_{Z\mid X}(z\mid x)$, $E[Z],$ and $E[X\mid Z=z] $
I'm currently working on $f_Z(z)$. I have $P(Z \leq z) = P(X \leq z)P(y \leq\ z)$
First question, is this even correct? I've been trying to figure out how to define the $\min(X,Y)$ requirement, and this is what I have seen repeated a few times. And if it is correct... how do I evaluate it? I know it's an integral, but what am I integrating from? I could use some conceptual help on understanding what is being asked of me.

The minimum of $X$ and $Y$ is less than or equal to $z$ if at least one of the two random variables $X,Y$ is less than or equal to $z$, not only if both are less than or equal to $z$.  Therefore $\Pr(Z\le z)$ is not the same as $\Pr(X\le z\text{ and }Y\le z)$.
greater than $z$ precisely if both of $X,Y$ are greater than $z$.  Therefore $\Pr(Z>z) = \Pr(X>z\text{ and }Y>z)$.  If $X,Y$ are independent, that is the same as $\Pr(X>z)\Pr(Y>z)$.  You didn't say anything about the joint distribution of $X$ and $Y$.  Often people neglect that, when what they should say is that they're independent. Later addendum to this paragraph: I see now that by "the region $0$ to $1$" you apparently meant the unit square.end of later addendum
Once you've found $\Pr(Z>z)$, you can deduce that $\Pr(Z\le z)=1-\Pr(Z>z)$.
