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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.

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Your way of using $\TeX$ is.....um.... quaint, but here's a more substantial issue that's about actually understanding the math. There are reasons why in expressions like $f_{Z\mid X}(z\mid x)$, the letters in the subscript are capital and the arguments are in lower case. For example $f_{Z\mid X}(3\mid 5)$ would be the value of the conditional density evaluated at $3$, given that $X=5$. The arguments are particular numbers; the subscripts identify random variables, not particular numbers. –  Michael Hardy Oct 23 '12 at 20:30
    
Given that the joint PDF is $f_{X,Y}(x,y)=1$, can you determine $f_X$ and $f_Y$? From there it should be easier to find the desired $f_Z$ (perhaps by finding the cumulative $F_Z(z) = \text{P}(Z\leq z)$ and differentiating, although there are probably better ways). –  Eric Stucky Oct 23 '12 at 20:39
    
@EricStucky : Your conclusion about the joint PDF relies on an assumption about independence that was not stated. (And, as noted in my answer, finding $\Pr(Z\le z)$ is most easily done in this case by finding $\Pr(Z>z)$.) –  Michael Hardy Oct 23 '12 at 21:04
    
@MichaelHardy: Really? I thought that $\int_{-\infty}^\infty f_{X,Y}(x,y)dy = f_X(x)$, regardless of dependence? Is there a better way to find Pr($Z>z$) for dependent $X$ and $Y$? –  Eric Stucky Oct 24 '12 at 15:23
    
It is true that $\int_{-\infty}^\infty f_{X,Y}(x,y)\,dy = f_X(x)$ regardless of dependence. BUT you asserted that the joint pdf is equal to $1$ at every point in the unit square. That's true ONLY if $X,Y$ are independent. At any rate, you have not proposed any particular way to find $F_Z$, so it makes no sense to speak of whether my way is "better". (I posted it as an answer.) –  Michael Hardy Oct 24 '12 at 16:42
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1 Answer

up vote 2 down vote accepted

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 $\Pr(Z\le z)=1-\Pr(Z>z)$.

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