What are the CDFs of $Z_1$ and $Z_2$ with $Z_1=max(X,Y)$ and $Z_2= min(X,Y)$?

Question

Let $X$ and $Y$ be i.i.d. random variables with common CDF(F) and PDF(f) and let $Z_1=\max(X,Y)$ and $Z_2= \min(X,Y)$, then what are the CDFs of $Z_1$ and $Z_2$?

I think both should be $F^2(z)$, but my friend disagreed with me in the CDF of $Z_2$. Is the CDF of Z a kind of joint distribution of X and Y? I cannot see clearly the relations between them.

Answer 1

Proceed by definition.

$Z_1$ is the maximum of $X$ and $Y$. Therefore, if it is smaller than some quantity, then both $X$ and $Y$ are smaller than that quantity. More rigorously, $$ \forall x \quad Z_1 \leq x \iff X \leq x \textrm{ and } Y \leq x $$ Hence, the probabilities of the LHS and RHS occuring are the same. But the LHS is true with probability $F_{Z_1}(x)$, and the RHS is true with probability $F_X(x)F_Y(x) = F^2(x)$. Hence, it follows that: $$ \bbox[yellow, 5px , border:2px solid red] {F_{Z_1}(x) = F^2(x)} $$

By differentiating the CDF(if possible), one obtains the PDF of $Z_1$.

Now, note that an inverse relationship occurs with $Z_2$: $$ \forall x \quad Z_1 \geq x \iff X \geq X \textrm{ and } Y \geq x $$

Now, the probability of the left hand side occuring is $1 - F_{Z_2}(x)$, and the probability of the right hand side occuring is $(1 - F(x))^2$.

Hence, it follows that: $$ \bbox[yellow,5px,border:2px solid red] {F_{Z_2}(x) = 1 - (1 - F(x))^2} $$

From the above, the PDF of $F_{Z_2}$ should again be calculable by differentiation.

Answer 2

It seems unlikely that $Z_1$ and $Z_2$ would have the same CDF, so I think your friend is right.

The CDF of $Z_1$ isn't a joint distribution of $X$ and $Y$ but since $Z_1$ is a function of $X$ and $Y,$ the distribution of $Z_1$ can be computed from the distribution of $X$ and $Y.$ Same goes for $Z_2.$

The trick is to write $$ P(Z_1\le z) = P(\max(X,Y) \le z) = P(X\le z,Y\le z) = F_{X,Y}(z,z)$$ where $F_{X,Y}(x,y) = P(X\le x,Y\le y)$ is the joint CDF for $X$ and $Y.$ Since in this case $X$ and $Y$ are independent and identically distributed with CDF $F$, we have $$P(X\le x, Y\le y) = P(X\le x)P(Y\le y) = F(x)F(y),$$ so $P(Z_1\le z) = F^2(z).$

For $Z_2$ the calculation is a bit different. To apply the parallel argument, it helps to instead start with $>$ rather than $\le,$ and we have $$ P(Z_2> z) = P(\min(X,Y) > z) = P(X > z,Y>z) = P(X> z)P(Y> z) = (1-F(z))^2.$$ so $$ P(Z_2\le z) = 1-P(Z_2> z) = 1-(1-F(z))^2. $$