# Cumulative density function of a maximum and minimum

It is given that both $Y$ and $X$ are $U(0,a)$, and that $Z_+ = \max(X,Y)$ and $Z_- = \min(X,Y)$. I need to find the CDF for both zetas.

I know that both $X$ and $Y$ have to be smaller or equal to $Z_+$ and bigger or equal to $Z_-$.

For $Z_+$:

$$F_{Z_+}(z) = P(Z \leq z)$$

And for $Z_-$:

$$F_{Z_-}(z) = P(Z \leq z)$$

This were I'm stuck, what is the next thing to do?

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Hint: $$P(Z_-<z)=P(X<z)\cdot P(Y<z)=\frac za\cdot\frac za$$ if $0\le z\le a$. And $$P(Z_+<z)=P(Z_->a-z).$$

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You are looking for the cumulative distribution function. We need to assume that $X$ and $Y$ are independent.

For the maximum, note that the maximum is $\le z$ if and only if $X$ and $Y$ are both $\le z$. If $0\lt z\lt a$, the probability that $X\le z$ is $\dfrac{z}{a}$. The probability that $Y\le z$ is the same. So the probability they are both $\le z$ is $\left(\dfrac{z}{a}\right)^2$.

We need to add that the cdf is $0$ for $z\le 0$, and $1$ if $z\ge 1$.

The same idea extends to the maximum of more than two uniforms on $(0,a)$.

For the minimum, note that the minimum is $\ge z$ if and only if both $X$ and $Y$ are $\ge z$.

If $z$ is in the interval $(0,a)$, the probability that $X\ge z$ is $\dfrac{a-z}{a}=1-\dfrac{z}{a}$.

It follows that the minimum is $\ge z$ with probability $\left(1-\dfrac{z}{a}\right)^2$. Thus the probability that the minimum is $\le z$ is equal to $$1-\left(1-\dfrac{z}{a}\right)^2.$$ For $z$ beween $0$ and $a$, this gives the cdf of the minimum. We need to add that the cdf is $0$ for $z\le 0$, and $1$ for $z\ge a$.

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