# Density function $Y= \max(X_1, X_2)$

If $X_1$ and $X_2$ are independent random variables each of which has density function of the form:

$$f(x)= \Bigg\{ \begin{array}{cc} 2x;&0<x<1\\ 0; & \text{otherwise} \end{array}$$

Let $Y = \max\{X_1, X_2\}$; show the density function of $Y$ is $4y^3$.

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what is $y_3$?${}{}$ – Nana Jun 11 '12 at 17:02
@Chas: Please check if I've changed your question unintentionally. – Gigili Jun 11 '12 at 17:10

We show how to find the density function of the random variable $Y$. First we find the cumulative distribution function $F_Y(y)$ of $Y$.

We have $Y\le y$ iff $X_1\le y$ and $X_2 \le y$. For $0\le y\le 1$, $$P(X_1 \le y)=\int_0^y 2x\,dx=y^2.$$ In the same way, we can see that $P(X_2\le y)=y^2$. Thus by independence $P(Y \le y)=y^4$.

We conclude that $F_Y(y)=y^4$ for $0\le y\le 1$. For completeness, note that $F_Y(y)=0$ if $y \lt 0$, and $F_Y(y)=1$ for $y \gt 1$.

Finally, differentiate $F_Y(y)$ to find the density function of $Y$.

Remark: You may run into a similar problem with $\min$ instead of $\max$. Let $W=\min(X_1,X_2)$. Then $P(W\le w)=1-P(X_1 \gt w)\cdot P(X_2\gt w)$. In our case, if $0\le w\le 1$, we get $$P(W\le w)=1-(1-w^2)(1-w^2)=2w^2-w^4,$$ and by differentiating we find the density function of $W$.

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