Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $4y^3$.

share|improve this question
    
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

1 Answer 1

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 $W$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.