Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X_1,X_2,X_3$ be mutually stochastically independent random variables and let each of them have the following density function:

$f(x)= 2x$ when $0\leq x\leq 1$ and $f(x)=0$ elsewhere.

Let Y be the random variable defined as, $Y=\max(X_1, X_2, X_3)$, find (a) the distribution function and (b) the probability function of the random variable $Y$.

I saw somewhere that the distribution function of such variable to be given by $F_Y(y)=P(Y\leq y)=P(\max (X_1,X_2,X_3)\leq y)$.

If this is true, how may I apply it in this question to find (a) and (b) above?

share|cite|improve this question
You are on the right track. $\mathbb{P}(Y \leq y) = \mathbb{P}(\max(X_1,X_2,X_3) \leq y)$. Note that $\max(X_1,X_2,X_3) \leq y$ is equivalent to $X_1 \leq y$ and $X_2 \leq y$ and $X_3 \leq y$ and make use of the fact that $X_1,X_2,X_3$ are independent. – user17762 Dec 22 '11 at 7:09

You were almost finished. The cumulative distribution function $F(y)$ of $Y$ is then the probability that all the $X_i$ are $\le y$. For any $i$, and $0\le y\le 1$, $$P(X_i \le y)=\int_0^y 2x\,dx=y^2.$$

So, for $0\le y \le 1$, the probability that all the $X_i$ are $\le y$ is, by independence, $(y^2)^3$.

We conclude that $F(y)=0$ if $y<0$, $F(y)=y^6$ if $0\le y\le 1$, and $F(y)=1$ if $y>1$. For the density function of $Y$, differentiate.

Comment: Let $W$ be the minimum of the $X_i$. Then $P(W \le w)$ is $1$ minus the probability that all the $X_i$ are $\ge w$. For any $i$, and $0 \le w\le 1$, $P(X_i>w)=1-w^2$. It follows that $$P(W \le w)=1-(1-w^2)^3.$$

share|cite|improve this answer

The answer follows just from what you have told us:

Let $F_Y$ be the distribution function of $Y$ and let $F$ be the distribution function of $X_1$ ( and hence that of $X_2, X_3$ as these are identically distributed )

$$F_Y(y) = P(Y \leq y) = P(\max(X_1, X_2, X_3) \leq y)$$ Since the maximum of three numbers is less than $y$, each of them must also be less than $y$. And, since the $X_1, X_2, X_3$ are independent random variables, $$F_Y(y) = P(X_1 \leq y, X_2 \leq y, X_3 \leq y)$$ $$F_Y(y) = P(X_1 \leq y)\cdot P(X_2 \leq y)\cdot P(X_2 \leq y)$$ $$F_Y(y)= (F(y))^3$$

I am sure you'll figure out what $F$ is, for yourself.

share|cite|improve this answer

Your Answer


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.