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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?

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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
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2 Answers

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 $$P(W \le w)=1-(1-w^2)^3.$$

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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.

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