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Let $X_1, X_2, X_3$ be three independent, identically distributed random variables each with density function $$f(x)= \begin{cases} 3x^2 & 0 \le x \le 1 \\ 0 & \text{otherwise} \end{cases} $$ Let $Y=\max\{X_1,X_2,X_3\}$. Find $P(Y\gt \frac{1}{2}).$ Answer key $\frac{511}{512}$

Now the author has not defined what identically distributed random variables are so I looked it up and found:

  • Random variables $X$ and $Y$ are identically distributed if for every set $A\in S$ $P(X\in A)=P(Y \in A)$.

Using this I used the following reasoning

  • WLOG let $Y=\max\{X_1,X_2,X_3\}=X_1$

  • because one of the $X_i$ for $i=1,2,3$ has to be the max

  • and since $P\left(X_1>\frac{1}{2}\right)=P\left(X_2>\frac{1}{2}\right)=P\left(X_3 > \frac{1}{2}\right)$

then $$P(Y >\frac{1}{2})=P(X_1>\frac{1}{2})=\int_{\frac{1}{2}}^1 3x^2 \, dx= \left.x^3 \vphantom{\frac 1 1} \right|_{\frac{1}{2}}^1=1-\frac{1}{8}=\frac{7}{8}.$$ I kind of had a suspicion of the answer being wrong since I did not use independence at all but I guess I do not understand what the question is actually trying to get me to figure out. Now after seeing the solution I have noticed that somehow they must have got $$P(Y>0.5)=\left.x^9\vphantom{\frac 1 1} \right|_{0.5}^1 = \frac{511}{512}$$ but why I do not understand.

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Think of a dice and $X_1$ (or $X_2,X_3$ it doesn't matter since they are iid) as the result of one toss. Then, it is clear that the result of $X_1$ is not the same as doing $X_1$ three times and taking the maximum result !

Now, mathematicaly, you need to calculate $P(Y>1/2)=P(\max(X_i)>1/2)$. But the saying that the max of three numbers is greather than a $1/2$ (or any quantity) doesn't give you much information, only that there is at least one of the $X_i$ greater than $1/2$, whereas the opposite event gives : $1-P(Y>1/2) = P(Y<1/2) = P(X_1 < 1/2, X_2 < 1/2, X_3 < 1/2)$. And then you can use independence to calculate your probability.

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  • $\begingroup$ thanks it makes sense now. $\endgroup$ – Andrew Jul 9 '16 at 1:59
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The probability the maximum is $\gt 1/2$ is $1$ minus the probability the maximum is $\le 1/2$.

And the maximum is $\le 1/2$ if and only if $X_1$, $X_2$, and $X_3$ are all $\le 1/2$.

By independence the probability they are all $\le 1/2$ is $(1/2^3)^3$. This is $\frac{1}{512}$.

Remark: The fact that the $X_i$ all have the same distribution does not play a crucial role. If they had different distributions, we could still find the probability that $Y\le 1/2$ by computing $\Pr(X_1\le 1/2)\Pr(X_2\le 1/2)\Pr(X_3\le 1/2)$. However, the independence is crucial, for that is what allows us to multiply.

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  • $\begingroup$ @user101388: You are welcome. $\endgroup$ – André Nicolas Jul 9 '16 at 2:01

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