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According to my book, for a random sample $(X_1, \ldots, X_n)$ from a continuous distribution with p.d.f. $f(x)$ and c.d.f. $F(x)$, the p.d.f. of the maximum of the sample is $g(z)=nf(z)[F(z)]^{n-1}$, where $z=\mathrm{max}(x_1, \ldots,x_n)$. The book gives the following question:

the random variable $X$ has p.d.f. $f(x)=12x^2 (1-x), 0≤x≤1$.

I'm assuming that I put this into the $g$ function, such that $$ g(z)=n \{12z^2 (1-z)\} \left[\int_{-\infty}^z 12t^2 (1-t)dt\right]^{n-1} . $$

However, I have no clue how to carry on from there to find the probability that the largest maximum is 1/2.

Mathmathmath
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    As this is a self-study question, could you please add the [`self-study`](http://stats.stackexchange.com/tags/self-study/info) tag? Thank you. – COOLSerdash May 17 '14 at 16:49
  • Question: Have you actually got the function $g$? For that, you'll need the CDF of $X$, which is $F(x)=\int_{0}^{x} 12x^{2}(1-x)\,\mathrm{dx} = 4x^3-3x^4$. With that, you get $g$ which is the PDF of the distribution of the maximum. Then, you have to integrate $g$ to get the CDF of the distribution of the maximum. Also, have a look at @Macro's answer [here](http://stats.stackexchange.com/a/32353/21054). – COOLSerdash May 17 '14 at 17:04
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    What do you mean by "find the probability that the largest maximum is 1/2"? – QuantIbex May 17 '14 at 17:35
  • Added the missing part. – Mathmathmath May 17 '14 at 19:19

1 Answers1

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I'd rewrite: $g_Z(z)=n[F_X(z)]^{n-1}f_X(z)$ and try a short recap. This happens because: $$\begin{align} F_Z(z) &= P(\max\{X_i\}<z)=P(X_1<z,\dots,X_n<z) \\ &\overset{ind}{=}P(X_1<z)\cdots P(X_n<z)\overset{i.d.}{=}[P(X<z)]^n=[F_X(z)]^n\end{align}$$ Then: $$g_Z(z)=\frac{\text{d}[F_X(z)]^n}{\text{d}z}=n[F_X(z)]^{n-1}\frac{\text{d}F_X(z)}{\text{d}z}=n[F_X(z)]^{n-1}f_X(z),\quad 0\le z\le 1$$ Now:

  • $F_X(z)=\int_0^z 12t^2(1-t)\text{d}t$;
  • $F_Z(z)=[F_X(z)]^n$;
  • probability that the largest maximum is 1/2: $P(Z\le1/2)=F_Z(1/2)$.
Sergio
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  • The "probability that the largest maximum is 1/2" is not $P(Z \leq 1/2)$. – QuantIbex May 17 '14 at 17:37
  • Sorry, if you ask what does "the probability that the largest maximum is 1/2" mean, how can you say that it is not "the probability that the maximum is not larger than 1/2"? – Sergio May 17 '14 at 18:32
  • I followed this through myself and reached that answer; I also had P(Z≤1/2)=G(1/2) where G(Z) is (4z^3 - 3z^4)^n. If I'm not mistaken my G(Z) is FZ(z)? @QuantIbex, I'm pretty sure that "the probability that the largest maximum is 1/2" is P(Z≤1/2) since it means that the probability that the maximum is 1/2 or less. – Mathmathmath May 17 '14 at 19:10
  • Yes, my $F_Z(z)$ is your $G(z)$. I apologize for my notation change. – Sergio May 17 '14 at 19:15
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    The wording "largest maximum is 1/2" suggests equality, not inequality as in your expression. Why not improve the last line of this nice answer and write "probability that the maximum is not larger than 1/2" which corresponds to the mathematical expression that follows (and probably also to probability seeked by the OP)? – QuantIbex May 17 '14 at 19:35
  • I thank you but... may I diagree? "Probability that the maximum is not larger than 1/2" is the only possibile interpretation of "probability that the largest maximum is 1/2" because a) $\max(z)=1/2\Leftrightarrow z\le 1/2$, b) if we follow the equality suggestion we get $P(Z=1/2)=0$. However I'm going to follow your kind advice. – Sergio May 17 '14 at 19:45
  • The edited version of the last bullet point in your answer adds confusion, and might even be wrong. What is $z$ in $\max(z) = 1/2 \Leftrightarrow z \leq 1/2$? A scalar, a random variable? How do you prove that $ z \leq 1/2 \Rightarrow \max(z) = 1/2$? – QuantIbex May 17 '14 at 23:21
  • Well. I'll restore my previous version and I'll edit it again if a) someone else has trouble with it, b) you show that "the probability that the largest maximum is 1/2" is not $P(Z\le 1/2)$ but something else (what?) – Sergio May 17 '14 at 23:37