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Problem of the Week #47 - February 18th, 2013

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Chris L T521

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Jan 26, 2012
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Thanks to those who participated in last week's POTW!! Here's this week's problem (going with another probability question)!

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Problem: Let $X_1,X_2,\ldots,X_n$ be independent random variables, each having a uniform distribution over $(0,1)$. Let $M=\max(X_1,X_2,\ldots,X_n)$. Show that the distribution function of $M$, $F_M(\cdot)$, is given by \[F_M(x)=x^n,\qquad 0\leq x\leq 1.\]
What is the probability density function of $M$?

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

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Jan 26, 2012
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No one answered this week's question. Here's my solution:

Observe that $\mathbb{P}(X_i\leq x)=\displaystyle\int_0^x 1dx=x$. So

$\hspace{.5in}\begin{aligned}F_M(x)&=\mathbb{P}(M \leq x)\\ &=\mathbb{P}\left[\bigcap\limits_{i=1}^n(X_i\leq x)\right]\\ &= \prod\limits_{i=1}^n\mathbb{P}(X_i\leq x)\,\,\text{ since $X_i$'s are independent}\\ &= \mathbb{P}(X_i\leq x)^n\\ &= x^n\end{aligned}$

The density function is the derivative of $F_M(x)$ which is $F_M^{\prime}(x)=f_M(x)=nx^{n-1}$.
 
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