I will first enuntiate the question and then explain what I'm not understanding.

Suppose $ X_1, X_2,\ldots, X_n $ iid with common distribution $ U(0,\theta)$. Define $M$ as follows:

$ M : =\max\{ X_1, X_2,\ldots, X_n \} $

Evaluate : $\Pr\left( M\leq \frac{\theta }{2} \right ) $

So, i have $ n $ random variables that are independent and identically distributed; I have a another random variable that will yield the maximum value of the mentioned sequence of random variables and then i have to evaluate the probability that this random variable is 'bounded' by $\theta$ divided by two. I understand all this informations separately, but i don't know how to connect them. How in the world, the fact that I know that they are iid and have a uniform distribution with a parameter $ a = 0 $ and $ b = \theta $ will help me finding out the probability of $M$? How even would I know what's the maximum value of the sequence? Thanks!


Hint: The probability that $M\leq \theta/2$ is the same as the probability that all $X_i$s are less than $\theta/2$.

  • $\begingroup$ Also note that if you normalize this to $\theta=1$, it does not affect your answer. $\endgroup$
    – parsiad
    Oct 11 '14 at 20:09
  • $\begingroup$ Sorry but the maximum does not do what you say. $\endgroup$
    – Did
    Oct 11 '14 at 20:16
  • $\begingroup$ Woops, thank you for catching that. Fixed. $\endgroup$
    – parsiad
    Oct 11 '14 at 20:20
  • $\begingroup$ With that in mind can I affirm that $ Pr\left ( X_{1} , X_{2}... X_{n} \leq \frac{\theta }{2} \right ) $ = $ \int_{0}^{\frac{\theta }{2}}\frac{1}{\theta - 0} = 1 $ ? $\endgroup$
    – matt_zarro
    Oct 11 '14 at 20:35
  • 2
    $\begingroup$ @matt_zarro : $\Pr(X_1\le\theta/2\ \&\ X_2\le\theta/2\ \&\ \cdots\ \&\ X_n\le\theta/2)$ $=\Pr(X_1\le\theta/2)\cdot\Pr(X_2\le\theta/2)\cdots\Pr(X_n\le\theta/)$ $=(\theta/2)(\theta/2)\cdots(\theta/2)$. ${}\qquad{}$ $\endgroup$ Oct 11 '14 at 21:32

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