1
$\begingroup$

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!

$\endgroup$
1
$\begingroup$

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

$\endgroup$
12
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.