2
$\begingroup$

I have troubles solving this task:

Let $U_1,U_2,\dots$ be an i.i.d. sequence of random variables with uniform distribution on $[0,1]$. We set for every integer $n\geq 1$ $$M_n=\max\{1/\sqrt{U_1},\dots,1/\sqrt{U_n}\}$$

Compute the distribution function $F_n(x)=P(M_n\leq x)\quad \text{for } x\geq 0$

What I have so far:

Since $U_i$ are i.i.d. we have $$P(M_n\leq x)=P(1/\sqrt{U_1}\leq x)^n=P(\frac{1}{x^2}\leq U_1)^n=\int_{x^{-2}}^\infty \mathbb{1}_{[0,1]}\mathrm{d}x$$ but here I get stucked. I don't know what to do. I think I have to do a substitution but this confuses me since I do not have a variable in the integrand?

$\endgroup$
1
  • $\begingroup$ You should not integrate with respect to $x$. Integrate with respect to another variable. $\endgroup$
    – Ken Duna
    Commented Jun 30, 2016 at 14:36

1 Answer 1

Reset to default
2
$\begingroup$

The error comes right at the last step.

$$P\left(\frac{1}{x^2}\leq U_1\right)^n = \left(\int_{\frac{1}{x^2}}^1 dy\right)^n = \left(1-\frac{1}{x^2}\right)^n$$

Note that this is for $x \geq 1$.

For $x<1$, $P(M_n\leq x) = 0$.

$\endgroup$
1
  • $\begingroup$ Hi, thanks a lot for the answer. Now it's clear. $\endgroup$
    – MarcE
    Commented Jun 30, 2016 at 15:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .