Convergence of sequence of continuous random variables

**funnyinga** · August 29th 2009, 06:04 AM

Let $X_1, X_2, ..$ be a sequence of absolutely continuous random variables such that $X_n$ has pdf

$f_n(x) = \begin{cases} 1 - \cos (2 \pi n x) & x \in [0,1], \\ 0 & otherwise \end{cases}$ .

Show that the sequence converges in distribution. What happens when $f_n \rightarrow \infty$ ?

**Moo** · August 29th 2009, 06:33 AM

Hello,

to check for the convergence of a sequence of rv, you mustn't see the convergence of the pdf, but the convergence of the mgf, cdf, or characteristic function.

so for example, let's find the cdf :
$F_n(x)=\begin{cases} 0 & \text{if } x<0 \\ x-\frac{\sin(2\pi n x)}{2\pi n} & \text{if } x \in[0,1] \\ 1 &\text{if } x>1 \end{cases}$

by using the squeeze (or sandwich) theorem, we can show that $\lim_{n\to\infty} \frac{\sin(2\pi n x)}{2\pi n}=0$

so the limiting cdf is $F(x)=\begin{cases} 0 & \text{if } x<0 \\ x & \text{if } x \in[0,1] \\ 1 &\text{if } x>1 \end{cases}$
which is the cdf of a uniform distribution over $[0,1]$

looks clear to you ?

**funnyinga** · August 30th 2009, 03:56 AM

Originally Posted by Moo

looks clear to you ?

Yes I think so.

Since we have the cdf, we can show that the sequence converges in distribution if the cdf converges weakly i.e.

$\lim_{n\to\infty} F_n = F(x)$

Is that essentially what you did here?

**Moo** · August 30th 2009, 04:43 AM

Yes, exactly

converges weakly towards a uniform distribution.

**funnyinga** · August 31st 2009, 05:26 AM

Im now only having trouble picturing what happens when $f_n \rightarrow \infty$ . Can anyone please explain?

**Moo** · September 18th 2009, 07:25 AM

There is no point finding the convergence of $f_n$ . I mean it's not useful for the convergence of random variables here.

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