0
$\begingroup$

$X_n$ converges in distribution to $X$ with a uniform distribution on $[0,2]$. Does it imply that $X_n^2$ converges in distribution?

My attempt:

I want to show the convergence of cdfs, so:

$\Pr\left(X_n^2 \le t\right)=\Pr(-\sqrt{t}\le X_n \le \sqrt{t})\rightarrow \int_{-\sqrt{t}}^{\sqrt{t}} \frac{1}{2}\chi_{[0,2]}(x)dx=\frac{1}{2}\int_{0}^{\sqrt{t}}\chi_{[0,2]}(x)dx= 0$ for $t \le0$, $\frac{\sqrt{t}}{2}$ for $0 <t \le4$ and $1$ for $t\ge 4$.

Could you help me to determine what distribution stands for the limit cdf?

It seems to be $\mathrm{Beta}\left(\frac{1}{2},1\right)$, but I'm not sure...

$\endgroup$
1
$\begingroup$

Yes, $X_n^2 \overset{d}{\rightarrow}X^2$ by the Continuous Mapping Theorem.

$\endgroup$

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.