1. ## Convergence in probability (Uniform Distribution)

let $X_1,...,X_n$ are iid $U(0, \theta)$. I am trying to show that $\sqrt{T_n} \rightarrow^{p} \sqrt{\theta}$(converges in probability to), where $T_n$ is the $largest\;order\; statistic$

what I did was using the pdf of max order statistic find

$E(T_n)\;=\;\dfrac{n\theta}{n+1}$

$E({T_n}^2)=\dfrac{n\theta^2}{n+2}$

so, $E\{(T_{n}-\theta)^2\} = \dfrac{2{\theta}^2}{(n+1)(n+2)}$, which converges to 0, as n goes to infinity,

so we can conclude by weak law of large numbers that $T_n \rightarrow^{p} \theta \;as\; n \to \infty$

and this $\sqrt{T_n} \rightarrow^{p} \sqrt{\theta}$ follows from the theorem of continuous mapping....

is my approach to the proof correct?

thank you.

2. I didn't check your calculations, but by Markov all you need is

$P\left(|T_n-\theta|\ge \epsilon\right)\le {E(T_n-\theta)^2\over \epsilon^2}\to 0$

Then you can use the fact that the square root is a continuous function giving

$\sqrt{T_n}\to \sqrt{\theta}$

So, you may need to add and subtract your mean in that second moment and expand
And this is not the Weak Law of Large Numbers, this is not a sum.

Like $E(T_n-\theta)^2=E[(T_n- {n\theta\over n+1}) +({n\theta\over n+1}-\theta)]^2$

3. thank you matheagle..

actually what I have done is:

$P(|T_n-\theta|\geq \epsilon)=P(|T_n - \theta|^2 \geq \epsilon^2) \leq \dfrac{E[(T_n-\theta)]^2}{\epsilon^2}=\dfrac{2\theta^2}{\epsilon^2(n+1)( n+2)}$ which goes to 0 as n goes to infty

4. The idea is right, but I didn't check your calculations. I do see a power missing on epsilon though.

5. Thank you. I was typing too fast i guess.. thats an epsilon squared in the den..

if $X_n$ is a Poisson(n) RV with positive integer rate parameter, show that $as\; n\to \infty, \dfrac{X_n-E(X_n)}{\sqrt{Var(X_n)}}} \rightarrow^d \; N(0,1)$

I think this involves CLT, but I am confused on how to solve this.. could you help?

thanks.

6. Use the fact that the sum of i.i.d. Poisson's with mean one is a Poisson with mean n.
NOW use the classical CLT on the sum.
Just multiply the sample mean by n to get a CLT for the sum.

${\bar X_n- E(\bar X_n)\over \sqrt{V(\bar X_n)}} \to N(0,1)$

${\bar X_n- E(\bar X_n)\over \sqrt{V(\bar X_n)}} ={n\bar X_n- nE(\bar X_n)\over n\sqrt{V(\bar X_n)}}$

This $n\bar X_n=\sum_{i=1}^nY_i$ where each $Y_i$ is a Poisson(1)