# Plz Help is required in expectation

• Sep 23rd 2008, 05:09 PM
UMD
Plz Help is required in expectation
I shall be very grateful to anyone who can guide me regarding the solution of the following problems.

Show that if Xn converges to X in rth mean ( r>=1), then E ( |Xn^r| )converges to E ( | X ^r |)

Show that if Xn converges to X in rth mean with r=1, then E(Xn) Converges to E(X). Give an example to show that the converse is false
• Sep 24th 2008, 02:13 PM
Laurent
For the first question: rewrite the conclusion and hypothesis in terms of $L^r$ norm, and it should become straightforward.

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Don't read the following if you want to find it out by yourself
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By the triangle inequality, $|\|X_n\|_r-\|X\|_r|\leq \|X_n-X\|_r\to_n 0$, so that $\|X_n\|_r\to_n\|X\|_r$, which directly implies $E[|X_n|^r]\to_n E[|X|^r]$.

For the second one, I think you can prove the first part by yourself. And for the reverse, almost every example works: $X_n=\pm1$ with probability 1/2, and $X=0$ may be the easiest.
• Sep 27th 2008, 11:50 AM
UMD
@ Laurant
Thanks alot Brother. U r providing great help for which i am grateful
• Sep 27th 2008, 12:17 PM
UMD
@ Laurant
one last question

Show that if Xn converges to X in probability, then Xn converges mutually in probability .

• Sep 27th 2008, 02:03 PM
Laurent
Quote:

Originally Posted by UMD
Show that if Xn converges to X in probability, then Xn converges mutually in probability .

I don't know what it means to "converge mutually in probability". It seems to be an unusual expression. Could you define that?
• Sep 29th 2008, 10:57 AM
Player1
I think "mutual convergence in probability" means $P(|X_m - X_n| > \varepsilon) \rightarrow_{m,n \rightarrow \infty} 0$

They use the triangle inequality to argue something related here. However,
1) I don't know what they mean by norm of a random variable, and
2) I don't know if or how you can use that here...
• Sep 29th 2008, 01:20 PM
Laurent
Quote:

Originally Posted by Player1
I think "mutual convergence in probability" means $P(|X_m - X_n| > \varepsilon) \rightarrow_{m,n \rightarrow \infty} 0$

Thank you, I was suspecting something like this as well but I was pretty unsure. So "mutually" must refer to a "Cauchy sequence"-like behaviour.

Suppose $(X_n)_n$ converges to $X$ in probability. Let $\varepsilon>0$. For any $m,n$, $P(|X_n-X_m|>\varepsilon)=P(|(X_n-X)-(X_m-X)|>\varepsilon)$ $\leq P(|X_n-X|+|X_m-X|>\varepsilon)\leq P(|X_n-X|>\varepsilon/2\mbox{ or }|X_m-X|>\varepsilon/2)$ $\leq P(|X_n-X|>\varepsilon/2)+P(|X_m-X|>\varepsilon/2)$. And you can easily conclude from here.
• Oct 4th 2008, 04:23 PM
UMD
@ Laurant
Thanks for ur reply. I am sorry for not replying on time. Yes u were right that was the thing that i exactly needed. But i submitted my assignment without it. Anyways many many thanks for ur reply. i hav a few more problems to solve if u have time to think over it, I shall be grateful. I have to submit this assignment the day after tomorrow. Lastly how to use mathematical symbols in this forum ?

1) If Xn converges in quadratic mean to X and Xn is a guassian random variable with mean Un and variance ( the symbol of variance), and if variance = lim ( n approaches to infinity) variance is non zero, show that X is Guassian

2) the second question i cannot write bcz of lack of mathematical symbols

• Oct 4th 2008, 04:46 PM
Player1
Hi!

$X_n$ converges in the second mean is telling you
1) The variance is finite
2) The mean is finite
3) $X_n$ converges in probability and in distribution to $X$

Then write the PDF of $X_n$. As you have convergence in distribution, and 1) and 2), then $\mu_n \rightarrow \mu$ and $\sigma_n^2 \rightarrow \sigma^2$ as $n \rightarrow \infty$. And that's gaussian too.

Please somebody correct my post if I am wrong. Thanks.

To write math symbols, type "latex math symbols" in google. Then, enclose your symbols between opensquarebracket math closesquarebracket and opensquarebracket /math closesquarebracket. You get me, just replace the brackets with real brackets ;)
• Oct 4th 2008, 05:10 PM
UMD
@player1
(Sleepy)Thanks. U said take its pdf. can u elaborate this .
• Oct 4th 2008, 05:57 PM
Player1
Each of $X_n$ has probability density function (or PDF)

$f_{X_n}(x) = \frac{1}{\sqrt{2 \pi \sigma_n ^2}} exp\{-\frac{(x-\mu_n)^2}{2 \sigma_n^2}\}$

and by 1), 2) and 3), the PDF of $X$ is

$f_{X_n}(x) = \frac{1}{\sqrt{2 \pi \sigma ^2}} exp\{-\frac{(x-\mu)^2}{2 \sigma^2}\}$

which is gaussian too.
• Oct 4th 2008, 06:02 PM
UMD
@laurant,Player1
Please see the attached word document for question. I shall be very grateful to anyone who can guide me regarding its solution
• Oct 4th 2008, 06:05 PM
UMD
@player1
Plz if u can see the second one , the attached word file
• Oct 5th 2008, 05:18 AM
Laurent
Quote:

Originally Posted by Player1
Each of $X_n$ has probability density function (or PDF)

$f_{X_n}(x) = \frac{1}{\sqrt{2 \pi \sigma_n ^2}} exp\{-\frac{(x-\mu_n)^2}{2 \sigma_n^2}\}$

and by 1), 2) and 3), the PDF of $X$ is

$f_{X_n}(x) = \frac{1}{\sqrt{2 \pi \sigma ^2}} exp\{-\frac{(x-\mu)^2}{2 \sigma^2}\}$

which is gaussian too.

Beware! It is common misbelief that the probability density function converges if there is convergence in distribution. What is for sure is that for instance the cumulative distribution function converges, as well as the characteristic function.

Because $\|X_n-X\|_2\to_n 0$, you know (by a previous question) that $E[X_n^2]\to_n E[X^2]$.

Because $\| X_n-X\|_1\leq \|X_n-X\|_2$ (as Cauchy-Schwarz inequality shows), and because of a previous question of yours, you know that the mean $\mu_n$ of $X_n$ converges to $\mu=E[X]$. And hence the same with variances using the previous paragraph: $\sigma_n^2\to_n \sigma^2={\rm Var}(X)$.

As for the distribution of the limit, it results from considering the characteristic function: because of the convergence in distribution, for all $t\in\mathbb{R}$, $E[e^{it X_n}]\to_n E[e^{itX}]$. However, you know that $E[e^{itX_n}]=e^{it\mu_n-t^2\sigma_n^2/2}\to_n e^{it\mu-t^2\sigma^2/2}$ because $X_n$ is Gaussian and because of what we said about the convergence of the mean and variance. So the characteristic function of $X$ is $e^{i t\mu-t^2\sigma^2/2}$, which is the characteristic function of a Gaussian random variable of mean $\mu$ and variance $\sigma^2$. The fact that the characteristic function characterizes the distribution allows to conclude.