# Thread: Plz Help is required in expectation

1. ## 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

2. For the first question: rewrite the conclusion and hypothesis in terms of $\displaystyle L^r$ 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, $\displaystyle |\|X_n\|_r-\|X\|_r|\leq \|X_n-X\|_r\to_n 0$, so that $\displaystyle \|X_n\|_r\to_n\|X\|_r$, which directly implies $\displaystyle E[|X_n|^r]\to_n E[|X|^r]$.

For the second one, I think you can prove the first part by yourself. And for the reverse, almost every example works: $\displaystyle X_n=\pm1$ with probability 1/2, and $\displaystyle X=0$ may be the easiest.

3. ## @ Laurant

Thanks alot Brother. U r providing great help for which i am grateful

4. ## @ Laurant

one last question

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

5. 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?

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

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...

7. Originally Posted by Player1
I think "mutual convergence in probability" means $\displaystyle P(|X_m - X_n| > \varepsilon) \rightarrow_{m,n \rightarrow \infty} 0$
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 $\displaystyle (X_n)_n$ converges to $\displaystyle X$ in probability. Let $\displaystyle \varepsilon>0$. For any $\displaystyle m,n$, $\displaystyle P(|X_n-X_m|>\varepsilon)=P(|(X_n-X)-(X_m-X)|>\varepsilon)$$\displaystyle \leq P(|X_n-X|+|X_m-X|>\varepsilon)\leq P(|X_n-X|>\varepsilon/2\mbox{ or }|X_m-X|>\varepsilon/2)$$\displaystyle \leq P(|X_n-X|>\varepsilon/2)+P(|X_m-X|>\varepsilon/2)$. And you can easily conclude from here.

8. ## @ 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

9. Hi!

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

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

10. ## @player1

Thanks. U said take its pdf. can u elaborate this .

11. Each of $\displaystyle X_n$ has probability density function (or PDF)

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

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

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

which is gaussian too.

12. ## @laurant,Player1

Please see the attached word document for question. I shall be very grateful to anyone who can guide me regarding its solution

13. ## @player1

Plz if u can see the second one , the attached word file

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

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

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

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

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

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

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