| Line 71: | Line 71: | ||
==Solution 2== | ==Solution 2== | ||
| − | <math>\ | + | Since |
| − | + | ||
| − | + | <math>\lim_{x\to \infty}f_n(x)=\frac{1}{\sqrt{2\pi}\sigma}exp[-\frac{1}{2\sigma^2}(x-\sigma)^2] \sim N(\sigma,\sigma^2)</math> | |
| − | + | ||
| − | + | Guess it converges to <math>k\sigma</math> | |
| − | \ | + | |
<math>\begin{align} | <math>\begin{align} | ||
| − | + | &\lim_{x\to \infty}E(|X_n-k\sigma|^2)\\ | |
| − | &=\ | + | &=\lim_{x\to \infty}E(X_n^2-2k\sigma x_n+k^2\sigma^2)\\ |
| − | &= | + | &=\lim_{x\to \infty}E(X_n^2)-2\sigma k \lim_{x\to \infty}E(X_n) + k^2\sigma^2\\ |
| − | &=\ | + | &=2\sigma^2-2\sigma^2k+k^2\sigma^2=0 |
\end{align}</math> | \end{align}</math> | ||
| − | + | So we need to solve <math>k^2-2k+2=0</math>. | |
| − | + | Since there is no solution, this sequence doesn't converge. | |
<font color="red"><u>'''Critique on Solution 2:'''</u> | <font color="red"><u>'''Critique on Solution 2:'''</u> | ||
| − | + | This solution uses wrong logic. If we consider <math>\lim_{x\to \infty}f_n(x)</math>, we lose the information of <math>n</math> and <math>n+m</math>. The result is we can no longer use the Cauchy criterion. The solution then guesses the convergence, which is highly unreliable. As it turns out, it fails to find the convergence. | |
</font> | </font> | ||
Revision as of 13:03, 4 November 2014
Communication, Networking, Signal and Image Processing (CS)
Question 1: Probability and Random Processes
August 2013
Part 4
Consider a sequence of independent random variables $ X_1,X_2,... $, where $ X_n $ has pdf
$ \begin{align}f_n(x)=&(1-\frac{1}{n})\frac{1}{\sqrt{2\pi}\sigma}exp[-\frac{1}{2\sigma^2}(x-\frac{n-1}{n}\sigma)^2]\\ &+\frac{1}{n}\sigma exp(-\sigma x)u(x)\end{align} $.
Does this sequence converge in the mean-square sense? Hint: Use the Cauchy criterion for mean-square convergence, which states that a sequence of random variables $ X_1,X_2,... $ converges in mean-square if and only if $ E[|X_n-X_{n+m}|] \to 0 $ as $ n \to \infty $, for every $ m>0 $.
Solution 1
$ E(X_n)=\frac{n-1}{n}E(Y)+\frac{1}{n}E(Z) $
Where
$ Y \sim N(\frac{n-1}{n}\sigma, \sigma^2) $
$ Z \sim EXP(\sigma) $
From the property of Normal distribution and exponential distribution,
$ E(Y)=\frac{n-1}{n}\sigma $
$ E(Z)=\frac{1}{\sigma} $.
Therefore,
$ \lim_{x\to \infty}E(X_n)=\lim_{x\to \infty}(\frac{n-1}{n})^2\sigma+\frac{1}{n}\frac{1}{\sigma}=\sigma $.
Also,
$ \lim_{x\to \infty}E(X_{n+m})=\lim_{x\to \infty}(\frac{n+m-1}{n+m})^2\sigma+\frac{1}{n+m}\frac{1}{\sigma}=\sigma $.
Thus,
$ \lim_{x\to \infty}E(X_n-X_{n+m})=\lim_{x\to \infty}E(X_n)-\lim_{x\to \infty}E(X_{n+m})=0 $,
$ \lim_{x\to \infty}E(X_{n+m}-X_n)=\lim_{x\to \infty}E(X_{n+m})-\lim_{x\to \infty}E(X_n)=0 $.
So we have
$ \lim_{x\to \infty}E(|X_{n+m}-X_n|)=0 $
for every m.
From the Cauchy criterion for mean-square convergence, this sequence converges int he mean-square sense
Solution 2
Since
$ \lim_{x\to \infty}f_n(x)=\frac{1}{\sqrt{2\pi}\sigma}exp[-\frac{1}{2\sigma^2}(x-\sigma)^2] \sim N(\sigma,\sigma^2) $
Guess it converges to $ k\sigma $
$ \begin{align} &\lim_{x\to \infty}E(|X_n-k\sigma|^2)\\ &=\lim_{x\to \infty}E(X_n^2-2k\sigma x_n+k^2\sigma^2)\\ &=\lim_{x\to \infty}E(X_n^2)-2\sigma k \lim_{x\to \infty}E(X_n) + k^2\sigma^2\\ &=2\sigma^2-2\sigma^2k+k^2\sigma^2=0 \end{align} $
So we need to solve $ k^2-2k+2=0 $.
Since there is no solution, this sequence doesn't converge.
Critique on Solution 2:
This solution uses wrong logic. If we consider $ \lim_{x\to \infty}f_n(x) $, we lose the information of $ n $ and $ n+m $. The result is we can no longer use the Cauchy criterion. The solution then guesses the convergence, which is highly unreliable. As it turns out, it fails to find the convergence.
