Revision as of 06:24, 4 November 2014

Communication, Networking, Signal and Image Processing (CS)

Question 1: Probability and Random Processes

August 2013

Part 3

Let $$ X $$ be an exponential random variable with parameter $\lambda$ , so that $f_X(x)=\lambda{exp}(-\lambda{x})u(x)$ . Find the variance of $$ X $$ . You must show all of your work.

Solution 1

$$ Var(X)=E(X^2)-E(X)^2 $$

First,

$E(X^2)=\int_0^{\infty}x^2\lambda{e}^{-\lambda{x}}dx$

Since

$\begin{array}{l}\int{x}^2\lambda{e}^{-\lambda{x}}dx\\ =\int -x^2 de^{-\lambda x}\\ =-x^2e^{-{\lambda}x}+{\int}2xe^{-{\lambda}x}dx\\ =-x^2e^{-{\lambda}x}-\frac{2x}{\lambda}e^{\lambda x}+{\int}\frac{e^{-{\lambda}x}}{\lambda}2dx\\ =-x^2e^{-{\lambda}x}-\frac{2x}{\lambda}e^{\lambda x}-\frac{2}{\lambda^2}e^{\lambda x} \end{array}$ ,

We have

$E(X^2)=-x^2e^{-\lambda x}-\frac{2x}{\lambda}e^{\lambda x}-\frac{2}{\lambda^2}e^{\lambda x}|_0^\infty$

By L'Hospital's rule, we have

$\lim_{x\to \infty}x^2e^{-\lambda x} = \lim_{x\to \infty}\frac{x^2}{e^{-\lambda x}}=\lim_{x\to \infty}\frac{2x}{\lambda e^{\lambda x}}=\lim_{x\to \infty}\frac{2}{\lambda^2e^{\lambda x}}=0$

and

$\lim_{x\to \infty}xe^{\lambda x} = \lim_{x\to \infty} \frac{x}{e^{\lambda x}}=\lim_{x\to \infty} \frac{1}{\lambda e^{\lambda x}} = 0$ .

Therefore,

$E(X) = \frac{2}{\lambda^2}$ .

Then we take a look at $$ E(X) $$ .

$E(X)=\int_0^{\infty}x\lambda{e}^{-\lambda{x}}dx$

$\begin{array}{l} \int x\lambda{e}^{-\lambda{x}}dx\\ =\int xde^(\lambda x)\\ =-xe^{-\lambda x}+\int e^{\lambda x}dx\\ =-xe^{-\lambda x}-\frac{1}{x}e^{\lambda x}\\ \end{array}$

Similar to the calculation of $$ E(X^2) $$ ,

$E(X)=\frac{1}{\lambda}$ .

Therefore,

$Var(X)=E(X^2)-E(X)^2=\frac{2}{\lambda^2}-\frac{1}{\lambda^2}=\frac{1}{\lambda^2}$ .

Solution 2

Assume

$Y=\left(\begin{array}{c}Y_i \\ Y_j\end{array} \right)=A\left(\begin{array}{c}X_i \\ X_j\end{array} \right)=\left(\begin{array}{c}a_{11}X_i+a_{12}X_j \\ a_{21}X_i+a_{22}X_j\end{array} \right)$ .

Then

$\begin{array}{l}E(Y_iY_j)=E[(a_{11}X_i+a_{12}X_j)(a_{21}X_i+a_{22}X_j)]\\ =a_{11}a_{21}\sigma^2+a_{12}a_{22}\sigma^2+(a_{11}a_{21}+a_{22}a_{11})E(X_iX_j) \end{array}$

For $|i-j|\geq1$ , $$ E(X_i,X_j)=0 $$ . Therefore, $a_{11}a_{21}+a_{12}a_{22}=0$ .

One solution can be

$A=\left(\begin{array}{cc} 1 & -1\\ 1 & 1 \end{array} \right)$ .

Critique on Solution 2:

1. $$ E(Y_iY_j)=0 $$ is not the condition for the two random variables to be independent.

2. "For $|i-j|\geq1$ , $$ E(X_i,X_j)=0 $$ " is not supported by the given conditions.

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@@ Line 27: / Line 27: @@
 =Solution 1=
-Suppose
+<math>Var(X)=E(X^2)-E(X)^2</math>
-<math>A=\left(\begin{array}{cc}
+First,
-a & b\\
-c & d
-\end{array} \right)</math>.
-Then the new 2-D random vector can be expressed as
+<math>E(X^2)=\int_0^{\infty}x^2\lambda{e}^{-\lambda{x}}dx</math>
-<math>Y=\left(\begin{array}{c}Y_1 \\ Y_2\end{array} \right)=A\left(\begin{array}{c}X_i \\ X_j\end{array} \right)=\left(\begin{array}{c}aX_i+bX_j \\ cX_i+dX_j\end{array} \right)</math>
+Since
+<math>\begin{array}{l}\int{x}^2\lambda{e}^{-\lambda{x}}dx\\
+=\int -x^2 de^{-\lambda x}\\
+=-x^2e^{-{\lambda}x}+{\int}2xe^{-{\lambda}x}dx\\
+=-x^2e^{-{\lambda}x}-\frac{2x}{\lambda}e^{\lambda x}+{\int}\frac{e^{-{\lambda}x}}{\lambda}2dx\\
+=-x^2e^{-{\lambda}x}-\frac{2x}{\lambda}e^{\lambda x}-\frac{2}{\lambda^2}e^{\lambda x}
+\end{array}</math>,
-Therefore,
+We have
-<math>\begin{array}{l}Cov(Y_1,Y_2)=E[(aX_i+bX_j-E(aX_i+bX_j))(cX_i+dX_j-E(cX_i+dX_j))] \\
+<math>E(X^2)=-x^2e^{-\lambda x}-\frac{2x}{\lambda}e^{\lambda x}-\frac{2}{\lambda^2}e^{\lambda x}|_0^\infty</math>
-=E[(aX_i+bX_j-aE(X_i)-bE(X_j))(cX_i+dX_j-cE(X_i)-dE(X_j))] \\
-=E[acX_i^2+adX_iX_j-acX_iE(X_i)-adX_iE(X_j)+bcX_iX_j+bdX_j^2-bcX_jE(X_i)\\
--bdX_jE(X_j)-acX_iE(X_i)-adX_jE(X_i)+acE(X_i)^2+adE(X_i)E(X_j)\\
--bcX_iE(X_j)-bdX_jE(X_j)+bcE(X_i)E(X_j)+bdE(X_i)^2]\\
-=E(ac(X_i-E(X_i))^2+(ad+bc)(X_i-E(X_i)(X_j-E(X_j))+bd(X_j-E(X_j))^2]\\
-=(ac)Cov(X_i,X_i)+(ad+bc)Cov(X-i,X_j)+(bd)Cov(X_j,X_j)\\
-=ac\sigma^2+(ad+bc)\rho\sigma^2+bd\sigma^2
-\end{array}</math>
-Let the above formula equal to 0 and <math>a=b=d=1</math>, we get <math>c=-1</math>.
+By L'Hospital's rule, we have
+<math>\lim_{x\to \infty}x^2e^{-\lambda x} = \lim_{x\to \infty}\frac{x^2}{e^{-\lambda x}}=\lim_{x\to \infty}\frac{2x}{\lambda e^{\lambda x}}=\lim_{x\to \infty}\frac{2}{\lambda^2e^{\lambda x}}=0</math>
+and
+<math>\lim_{x\to \infty}xe^{\lambda x} = \lim_{x\to \infty} \frac{x}{e^{\lambda x}}=\lim_{x\to \infty} \frac{1}{\lambda e^{\lambda x}} = 0</math>.
+Therefore,
+<math>E(X) = \frac{2}{\lambda^2}</math>.
+Then we take a look at <math>E(X)</math>.
+<math>E(X)=\int_0^{\infty}x\lambda{e}^{-\lambda{x}}dx</math>
+<math>\begin{array}{l}
+\int x\lambda{e}^{-\lambda{x}}dx\\
+=\int xde^(\lambda x)\\
+=-xe^{-\lambda x}+\int e^{\lambda x}dx\\
+=-xe^{-\lambda x}-\frac{1}{x}e^{\lambda x}\\
+\end{array}</math>
-Therefore, a solution is
+Similar to the calculation of <math>E(X^2)</math>,
-<math>A=\left(\begin{array}{cc}
+<math>E(X)=\frac{1}{\lambda}</math>.
-& 1\\
--1 & 1
-\end{array} \right)</math>.
+Therefore,
+<math>Var(X)=E(X^2)-E(X)^2=\frac{2}{\lambda^2}-\frac{1}{\lambda^2}=\frac{1}{\lambda^2}</math>.
 ----

Difference between revisions of "ECE PhD QE CNSIP 2013 Problem1.3" - Rhea

Revision as of 06:24, 4 November 2014

Part 3

Solution 1

Solution 2

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