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=Solution 1=
 
=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
  
−
Therefore, a solution is
+
<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>
  
−
<math>A=\left(\begin{array}{cc}
+
and
−
1 & 1\\
+
−
-1 & 1
+
−
\end{array} \right)</math>.
+
  
 +
<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>.
−
==Solution 2==
+
  
−
Assume
+
Then we take a look at <math>E(X)</math>.
  
−
<math>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)</math>.
+
<math>E(X)=\int_0^{\infty}x\lambda{e}^{-\lambda{x}}dx</math>
  
−
Then
+
<math>\begin{array}{l}
−
 
+
\int x\lambda{e}^{-\lambda{x}}dx\\
−
<math>\begin{array}{l}E(Y_iY_j)=E[(a_{11}X_i+a_{12}X_j)(a_{21}X_i+a_{22}X_j)]\\
+
=\int xde^(\lambda x)\\
−
=a_{11}a_{21}\sigma^2+a_{12}a_{22}\sigma^2+(a_{11}a_{21}+a_{22}a_{11})E(X_iX_j)
+
=-xe^{-\lambda x}+\int e^{\lambda x}dx\\
 +
=-xe^{-\lambda x}-\frac{1}{x}e^{\lambda x}\\
 
\end{array}</math>
 
\end{array}</math>
  
−
For <math>|i-j|\geq1</math>, <math>E(X_i,X_j)=0</math>. Therefore, <math>a_{11}a_{21}+a_{12}a_{22}=0</math>.
+
Similar to the calculation of <math>E(X^2)</math>,
  
−
One solution can be
+
<math>E(X)=\frac{1}{\lambda}</math>.
  
−
<math>A=\left(\begin{array}{cc}
+
Therefore,
−
1 & -1\\
+
−
1 & 1
+
−
\end{array} \right)</math>.
+
  
 +
<math>Var(X)=E(X^2)-E(X)^2=\frac{2}{\lambda^2}-\frac{1}{\lambda^2}=\frac{1}{\lambda^2}</math>.
  
−
<font color="red"><u>'''Critique on Solution 2:'''</u>
+
----
 +
==Solution 2==
  
−
1. <math>E(Y_iY_j)=0</math> is not the condition for the two random variables to be independent.
+
<math>\begin{align}
 +
E(X)&=\int_{-\infty}^{+\infty}xp(x)dx\\
 +
&=\int_{0}^{\infty}x\lambda e^{-\lambda x}dx\\
 +
&=-(xe^{-\lambda x}|_0^{\infty}-\int_0^{\infty}e^{-\lambda x}dx)\\
 +
&=\frac{1}{x}
 +
\end{align}</math>
 +
 
 +
<math>\begin{align}
 +
E(X^2)&=\int_{-\infty}^{+\infty}x^2p(x)dx\\
 +
&=\int_{0}^{\infty}x^2 \lambda e^{-\lambda x}dx\\
 +
&=-(x^2e^{-\lambda x}|_0^{\infty}-\int_0^{\infty}2xe^{-\lambda x}dx)\\
 +
&=\frac{2}{x^2}
 +
\end{align}</math>
 +
 
 +
Therefore,
 +
 
 +
<math>Var(X)=E(X^2)-E(X)^2=\frac{1}{\lambda^2}</math>
 +
 
 +
<font color="red"><u>'''Critique on Solution 2:'''</u>
  
−
2. "For <math>|i-j|\geq1</math>, <math>E(X_i,X_j)=0</math>" is not supported by the given conditions.
+
Solution 2 is correct. In addition, calculating <math>E(X)</math> first is better since the result can be used in calculating <math>E(X^2)</math>.
  
 
</font>
 
</font>

Latest revision as of 20:02, 5 August 2018


ECE Ph.D. Qualifying Exam

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

$ \begin{align} E(X)&=\int_{-\infty}^{+\infty}xp(x)dx\\ &=\int_{0}^{\infty}x\lambda e^{-\lambda x}dx\\ &=-(xe^{-\lambda x}|_0^{\infty}-\int_0^{\infty}e^{-\lambda x}dx)\\ &=\frac{1}{x} \end{align} $

$ \begin{align} E(X^2)&=\int_{-\infty}^{+\infty}x^2p(x)dx\\ &=\int_{0}^{\infty}x^2 \lambda e^{-\lambda x}dx\\ &=-(x^2e^{-\lambda x}|_0^{\infty}-\int_0^{\infty}2xe^{-\lambda x}dx)\\ &=\frac{2}{x^2} \end{align} $

Therefore,

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

Critique on Solution 2:

Solution 2 is correct. In addition, calculating $ E(X) $ first is better since the result can be used in calculating $ E(X^2) $.

Back to QE CS question 1, August 2013

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