Difference between revisions of "ECE PhD QE CNSIP 2015 Problem1.3" - Rhea

Latest revision as of 21:00, 7 March 2016

Computer Engineering(CE)

Question 1: Algorithms

August 2015

Solution 3

For this problem, it is very useful to note that for any independent random variables $$ X $$ and $$ Y $$ and their characteristic functions $\phi_X(\omega),\,\phi_Y(\omega)$ we have the following property:

$\phi_{X+Y}(\omega) = \phi_X(\omega)\phi_Y(\omega)$

We then note that the characteristic function of an exponential random variable $$ Z $$ is written as

$\phi_Z (\omega) = \frac{\lambda}{\lambda - i\omega}$

where $\lambda$ parameterizes the exponential distribution. As such, we can write the characteristic function of $$ X+Y $$ as

$\phi_{X+Y}(\omega) = \phi_X(\omega)\phi_Y(\omega) \\ = \left(\frac{\lambda}{\lambda-i\omega}\right)^2$

Next, we recall that the mean of an exponential random variable is equal to the inverse of its parameter, i.e. $\frac{1}{\lambda}$ . Then the above expression becomes

$\phi_{X+Y}(\omega) = \left(\frac{\frac{1}{\mu}}{\frac{1}{\mu}-i\omega}\right)^2$

Multiplying by $\frac{\mu^2}{\mu^2}$ gives

$\phi_{X+Y}(\omega) = \left(\frac{1}{1-i\omega\mu}\right)^2$

Back to QE CE question 1, August 2015

Back to ECE Qualifying Exams (QE) page

@@ Line 1: / Line 1: @@
 [[Category:ECE]]
 [[Category:QE]]
-[[Category:CNSIP]]
+[[Category:CE]]
 [[Category:problem solving]]
-[[Category:random variables]]
+[[Category:algorithms]]
-[[Category:probability]]
@@ Line 13: / Line 12: @@
 <font size= 4>
-Communication, Networking, Signal and Image Processing (CS)
+Computer Engineering(CE)
-Question 1: Probability and Random Processes
+Question 1: Algorithms
 </font size>
@@ Line 21: / Line 20: @@
 </center>
 ----
-===Solution 1===
-Let <math>\lambda = \frac{1}{\mu}</math>, then <math>E(X)=E(Y)=\frac{1}{\lambda}</math>.
-<math>
-\phi_{X+Y}=E[e^{it(X+Y)}]=\int_{X}\int_{Y}e^{it(X+Y)}p(x,y)dxdy
-</math>
-As X and Y are independent
-<math>
-\phi_{X+Y}=\int_{X}\int_{Y}e^{it(x+y)}p(x)p(y)dxdy = \int_{X}e^{itx}p(x)dx\int_{Y}e^{ity}p(y)dy=\phi_{X}\phi_{Y}
-</math>
-And
-<math>
-\phi_{X}=E[e^{itX}]=\int_{-\infty}^{\infty}e^{itx}\lambda e^{-\lambda x} dx \\
-= \lambda \int_{-\infty}^{\infty}e^{-(\lambda -iu)x} dx = -\frac{\lambda}{\lambda-iu}e^{-(\lambda-iu)x}|_0^\infty\\
-=\frac{\lambda}{\lambda-iu}
-</math>
-So
-<math>
-\phi_{X+Y}=E[e^{it(X+Y)}]=\phi_{X}\phi_{Y} =( \frac{\lambda}{\lambda-iu})^2=\frac{1}{(1+iu\mu)^2}
-</math>
-===Solution 2===
-<math>
-\phi_X(w)=E[e^{iwX}]=\int_0^{+\infty}e^{iwX}\frac{1}{\mu}e^{-\frac{x}{\mu}}dx=e^{X(iw-\frac{1}{\mu})}\frac{1}{\mu}\frac{1}{iw-\frac{1}{\mu}}|_0^{+\infty}\\
-= 0 - \frac{1}{\mu}\cdot\frac{1}{iw-\frac{1}{\mu}}=\frac{1}{1-iw\mu}\\
-\phi_{X+Y}(w)=E[e^{iw(X+Y)}]\\
-=\int\int e^{iw(X+Y)}f_X(x)f_Y(y)dxdy = \int e^{iwx}f_X(x)dx \cdot \int e^{iwy}f_Y(y)dy=\phi_X(w)\phi_Y(w)\\
-=\frac{1}{(1+iw\mu)^2}
-</math>
 ===Solution 3===
-For this problem, it is very useful to note that for any independent random variables <math>X</math> and <math>Y</math> and their characteristic functions <math>\phi_X(\omega), \phi_Y(\omega)</math> we have the following property:
+For this problem, it is very useful to note that for any independent random variables <math>X</math> and <math>Y</math> and their characteristic functions <math>\phi_X(\omega),\,\phi_Y(\omega)</math> we have the following property:
 <math>
 \phi_{X+Y}(\omega) = \phi_X(\omega)\phi_Y(\omega)
-</math>.
+</math>
 We then note that the characteristic function of an exponential random variable <math>Z</math> is written as
 <math>
-\phi_{Z}(\omega) = \frac{\lambda}{\lambda - i\omega}
+\phi_Z (\omega) = \frac{\lambda}{\lambda - i\omega}
 </math>
-where <math>\lambda</math> parameterizes the exponential distribution. As such, we can write the characteristic function of <math>X + Y</math> as
+where <math>\lambda</math> parameterizes the exponential distribution. As such, we can write the characteristic function of <math>X+Y</math> as
 <math>
-\phi_{X+Y}(\omega) = \phi_X(\omega)\phi_Y(\omega) = \left(\frac{\lambda}{\lambda - i\omega}\right)^2
+\phi_{X+Y}(\omega) = \phi_X(\omega)\phi_Y(\omega) \\
-</math>.
+= \left(\frac{\lambda}{\lambda-i\omega}\right)^2
+</math>
 Next, we recall that the mean of an exponential random variable is equal to the inverse of its parameter, i.e. <math>\frac{1}{\lambda}</math>. Then the above expression becomes
 <math>
-\phi_{X+Y}(\omega) = \left(\frac{\frac{1}{\mu}}{\frac{1}{\mu} - i\omega}\right)^2.
+\phi_{X+Y}(\omega) = \left(\frac{\frac{1}{\mu}}{\frac{1}{\mu}-i\omega}\right)^2
-</math>.
+</math>
-Multiplying by <math>\frac{\mu}{\mu}</math> gives
+Multiplying by <math>\frac{\mu^2}{\mu^2}</math> gives
 <math>
@@ Line 86: / Line 52: @@
 </math>
-===Similar Problem===
-Let <math>X(t)</math> be a Poisson random process with mean function <math>\mu(t)</math> and covariance function <math>C_{xx}(t_1,t_2)</math>. Find the <math>n^{th}</math>-order characteristic function of <math>X(t)</math>.
+[[ECE-QE_CE1-2015|Back to QE CE question 1, August 2015]]
-----
-[[ECE-QE_CS1-2015|Back to QE CS question 1, August 2015]]
 [[ECE_PhD_Qualifying_Exams|Back to ECE Qualifying Exams (QE) page]]

Difference between revisions of "ECE PhD QE CNSIP 2015 Problem1.3" - Rhea

Latest revision as of 21:00, 7 March 2016

Solution 3

Alumni Liaison