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'''Part 3.''' | '''Part 3.''' | ||
| − | Let <math>X</math> be an exponential random variable with parameter <math>\lambda</math>, so that <math>f_X(x)= | + | Let <math>X</math> be an exponential random variable with parameter <math>\lambda</math>, so that <math>f_X(x)=\lambda{exp(-\lambda{x}u(x)</math>. Find the variance of <math>X</math>. You must show all of your work. |
| − | lambda{exp(-\lambda{x}u(x)</math>. Find the variance of <math>X</math>. You must show all of your work. | + | |
:'''Click [[ECE_PhD_QE_CNSIP_2013_Problem1.3|here]] to view student [[ECE_PhD_QE_CNSIP_2013_Problem1.3|answers and discussions]]''' | :'''Click [[ECE_PhD_QE_CNSIP_2013_Problem1.3|here]] to view student [[ECE_PhD_QE_CNSIP_2013_Problem1.3|answers and discussions]]''' | ||
Revision as of 18:54, 3 November 2014
Communication, Networking, Signal and Image Processing (CS)
Question 1: Probability and Random Processes
August 2013
Question
Part 1.
Consider $ n $ independent flips of a coin having probability $ p $ of landing on heads. Say that a changeover occurs whenever an outcome differs from the one preceding it. For instance, if $ n=5 $ and the sequence $ HHTHT $ is observed, then there are 3 changeovers. Find the expected number of changeovers for $ n $ flips. Hint: Express the number of changeovers as a sum of Bernoulli random variables.
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Part 2.
Let $ X_1,X_2,... $ be a sequence of jointly Gaussian random variables with covariance
$ Cov(X_i,X_j) = \left\{ \begin{array}{ll} {\sigma}^2, & i=j\\ \rho{\sigma}^2, & |i-j|=1\\ 0, & otherwise \end{array} \right. $
Suppose we take 2 consecutive samples from this sequence to form a vector $ X $, which is then linearly transformed to form a 2-dimensional random vector $ Y=AX $. Find a matrix $ A $ so that the components of $ Y $ are independent random variables You must justify your answer.
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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.
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Part 4.
Write question here.
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