View source for ECE PhD QE CNSIP 2013 Problem1.1 - Rhea

[[Category:ECE]]
[[Category:QE]]
[[Category:CNSIP]]
[[Category:problem solving]]
[[Category:random variables]]
[[Category:probability]]


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[[ECE_PhD_Qualifying_Exams|ECE Ph.D. Qualifying Exam]]
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Communication, Networking, Signal and Image Processing (CS)

Question 1: Probability and Random Processes
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August 2013
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=Part 1=
Consider <math class="inline">n</math> independent flips of a coin having probability <math class="inline">p</math> of landing on heads. Say that a changeover occurs whenever an outcome differs from the one preceding it. For instance, if <math class="inline">n=5</math> and the sequence <math class="inline">HHTHT</math> is observed, then there are 3 changeovers. Find the expected number of changeovers for <math class="inline">n</math> flips. ''Hint'': Express the number of changeovers as a sum of Bernoulli random variables.
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=Solution 1=

The number of changeovers <math>Y</math> can be expressed as the sum of n-1 Bernoulli random variables:

<math>Y=\sum_{i=1}^{n-1}X_i</math>.

Therefore,

<math>E(Y)=E(\sum_{i=1}^{n-1}X_i)=\sum_{i=1}^{n-1}E(X_i)</math>.

For Bernoulli random variables,

<math>E(X_i)=p(E_i=1)=p(1-p)+(1-p)p=2p(1-p)</math>.

Thus

<math>E(Y)=2(n-1)p(1-p)</math>.

<math class="inline">\Phi_{\mathbf{X}}\left(\omega\right)=E\left[e^{i\omega\mathbf{X}}\right]=\int_{-\infty}^{\infty}\frac{A}{2}e^{-A\left|x\right|}\cdot e^{i\omega x}dx=\frac{A}{2}\left[\int_{-\infty}^{0}e^{x\left(A+i\omega\right)}dx+\int_{0}^{\infty}e^{x\left(-A+i\omega\right)}dx\right]</math><math class="inline">=\frac{A}{2}\left[\frac{e^{x\left(A+i\omega\right)}}{A+i\omega}\biggl|_{-\infty}^{0}+\frac{e^{x\left(-A+i\omega\right)}}{-A+i\omega}\biggl|_{0}^{\infty}\right]=\frac{A}{2}\left[\frac{1}{A+i\omega}-\frac{1}{-A+i\omega}\right]</math><math class="inline">=\frac{A}{2}\cdot\frac{A-i\omega+A+i\omega}{A^{2}+\omega^{2}}=\frac{A^{2}}{A^{2}+\omega^{2}}.</math> 

'''(b)'''

<math class="inline">P\left(\left\{ \left|\mathbf{X}-\overline{\mathbf{X}}\right|\leq2\sigma\right\} \right)=1-P\left(\left\{ \left|\mathbf{X}-\overline{\mathbf{X}}\right|>2\sigma\right\} \right).</math>  By [[ECE 600 Chebyshev Inequality|Chebyshev Inequality]], <math class="inline">P\left(\left\{ \left|\mathbf{X}-\overline{\mathbf{X}}\right|>2\sigma\right\} \right)\leq\frac{\sigma^{2}}{\left(2\sigma\right)^{2}}=\frac{1}{4}</math> .

<math class="inline">P\left(\left\{ \left|\mathbf{X}-\overline{\mathbf{X}}\right|\leq2\sigma\right\} \right)\geq\frac{3}{4}.</math> 
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==Solution 2==
Write it here.
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[[ECE_PhD_QE_CNSIP_2000_Problem1|Back to QE CS question 1, August 2000]]

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