(New page: Category:ECE Category:QE Category:CNSIP Category:problem solving Category:random variables Category:probability <center> <font size= 4> [[ECE_PhD_Qualifying_Exams...)
 
 
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The number of changeovers <math>Y</math> can be expressed as the sum of n-1 Bernoulli random variables:
 
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>
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<math>Y=\sum_{i=1}^{n-1}X_i</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>
+
Therefore,
  
'''(b)'''
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<math>E(Y)=E(\sum_{i=1}^{n-1}X_i)=\sum_{i=1}^{n-1}E(X_i)</math>.
  
<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> .
+
For Bernoulli random variables,
  
<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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<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>.
 +
 
 +
<font color="red"><u>'''Comments on Solution 1:'''</u>
 +
 
 +
Good solution with appropriate explanation.
 +
</font>
 
----
 
----
 
==Solution 2==
 
==Solution 2==
Write it here.
+
 
 +
For n flips, there are n-1 changeovers at most. Assume random variable <math>k_i</math> for changeover,
 +
 
 +
<math>P(k_i=1)=p(1-p)+(1-p)p=2p(1-p)</math>
 +
 
 +
<math>E(k)=\sum_{i=1}^{n-1}P(k_i=1)=2(n-1)p(1-p)</math>
 +
 
 +
<font color="red"><u>'''Critique on Solution 2:'''</u>
 +
 
 +
The solution is correct. However, it's better to explicitly express <math>k_i</math> as a Bernoulli random variable. This makes it easier for readers to understand.
 +
 
 +
</font>
 +
 
 +
----
 +
==Solution 3==
 +
 
 +
<p>
 +
First, we define a Bernoulli random variable
 +
</p>
 +
 
 +
<p><span class="mwe-math-fallback-source-inline tex" dir="ltr">$ X = \left\{ \begin{array}{ll}  0, &amp; the&nbsp;change&nbsp;over&nbsp;does&nbsp;not&nbsp;occur\\  1, &amp; the&nbsp;change&nbsp;over&nbsp;occurs  \end{array} \right. $</span>
 +
 
 +
</p><p>Then we can compute
 +
 
 +
</p><p><span class="mwe-math-fallback-source-inline tex" dir="ltr">$P(X = 1) = P(1-P)+(1-P)P = P-{P}^{2}+P-{P}^2=2P-2{P}^{2} $</span>
 +
</p><p><span class="mwe-math-fallback-source-inline tex" dir="ltr">$P(X = 0) = P•P+(1-P)(1-P) = {P}^{2}+1-2P+{P}^2 $</span>
 +
</p><p>Define Y as the number of changes occurred in n flips, there exists at most n-1 changes
 +
</p><p><span class="mwe-math-fallback-source-inline tex" dir="ltr">$E(Y)=E(\sum_{i=1}^{n-1}X_i)=\sum_{i=1}^{n-1}E(X_i) $</span>
 +
</p><p><span class="mwe-math-fallback-source-inline tex" dir="ltr">$E(X_i)=p(X_i=1)=p(1-p)+(1-p)p=2p(1-p) $</span>
 +
</p><P>Therefore, we have a final solution as
 +
</p><p><span class="mwe-math-fallback-source-inline tex" dir="ltr">$ E(Y)=2(n-1)p(1-p) $</span>.
 +
</p>
 +
 
 +
 
 +
==Similar Question==
 +
Bits are sent over a communications channel in packets of 12. If
 +
the probability of a bit being corrupted over this channel is 0.1 and
 +
such errors are independent, what is the probability that no more
 +
than 2 bits in a packet are corrupted?
 +
If 6 packets are sent over the channel, what is the probability that
 +
at least one packet will contain 3 or more corrupted bits?
 +
 
 
----
 
----
[[ECE_PhD_QE_CNSIP_2000_Problem1|Back to QE CS question 1, August 2000]]
+
[[ECE-QE_CS1-2013|Back to QE CS question 1, August 2013]]
  
 
[[ECE_PhD_Qualifying_Exams|Back to ECE Qualifying Exams (QE) page]]
 
[[ECE_PhD_Qualifying_Exams|Back to ECE Qualifying Exams (QE) page]]

Latest revision as of 17:28, 23 February 2017


ECE Ph.D. Qualifying Exam

Communication, Networking, Signal and Image Processing (CS)

Question 1: Probability and Random Processes

August 2013



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.


Solution 1

The number of changeovers $ Y $ can be expressed as the sum of n-1 Bernoulli random variables:

$ Y=\sum_{i=1}^{n-1}X_i $.

Therefore,

$ E(Y)=E(\sum_{i=1}^{n-1}X_i)=\sum_{i=1}^{n-1}E(X_i) $.

For Bernoulli random variables,

$ E(X_i)=p(E_i=1)=p(1-p)+(1-p)p=2p(1-p) $.

Thus

$ E(Y)=2(n-1)p(1-p) $.

Comments on Solution 1:

Good solution with appropriate explanation.


Solution 2

For n flips, there are n-1 changeovers at most. Assume random variable $ k_i $ for changeover,

$ P(k_i=1)=p(1-p)+(1-p)p=2p(1-p) $

$ E(k)=\sum_{i=1}^{n-1}P(k_i=1)=2(n-1)p(1-p) $

Critique on Solution 2:

The solution is correct. However, it's better to explicitly express $ k_i $ as a Bernoulli random variable. This makes it easier for readers to understand.


Solution 3

First, we define a Bernoulli random variable

$ X = \left\{ \begin{array}{ll} 0, & the change over does not occur\\ 1, & the change over occurs \end{array} \right. $

Then we can compute

$P(X = 1) = P(1-P)+(1-P)P = P-{P}^{2}+P-{P}^2=2P-2{P}^{2} $

$P(X = 0) = P•P+(1-P)(1-P) = {P}^{2}+1-2P+{P}^2 $

Define Y as the number of changes occurred in n flips, there exists at most n-1 changes

$E(Y)=E(\sum_{i=1}^{n-1}X_i)=\sum_{i=1}^{n-1}E(X_i) $

$E(X_i)=p(X_i=1)=p(1-p)+(1-p)p=2p(1-p) $

Therefore, we have a final solution as

$ E(Y)=2(n-1)p(1-p) $.


Similar Question

Bits are sent over a communications channel in packets of 12. If the probability of a bit being corrupted over this channel is 0.1 and such errors are independent, what is the probability that no more than 2 bits in a packet are corrupted? If 6 packets are sent over the channel, what is the probability that at least one packet will contain 3 or more corrupted bits?


Back to QE CS question 1, August 2013

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