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<math>E(Y)=2(n-1)p(1-p)</math>.
 
<math>E(Y)=2(n-1)p(1-p)</math>.
  
<font color="red"><u>'''Comments Solution 1:'''</u>  
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Good solution with a appropriate explanation.
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Good solution with appropriate explanation.
 
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==Similar Question==
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Bits are sent over a communications channel in packets of 12. If
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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
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than 2 bits in a packet are corrupted?
 +
If 6 packets are sent over the channel, what is the probability that
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at least one packet will contain 3 or more corrupted bits?
  
 
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Latest revision as of 16:28, 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?


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