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[[Category:problem solving]]
  
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= [[ECE_PhD_Qualifying_Exams|ECE Ph.D. Qualifying Exam]]: Automatic Control (AC)- Question 1, August 2007=
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X and Y are iid random variable with
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[[ECE_PhD_Qualifying_Exams|ECE Ph.D. Qualifying Exam]]
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<math> P(X=i) = P(Y=i) = \frac {1}{2^i}\ ,i = 1,2,3,... </math>
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Automatic Control (AC)
  
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a) Find <math> P(min(X,Y)=k)\ </math>.
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Question 1: Feedback Control Systems
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b) Find <math> P(X=Y)\ </math>.
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August 2007
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</center>
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c) Find <math> P(Y>X)\ </math>.
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d) Find <math> P(Y=kX)\ </math>.
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=Solution 1 (retrived from [[Automatic_Controls:Linear_Systems_(HKNQE_August_2007)_Problem_1|here]])=
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*To find <math> P(min(X,Y)=k)\ </math>, let  <math> Z = min(X,Y)\ </math>. Then finding the pmf of Z uses the fact that X and Y are iid
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        <math> P(Z=k) = P(X \ge k,Y \ge k) = P(X \ge k)P(Y \ge k) = P(X \ge k)^2 </math>
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        <math> P(Z=k) = \left ( \sum_{i=k}^N \frac {1}{2^i} \right )^2 = \left ( \frac {1}{2^k} \right )^2 = \frac {1}{4^k} </math>
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*To find <math> P(X=Y)\ </math>, note that  X and Y are iid and summing across all possible i,
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        <math> P(X=Y) = \sum_{i=1}^\infty P(X=i, Y=i) = \sum_{i=1}^\infty P(X=i)P(Y=i) = \sum_{i=1}^\infty \frac {1}{4^i} = \frac {1}{3} </math>
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*To find <math> P(Y>X)\ </math>, again note that X and Y are iid and summing across all possible i,
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        <math> P(Y>X) = \sum_{i=1}^\infty P(Y>i, X=i) = \sum_{i=1}^\infty P(Y>i)P(x=i)</math>
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:Next, find <math> P(Y<i)\ </math>
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        <math> P(Y>i) = 1 - P(Y \le i) </math>
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        <math> P(Y \le i) = \sum_{i=1}^\infty \frac {1}{2^i} = 1 + \frac {1}{2^i} </math>
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      <math> \therefore P(Y>i) = \frac {1}{2^i} </math>
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:Plugging this result back into the original expression yields
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        <math> P(Y<X) = \sum_{i=1}^\infty \frac {1}{4^i} = \frac {1}{3} </math>
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*To find <math> P(Y=kX)\ </math>, note that X and Y are iid and summing over all possible combinations one arrives at
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        <math> P(Y=kX) = \sum_{i=1}^\infty i = 1^\infty P(Y=ki, X=i) = \sum_{i=1}^\infty P(Y=ki)P(X=i) </math>
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:Thus,
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        <math> P(Y=kX) = \sum_{i=1}^\infty \frac {1}{2^{ki}} \frac {1}{2^i} = \sum_{i=1}^\infty \frac {1}{2^{(k+1)i}} = \frac {1}{2^{(k+1)}-1} </math>
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==Solution 2==
 
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Write it here.
 
 
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[[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 21:59, 6 August 2018


ECE Ph.D. Qualifying Exam

Automatic Control (AC)

Question 1: Feedback Control Systems

August 2007

Back to ECE Qualifying Exams (QE) page

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