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ECE Ph.D. Qualifying Exam: Automatic Control (AC)- Question 1, August 2007

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 = [[ECE_PhD_Qualifying_Exams|ECE Ph.D. Qualifying Exam]]: Automatic Control (AC)- Question 1, August 2007=
-X and Y are iid random variable with
-<math> P(X=i) = P(Y=i) = \frac {1}{2^i}\ ,i = 1,2,3,... </math>
-a) Find <math> P(min(X,Y)=k)\ </math>.
-b) Find <math> P(X=Y)\ </math>.
-c) Find <math> P(Y>X)\ </math>.
-d) Find <math> P(Y=kX)\ </math>.
 ----
-=Solution 1 (retrived from [[Automatic_Controls:Linear_Systems_(HKNQE_August_2007)_Problem_1|here]])=
+==Question==
+Write question here
-*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
+----
-        <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>
+=Solution 1=
+write it here
-        <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>
-*To find <math> P(X=Y)\ </math>, note that  X and Y are iid and summing across all possible i,
-        <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>
-*To find <math> P(Y>X)\ </math>, again note that X and Y are iid and summing across all possible i,
-        <math> P(Y>X) = \sum_{i=1}^\infty P(Y>i, X=i) = \sum_{i=1}^\infty P(Y>i)P(x=i)</math>
-:Next, find <math> P(Y<i)\ </math>
-        <math> P(Y>i) = 1 - P(Y \le i) </math>
-        <math> P(Y \le i) = \sum_{i=1}^\infty \frac {1}{2^i} = 1 + \frac {1}{2^i} </math>
-      <math> \therefore P(Y>i) = \frac {1}{2^i} </math>
-:Plugging this result back into the original expression yields
-        <math> P(Y<X) = \sum_{i=1}^\infty \frac {1}{4^i} = \frac {1}{3} </math>
-*To find <math> P(Y=kX)\ </math>, note that X and Y are iid and summing over all possible combinations one arrives at
-        <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>
-:Thus,
-        <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>
 ----
 ==Solution 2==

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ECE Ph.D. Qualifying Exam: Automatic Control (AC)- Question 1, August 2007

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