(New page: Category: ECE Category: HKN Category: QE Category: Automatic Controls Category: Linear Systems =Problem 1= X and Y are iid <math> P(X=i) = P(Y=i) = \frac {1}{2^i}...) |
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==Part A== | ==Part A== | ||
| − | + | *Find <math> P(min(X,Y)=k)\ </math> | |
Let <math> Z = min(X,Y)\ </math> | Let <math> Z = min(X,Y)\ </math> | ||
| − | Then finding the pmf of Z uses the fact that X and Y are iid | + | 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> | |
| − | + | <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> | |
| + | |||
| + | ==Part B== | ||
| + | *Find <math> P(X=Y)\ </math> | ||
| + | |||
| + | Noting 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> | ||
| + | |||
| + | ==Part C== | ||
| + | *Find <math> P(Y>X)\ </math> | ||
| + | |||
| + | Again noting 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> | ||
Revision as of 10:20, 3 December 2008
Contents
Problem 1
X and Y are iid
$ P(X=i) = P(Y=i) = \frac {1}{2^i}\ ,i = 1,2,3,... $
Part A
- Find $ P(min(X,Y)=k)\ $
Let $ Z = min(X,Y)\ $
Then finding the pmf of Z uses the fact that X and Y are iid
$ P(Z=k) = P(X \ge k,Y \ge k) = P(X \ge k)P(Y \ge k) = P(X \ge k)^2 $
$ 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} $
Part B
- Find $ P(X=Y)\ $
Noting that X and Y are iid and summing across all possible i,
$ 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} $
Part C
- Find $ P(Y>X)\ $
Again noting that X and Y are iid and summing across all possible i,
$ P(Y>X) = \sum_{i=1}^\infty P(Y>i, X=i) = \sum_{i=1}^\infty P(Y>i)P(x=i) $
Next, find $ P(Y<i)\ $ $ P(Y>i) = 1 - P(Y \le i) $
$ P(Y \le i) = \sum_{i=1}^\infty \frac {1}{2^i} = 1 + \frac {1}{2^i} $
$ \therefore P(Y>i) = \frac {1}{2^i} $
