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===== <math>\color{blue}\text{Solution 2:}</math> ===== | ===== <math>\color{blue}\text{Solution 2:}</math> ===== | ||
| + | '''(a)''' | ||
| + | <br> | ||
| + | Since <math class="inline">\mathbf{Y}_{n} = min \,\{{ \mathbf{X}_1, \mathbf{X}_2, \dots \mathbf{X}_n} \},</math> | ||
| + | <br> | ||
| + | We can have the CDF of <math class="inline">\mathbf{Y}_{n}</math> | ||
| + | <br> | ||
| + | <math> P(\mathbf{Y}_{n} \leq y) = P(\,\{{ \mathbf{X}_1, \mathbf{X}_2, \dots \mathbf{X}_n} \} \leq y) </math> <br> | ||
| + | <math> = 1-P(\,\{{ \mathbf{X}_1, \mathbf{X}_2, \dots \mathbf{X}_n} \} > y)</math><br> | ||
| + | <math> = 1-P(\mathbf{X}_1 > y)P(\mathbf{X}_2 > y) \dots P(\mathbf{X}_n > y)</math><br> | ||
| + | <math> = 1-(1-F_{X_1}(y))(1-F_{X_2}(y)) \dots (1-F_{X_n}(y))=1-(1-F_{X_1}(y))^n</math><br> | ||
| + | |||
| + | Since <math class="inline">\mathbf{Y}_{n} </math> is also uniform distributed on the interval [0,1] <br> | ||
| + | When <math class="inline"> y <0 </math>, <math class="inline">P(\{\mathbf{Y}_{n}<0 \}) =0 </math> since <math class="inline">F_{X_1}(y) =0 </math><br> | ||
Revision as of 18:41, 25 January 2014
Communication, Networking, Signal and Image Processing (CS)
Question 1: Probability and Random Processes
August 2012
Jump to Problem 2,3
Problem 3
$ \color{blue}\text{Solution 1:} $
$ \color{blue}\text{Solution 2:} $
(a)
Since $ \mathbf{Y}_{n} = min \,\{{ \mathbf{X}_1, \mathbf{X}_2, \dots \mathbf{X}_n} \}, $
We can have the CDF of $ \mathbf{Y}_{n} $
$ P(\mathbf{Y}_{n} \leq y) = P(\,\{{ \mathbf{X}_1, \mathbf{X}_2, \dots \mathbf{X}_n} \} \leq y) $
$ = 1-P(\,\{{ \mathbf{X}_1, \mathbf{X}_2, \dots \mathbf{X}_n} \} > y) $
$ = 1-P(\mathbf{X}_1 > y)P(\mathbf{X}_2 > y) \dots P(\mathbf{X}_n > y) $
$ = 1-(1-F_{X_1}(y))(1-F_{X_2}(y)) \dots (1-F_{X_n}(y))=1-(1-F_{X_1}(y))^n $
Since $ \mathbf{Y}_{n} $ is also uniform distributed on the interval [0,1]
When $ y <0 $, $ P(\{\mathbf{Y}_{n}<0 \}) =0 $ since $ F_{X_1}(y) =0 $
