(Created page with "Category:ECE Category:QE Category:CNSIP Category:problem solving Category:random variables Category:probability <center> <font size= 4> ECE_PhD_Qual...") |
|||
| Line 57: | Line 57: | ||
4. If <math class="inline">A_{1},A_{2},\cdots,A_{n},\cdots\in\mathcal{F}\left(\mathcal{S}\right)</math> is countable collection of disjoint events, then <math class="inline">P\left(\cup_{i=0}^{\infty}A_{i}\right)=\alpha_{1}P_{1}\left(\cup_{i=0}^{\infty}A_{i}\right)+\alpha_{2}P_{2}\left(\cup_{i=0}^{\infty}A_{i}\right)=\alpha_{1}\sum_{i=1}^{\infty}P_{1}\left(A_{i}\right)+\alpha_{2}\sum_{i=1}^{\infty}P_{2}\left(A_{i}\right)</math><math class="inline">=\sum_{i=1}^{\infty}\left\{ \alpha_{1}P_{1}\left(A_{i}\right)+\alpha_{2}P_{2}\left(A_{i}\right)\right\} =\sum_{i=1}^{\infty}P\left(A_{i}\right).</math> | 4. If <math class="inline">A_{1},A_{2},\cdots,A_{n},\cdots\in\mathcal{F}\left(\mathcal{S}\right)</math> is countable collection of disjoint events, then <math class="inline">P\left(\cup_{i=0}^{\infty}A_{i}\right)=\alpha_{1}P_{1}\left(\cup_{i=0}^{\infty}A_{i}\right)+\alpha_{2}P_{2}\left(\cup_{i=0}^{\infty}A_{i}\right)=\alpha_{1}\sum_{i=1}^{\infty}P_{1}\left(A_{i}\right)+\alpha_{2}\sum_{i=1}^{\infty}P_{2}\left(A_{i}\right)</math><math class="inline">=\sum_{i=1}^{\infty}\left\{ \alpha_{1}P_{1}\left(A_{i}\right)+\alpha_{2}P_{2}\left(A_{i}\right)\right\} =\sum_{i=1}^{\infty}P\left(A_{i}\right).</math> | ||
| + | |||
| + | ---- | ||
Latest revision as of 00:03, 10 March 2015
Communication, Networking, Signal and Image Processing (CS)
Question 1: Probability and Random Processes
January 2004
1. (30 pts.)
This question consists of two separate short questions relating to the structure of probability space:
(a)
Assume that $ \mathcal{S} $ is the sample space of a random experiment and that $ \mathcal{F}_{1} $ and $ \mathcal{F}_{2} $ are $ \sigma $ -fields (valid event spaces) on $ \mathcal{S} $ . Prove that $ \mathcal{F}_{1}\cap\mathcal{F}_{2} $ is also a $ \sigma $ -field on $ S $ .
(b)
Consider a sample space $ \mathcal{S} $ and corresponding event space $ \mathcal{F} $ . Suppose that $ P_{1} $ and $ P_{2} $ are both balid probability measures defined on $ \mathcal{F} $ . Prove that $ P $ defined by $ P\left(A\right)=\alpha_{1}P_{1}\left(A\right)+\alpha_{2}P_{2}\left(A\right),\qquad\forall A\in\mathcal{F} $ is also a valid probability measure on $ \mathcal{F} $ if $ \alpha_{1},\;\alpha_{2}\geq0 $ and $ \alpha_{1}+\alpha_{2}=1 $ .
Answer
• Because $ P_{1} $ and $ P_{2} $ are valid probability measures, we know that they satisfy the axioms of probability:
1. $ P_{1}\left(A\right)\geq0 $ and $ P_{2}\left(A\right)\geq0 $ , $ \forall A\in\mathcal{F}\left(\mathcal{S}\right) $ .
2. $ P_{1}\left(\mathcal{S}\right)=1 $ and $ P_{2}\left(\mathcal{S}\right)=1 $ .
3. If $ A_{1} $ and $ A_{2}\in\mathcal{F}\left(\mathcal{S}\right) $ are disjoint events, then $ P_{1}\left(A_{1}\cup A_{2}\right)=P_{1}\left(A_{1}\right)+P_{1}\left(A_{2}\right) $ and $ P_{2}\left(A_{1}\cup A_{2}\right)=P_{2}\left(A_{1}\right)+P_{2}\left(A_{2}\right) $ .
4. If $ A_{1},A_{2},\cdots,A_{n},\cdots\in\mathcal{F}\left(\mathcal{S}\right) $ is countable collection of disjoint events, then $ P_{1}\left(\cup_{i=1}^{\infty}A_{i}\right)=\sum_{i=1}^{\infty}P_{1}\left(A_{i}\right) $ and $ P_{2}\left(\cup_{i=1}^{\infty}A_{i}\right)=\sum_{i=1}^{\infty}P_{2}\left(A_{i}\right) $ .
• Now, we check each condition to become a valid probability measure:
1. $ P\left(A\right)=\alpha_{1}P_{1}\left(A\right)+\alpha_{2}P_{2}\left(A\right)\geq0 , \forall A\in\mathcal{F}\left(\mathcal{S}\right) $ .
– $ \because\alpha_{1}\geq0,\;\alpha_{2}\geq0,\; P_{1}\left(A\right)\geq0,\text{ and }P_{2}\left(A\right)\geq0 $ .
2. $ P\left(S\right)=\alpha_{1}P_{1}\left(A\right)+\alpha_{2}P_{2}\left(A\right)=\alpha_{1}+\alpha_{2}=1 $ .
3. If $ A_{1} $ and $ A_{2}\in\mathcal{F}\left(\mathcal{S}\right) $ are disjoint events, then $ P\left(A_{1}\cup A_{2}\right)=\alpha_{1}P_{1}\left(A_{1}\cup A_{2}\right)+\alpha_{2}P_{2}\left(A_{1}\cup A_{2}\right)=\alpha_{1}\left\{ P_{1}\left(A_{1}\right)+P_{1}\left(A_{2}\right)\right\} +\alpha_{2}\left\{ P_{2}\left(A_{1}\right)+P_{2}\left(A_{2}\right)\right\} $$ =\alpha_{1}P_{1}\left(A_{1}\right)+\alpha_{2}P_{2}\left(A_{1}\right)+\alpha_{1}P_{1}\left(A_{2}\right)+\alpha_{2}P_{2}\left(A_{2}\right)=P\left(A_{1}\right)+P\left(A_{2}\right). $
4. If $ A_{1},A_{2},\cdots,A_{n},\cdots\in\mathcal{F}\left(\mathcal{S}\right) $ is countable collection of disjoint events, then $ P\left(\cup_{i=0}^{\infty}A_{i}\right)=\alpha_{1}P_{1}\left(\cup_{i=0}^{\infty}A_{i}\right)+\alpha_{2}P_{2}\left(\cup_{i=0}^{\infty}A_{i}\right)=\alpha_{1}\sum_{i=1}^{\infty}P_{1}\left(A_{i}\right)+\alpha_{2}\sum_{i=1}^{\infty}P_{2}\left(A_{i}\right) $$ =\sum_{i=1}^{\infty}\left\{ \alpha_{1}P_{1}\left(A_{i}\right)+\alpha_{2}P_{2}\left(A_{i}\right)\right\} =\sum_{i=1}^{\infty}P\left(A_{i}\right). $
