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==Question from [[ECE_PhD_QE_CNSIP_Jan_2001_Problem1|ECE QE January 2001]]== | ==Question from [[ECE_PhD_QE_CNSIP_Jan_2001_Problem1|ECE QE January 2001]]== | ||
− | + | '''(a) (7 pts)''' | |
+ | |||
+ | Let <math class="inline">A</math> and <math class="inline">B</math> be statistically independent events in the same probability space. Are <math class="inline">A</math> and <math class="inline">B^{C}</math> independent? (You must prove your result). | ||
+ | |||
+ | '''(b) (7 pts)''' | ||
+ | |||
+ | Can two events be statistically independent and mutually exclusive? (You must derive the conditions on A and B for this to be true or not.) | ||
+ | |||
+ | ''(c) (6 pts)''' | ||
+ | |||
+ | State the Axioms of Probability. | ||
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==Share and discuss your solutions below.== | ==Share and discuss your solutions below.== | ||
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=Solution 1 (retrived from [[ECE600_QE_2000_August|here]])= | =Solution 1 (retrived from [[ECE600_QE_2000_August|here]])= | ||
− | + | '''(a) ''' | |
+ | |||
+ | <math class="inline">P\left(A\right)=P\left(A\cap\left(B\cup B^{C}\right)\right)=P\left(\left(A\cap B\right)\cup\left(A\cap B^{C}\right)\right)=P\left(A\cap B\right)+P\left(A\cap B^{C}\right)=P\left(A\right)P\left(B\right)+P\left(A\cap B^{C}\right).</math> | ||
+ | |||
+ | <math class="inline">P\left(A\cap B^{C}\right)=P\left(A\right)-P\left(A\right)P\left(B\right)=P\left(A\right)\left(1-P\left(B\right)\right)=P\left(A\right)P\left(B^{C}\right).</math> | ||
+ | |||
+ | <math class="inline">\therefore A\text{ and }B^{C}\text{ are independent. }</math> | ||
+ | |||
+ | '''(b)''' | ||
+ | |||
+ | If <math class="inline">P\left(A\right)=0</math> or <math class="inline">P\left(B\right)=0</math> , then A and B are statistically independent and mutually exclusive. Prove this: | ||
+ | |||
+ | Without loss of generality, suppose that <math class="inline">P\left(A\right)=0</math> . <math class="inline">0=P\left(A\right)\geq P\left(A\cap B\right)\geq0\Longrightarrow P\left(A\cap B\right)=0\qquad\therefore\text{mutually excclusive}.</math> | ||
+ | |||
+ | <math class="inline">P\left(A\cap B\right)=0=P\left(A\right)P\left(B\right)\qquad\therefore\text{statistically independent.}</math> | ||
+ | |||
+ | '''(c) ''' | ||
+ | |||
+ | '''Axioms of probability'''= | ||
+ | |||
+ | • The probability measure <math class="inline">P\left(\cdot\right)</math> corresponding to <math class="inline">S</math> and <math class="inline">F\left(S\right)</math> is the assignment of a real number <math class="inline">P\left(A\right)</math> to each <math class="inline">A\in F\left(S\right)</math> satisfying following properties. Axioms of probability: | ||
+ | |||
+ | 1. <math class="inline">P\left(A\right)\geq0</math> , <math class="inline">\forall A\in F\left(S\right)</math> . | ||
+ | |||
+ | 2. <math class="inline">P\left(S\right)=1</math> . | ||
+ | |||
+ | 3. If <math class="inline">A_{1}</math> and <math class="inline">A_{2}</math> are disjoint events, then <math class="inline">P\left(A_{1}\cup A_{2}\right)=P\left(A_{1}\right)+P\left(A_{2}\right)</math> . If <math class="inline">A_{1},A_{2}\in F\left(S\right)</math> and <math class="inline">A_{1}\cap A_{2}=\varnothing</math> , then <math class="inline">A_{1}</math> and <math class="inline">A_{2}</math> are disjoint events. | ||
+ | |||
+ | 4. If <math class="inline">A_{1},A_{2},\cdots,A_{n},\cdots\in F\left(S\right)</math> is a countable collection of disjointed events, then <math class="inline">P\left(\bigcup_{i=1}^{\infty}A_{i}\right)=\sum_{i=1}^{\infty}P\left(A_{i}\right)</math> . | ||
+ | |||
+ | • <math class="inline">P\left(\cdot\right)</math> is a set function. <math class="inline">P\left(\cdot\right):F\left(S\right)\rightarrow\mathbf{R}</math> . | ||
+ | |||
+ | • If you want to talk about the probability of a single output <math class="inline">\omega_{0}\in S</math> , you do so by considering the single event | ||
+ | |||
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==Solution 2== | ==Solution 2== |
Revision as of 05:42, 17 July 2012
Contents
Question from ECE QE January 2001
(a) (7 pts)
Let $ A $ and $ B $ be statistically independent events in the same probability space. Are $ A $ and $ B^{C} $ independent? (You must prove your result).
(b) (7 pts)
Can two events be statistically independent and mutually exclusive? (You must derive the conditions on A and B for this to be true or not.)
(c) (6 pts)'
State the Axioms of Probability.
Solution 1 (retrived from here)
(a)
$ P\left(A\right)=P\left(A\cap\left(B\cup B^{C}\right)\right)=P\left(\left(A\cap B\right)\cup\left(A\cap B^{C}\right)\right)=P\left(A\cap B\right)+P\left(A\cap B^{C}\right)=P\left(A\right)P\left(B\right)+P\left(A\cap B^{C}\right). $
$ P\left(A\cap B^{C}\right)=P\left(A\right)-P\left(A\right)P\left(B\right)=P\left(A\right)\left(1-P\left(B\right)\right)=P\left(A\right)P\left(B^{C}\right). $
$ \therefore A\text{ and }B^{C}\text{ are independent. } $
(b)
If $ P\left(A\right)=0 $ or $ P\left(B\right)=0 $ , then A and B are statistically independent and mutually exclusive. Prove this:
Without loss of generality, suppose that $ P\left(A\right)=0 $ . $ 0=P\left(A\right)\geq P\left(A\cap B\right)\geq0\Longrightarrow P\left(A\cap B\right)=0\qquad\therefore\text{mutually excclusive}. $
$ P\left(A\cap B\right)=0=P\left(A\right)P\left(B\right)\qquad\therefore\text{statistically independent.} $
(c)
Axioms of probability=
• The probability measure $ P\left(\cdot\right) $ corresponding to $ S $ and $ F\left(S\right) $ is the assignment of a real number $ P\left(A\right) $ to each $ A\in F\left(S\right) $ satisfying following properties. Axioms of probability:
1. $ P\left(A\right)\geq0 $ , $ \forall A\in F\left(S\right) $ .
2. $ P\left(S\right)=1 $ .
3. If $ A_{1} $ and $ A_{2} $ are disjoint events, then $ P\left(A_{1}\cup A_{2}\right)=P\left(A_{1}\right)+P\left(A_{2}\right) $ . If $ A_{1},A_{2}\in F\left(S\right) $ and $ A_{1}\cap A_{2}=\varnothing $ , then $ A_{1} $ and $ A_{2} $ are disjoint events.
4. If $ A_{1},A_{2},\cdots,A_{n},\cdots\in F\left(S\right) $ is a countable collection of disjointed events, then $ P\left(\bigcup_{i=1}^{\infty}A_{i}\right)=\sum_{i=1}^{\infty}P\left(A_{i}\right) $ .
• $ P\left(\cdot\right) $ is a set function. $ P\left(\cdot\right):F\left(S\right)\rightarrow\mathbf{R} $ .
• If you want to talk about the probability of a single output $ \omega_{0}\in S $ , you do so by considering the single event
Solution 2
Write it here.