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</center> | </center> | ||
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| − | ===Solution=== | + | ===Solution 1=== |
<math> | <math> | ||
P((Z(t)=0) = P(Z(0)=0, N(t)=Even) + P(Z(0)=1, N(t)=Odd)\\ | P((Z(t)=0) = P(Z(0)=0, N(t)=Even) + P(Z(0)=1, N(t)=Odd)\\ | ||
| − | = pP( N(t)=Even) + (1-p)P( N(t)=Odd) | + | = pP( N(t)=Even) + (1-p)P( N(t)=Odd)\\ |
| − | =p\sum_{m= | + | =p\sum_{m=0,1, 2, ...}P(N(t) = 2m)+ (1-p)\sum_{n=0,1,2,...}P(N(t)=2n-1)\\ |
| + | =p\sum_{m=0,1,2,...}\frac{1}{1+\lambda t}(\frac{\lambda t}{1+\lambda t})^2m + (1-p)\sum_{n=0,1,2,...}\frac{1}{1+\lambda t}(\frac{\lambda t}{1+\lambda t})^{2n-1}\\ | ||
| + | =p\cdot\frac{1}{1+\lambda t}\cdot \frac{1}{1-(\frac{\lambda t}{1+\lambda t})^2} + (1-p)\cdot\frac{\lambda t}{1+\lambda t}\cdot\frac{1}{1+\lambda t}\cdot \frac{1}{1-(\frac{\lambda t}{1+\lambda t})^2}\\ | ||
| + | =\frac{p+\lambda t}{1+2\lambda t} | ||
</math> | </math> | ||
| + | |||
| + | <math> | ||
| + | P((Z(t)=1) = 1 - P((Z(t)=0) = \frac{1+\lambda t - p}{1+2\lambda t} | ||
| + | </math> | ||
| + | |||
| + | ===Solution 2=== | ||
| + | <math> | ||
| + | P(Z(t)=0)=P(Z(t)=0|N(t)=even)P(N(t)=even)+P(Z(t)=0|N(t)=odd)P(N(t)=odd) | ||
| + | </math> | ||
| + | |||
| + | Note that <math>\{Z(t)=0|N(t)=odd\}=\{Z(0)=1\} </math> and <math>\{Z(t)=0|N(t)=even\}=\{Z(0)=0\} </math>, therefore, | ||
| + | |||
| + | <math> | ||
| + | P(Z(t)=0)=P(Z(0)=0)P(N(t)=even)+P(Z(0)=1)P(N(t)=odd)\\ | ||
| + | =p\sum_{k=0}^{\infty}\frac{1}{1+\lambda t}\cdot (\frac{\lambda t}{1+\lambda t})^{2k}+(1-p)\sum_{k=0}^{\infty}\frac{1}{1+\lambda t}\cdot (\frac{\lambda t}{1+\lambda t})^{2k+1}\\ | ||
| + | =\frac{p}{1+\lambda t}\cdot \frac{1}{1-(\frac{\lambda t}{1+\lambda t})^2}+ \frac{(1-p)\lambda t}{(1+\lambda t)^2}\cdot \frac{1}{1-(\frac{\lambda t}{1+\lambda t})^2}\\ | ||
| + | =\frac{p+\lambda t}{1+2\lambda t}\\ | ||
| + | P(Z(t)=1) = 1- P(Z(t)=0) = 1-\frac{p+\lambda t}{1+2\lambda t} = \frac{1-p+\lambda t}{1+2\lambda t}\\ | ||
| + | P(Z(t)=k)=\left\{ | ||
| + | \begin{array}{cc} | ||
| + | \frac{p+\lambda t}{1+2\lambda t}, k =0 \\ | ||
| + | \frac{1-p+\lambda t}{1+2\lambda t}, k =1\\ | ||
| + | 0, else | ||
| + | \end{array} | ||
| + | \right. | ||
| + | </math> | ||
| + | |||
| + | <span style="color:green"> The solution is correct. However the else case is not necessary. K can only be 0 or 1. </span> | ||
| + | |||
| + | ===Solution 3=== | ||
| + | |||
| + | We know that <math>Z(t)</math> can only take on the values 0 and 1, so we set out to find the probability that <math>Z(t)</math> = 0; if we subtract this probability from 1, we will have found the probability that <math>Z(t) = 1</math>, and thus we will have described the entire pmf. We also know that if <math>Z(0)</math> = 0, <math>Z(t)</math> must be equal to 0 if <math>N(t)</math> is even (i.e., if <math>k</math> is even). Similarly, if <math>Z(0)\neq 0, Z(t)</math> must be equal to 0 if <math>k</math> is odd. As such, we can write the expression | ||
| + | |||
| + | <math> | ||
| + | P(Z(t) = 0) = P(Z(0) = 0, \,N(t)\,\,is\,\,even) + P(Z(0) = 1, \,N(t)\,\,is\,\,odd) = P(Z(0) = 0)P(N(t)\,\,is\,\,even) + P(Z(0) = 1)P(N(t)\,\,is\,\,odd) \\= p\cdot\sum_{i=0}^\infty\frac{1}{1+\lambda t}\left(\frac{\lambda t}{1+\lambda t}\right)^{2i} + (1-p)\cdot\sum_{j=0}^\infty\frac{1}{1+\lambda t}\left(\frac{\lambda t}{1+\lambda t}\right)^{2j+1}. | ||
| + | </math> | ||
| + | |||
| + | We now recall that the sum of an infinite geometric series can be expressed as | ||
| + | |||
| + | <math> | ||
| + | \sum_{k = 0}^\infty ar^k = \frac{a}{1-r}. | ||
| + | </math> | ||
| + | |||
| + | We can use this to simplify the preceding equation: | ||
| + | |||
| + | <math> | ||
| + | P(Z(t) = 0) = p\cdot\sum_{i=0}^\infty\frac{1}{1+\lambda t}\left(\frac{\lambda t}{1+\lambda t}\right)^{2i} + (1-p)\cdot\sum_{j=0}^\infty\frac{1}{1+\lambda t}\left(\frac{\lambda t}{1+\lambda t}\right)^{2j+1} \\ | ||
| + | = p\cdot\sum_{i=0}^\infty\frac{1}{1+\lambda t}\left(\frac{\lambda t}{1+\lambda t}\right)^{2i} + (1-p)\cdot\frac{\lambda t}{1 + \lambda t}\cdot\sum_{j=0}^\infty\frac{1}{1+\lambda t}\left(\frac{\lambda t}{1+\lambda t}\right)^{2j} \\ | ||
| + | = p\cdot\frac{1}{1+\lambda t}\cdot\frac{1}{1-\left(\frac{\lambda t}{1 + \lambda t}\right)^2} + (1-p)\cdot\frac{1}{1+\lambda t}\cdot\frac{\lambda t}{1 + \lambda t}\cdot\frac{1}{1-\left(\frac{\lambda t}{1 + \lambda t}\right)^2}. | ||
| + | </math> | ||
| + | |||
| + | We first combine terms: | ||
| + | |||
| + | <math> | ||
| + | P(Z(t) = 0) =p\cdot\frac{1}{1+\lambda t}\cdot\frac{1}{1-\left(\frac{\lambda t}{1 + \lambda t}\right)^2} + (1-p)\cdot\frac{1}{1+\lambda t}\cdot\frac{\lambda t}{1 + \lambda t}\cdot\frac{1}{1-\left(\frac{\lambda t}{1 + \lambda t}\right)^2} \\ | ||
| + | = \frac{p}{(1+\lambda t)\left(1-\left(\frac{\lambda t}{1+\lambda t}\right)^2\right)} + \frac{\lambda t}{(1+\lambda t)^2\left(1-\left(\frac{\lambda t}{1+\lambda t}\right)^2\right)} - \frac{p\lambda t}{(1+\lambda t)^2\left(1-\left(\frac{\lambda t}{1+\lambda t}\right)^2\right)} \\ | ||
| + | = \frac{p + p\lambda t}{(1+\lambda t)^2\left(1-\left(\frac{\lambda t}{1+\lambda t}\right)^2\right)} + \frac{\lambda t}{(1+\lambda t)^2\left(1-\left(\frac{\lambda t}{1+\lambda t}\right)^2\right)} - \frac{p\lambda t}{(1+\lambda t)^2\left(1-\left(\frac{\lambda t}{1+\lambda t}\right)^2\right)} \\ | ||
| + | = \frac{p + \lambda t}{(1+\lambda t)^2\left(1-\left(\frac{\lambda t}{1+\lambda t}\right)^2\right)}. | ||
| + | </math> | ||
| + | |||
| + | Then we simplify the denominator: | ||
| + | |||
| + | <math> | ||
| + | P(Z(t) = 0) = \frac{p + \lambda t}{(1+\lambda t)^2\left(1-\left(\frac{\lambda t}{1+\lambda t}\right)^2\right)} \\ | ||
| + | = \frac{p + \lambda t}{(1+\lambda t)^2 - (\lambda t)^2} \\ | ||
| + | = \frac{p + \lambda t}{1 + 2\lambda t}. | ||
| + | </math> | ||
| + | |||
| + | Now that we have found <math>P(Z(t) = 0)</math>, we can easily find <math>P(Z(t) = 1)</math> by subtracting our result from 1: | ||
| + | |||
| + | <math> | ||
| + | P(Z(t) = 1) = 1 - P(Z(t) = 0) \\ | ||
| + | = 1 - \frac{p + \lambda t}{1 + 2\lambda t}\\ | ||
| + | = \frac{1+2\lambda t}{1+ 2\lambda t} - \frac{p + \lambda t}{1 + 2\lambda t} \\ | ||
| + | = \frac{1 + \lambda t-p}{1 + 2\lambda t}. | ||
| + | </math> | ||
| + | |||
| + | ===Similar Problem=== | ||
| + | |||
| + | Find the mean function <math>\mu(t)</math> and covariance function <math>C_{zz}(t_1,t_2)</math> of the process <math>Z(t)</math>. | ||
---- | ---- | ||
[[ECE-QE_CS1-2015|Back to QE CS question 1, August 2015]] | [[ECE-QE_CS1-2015|Back to QE CS question 1, August 2015]] | ||
[[ECE_PhD_Qualifying_Exams|Back to ECE Qualifying Exams (QE) page]] | [[ECE_PhD_Qualifying_Exams|Back to ECE Qualifying Exams (QE) page]] | ||
Latest revision as of 23:57, 31 January 2016
Communication, Networking, Signal and Image Processing (CS)
Question 1: Probability and Random Processes
August 2015
Solution 1
$ P((Z(t)=0) = P(Z(0)=0, N(t)=Even) + P(Z(0)=1, N(t)=Odd)\\ = pP( N(t)=Even) + (1-p)P( N(t)=Odd)\\ =p\sum_{m=0,1, 2, ...}P(N(t) = 2m)+ (1-p)\sum_{n=0,1,2,...}P(N(t)=2n-1)\\ =p\sum_{m=0,1,2,...}\frac{1}{1+\lambda t}(\frac{\lambda t}{1+\lambda t})^2m + (1-p)\sum_{n=0,1,2,...}\frac{1}{1+\lambda t}(\frac{\lambda t}{1+\lambda t})^{2n-1}\\ =p\cdot\frac{1}{1+\lambda t}\cdot \frac{1}{1-(\frac{\lambda t}{1+\lambda t})^2} + (1-p)\cdot\frac{\lambda t}{1+\lambda t}\cdot\frac{1}{1+\lambda t}\cdot \frac{1}{1-(\frac{\lambda t}{1+\lambda t})^2}\\ =\frac{p+\lambda t}{1+2\lambda t} $
$ P((Z(t)=1) = 1 - P((Z(t)=0) = \frac{1+\lambda t - p}{1+2\lambda t} $
Solution 2
$ P(Z(t)=0)=P(Z(t)=0|N(t)=even)P(N(t)=even)+P(Z(t)=0|N(t)=odd)P(N(t)=odd) $
Note that $ \{Z(t)=0|N(t)=odd\}=\{Z(0)=1\} $ and $ \{Z(t)=0|N(t)=even\}=\{Z(0)=0\} $, therefore,
$ P(Z(t)=0)=P(Z(0)=0)P(N(t)=even)+P(Z(0)=1)P(N(t)=odd)\\ =p\sum_{k=0}^{\infty}\frac{1}{1+\lambda t}\cdot (\frac{\lambda t}{1+\lambda t})^{2k}+(1-p)\sum_{k=0}^{\infty}\frac{1}{1+\lambda t}\cdot (\frac{\lambda t}{1+\lambda t})^{2k+1}\\ =\frac{p}{1+\lambda t}\cdot \frac{1}{1-(\frac{\lambda t}{1+\lambda t})^2}+ \frac{(1-p)\lambda t}{(1+\lambda t)^2}\cdot \frac{1}{1-(\frac{\lambda t}{1+\lambda t})^2}\\ =\frac{p+\lambda t}{1+2\lambda t}\\ P(Z(t)=1) = 1- P(Z(t)=0) = 1-\frac{p+\lambda t}{1+2\lambda t} = \frac{1-p+\lambda t}{1+2\lambda t}\\ P(Z(t)=k)=\left\{ \begin{array}{cc} \frac{p+\lambda t}{1+2\lambda t}, k =0 \\ \frac{1-p+\lambda t}{1+2\lambda t}, k =1\\ 0, else \end{array} \right. $
The solution is correct. However the else case is not necessary. K can only be 0 or 1.
Solution 3
We know that $ Z(t) $ can only take on the values 0 and 1, so we set out to find the probability that $ Z(t) $ = 0; if we subtract this probability from 1, we will have found the probability that $ Z(t) = 1 $, and thus we will have described the entire pmf. We also know that if $ Z(0) $ = 0, $ Z(t) $ must be equal to 0 if $ N(t) $ is even (i.e., if $ k $ is even). Similarly, if $ Z(0)\neq 0, Z(t) $ must be equal to 0 if $ k $ is odd. As such, we can write the expression
$ P(Z(t) = 0) = P(Z(0) = 0, \,N(t)\,\,is\,\,even) + P(Z(0) = 1, \,N(t)\,\,is\,\,odd) = P(Z(0) = 0)P(N(t)\,\,is\,\,even) + P(Z(0) = 1)P(N(t)\,\,is\,\,odd) \\= p\cdot\sum_{i=0}^\infty\frac{1}{1+\lambda t}\left(\frac{\lambda t}{1+\lambda t}\right)^{2i} + (1-p)\cdot\sum_{j=0}^\infty\frac{1}{1+\lambda t}\left(\frac{\lambda t}{1+\lambda t}\right)^{2j+1}. $
We now recall that the sum of an infinite geometric series can be expressed as
$ \sum_{k = 0}^\infty ar^k = \frac{a}{1-r}. $
We can use this to simplify the preceding equation:
$ P(Z(t) = 0) = p\cdot\sum_{i=0}^\infty\frac{1}{1+\lambda t}\left(\frac{\lambda t}{1+\lambda t}\right)^{2i} + (1-p)\cdot\sum_{j=0}^\infty\frac{1}{1+\lambda t}\left(\frac{\lambda t}{1+\lambda t}\right)^{2j+1} \\ = p\cdot\sum_{i=0}^\infty\frac{1}{1+\lambda t}\left(\frac{\lambda t}{1+\lambda t}\right)^{2i} + (1-p)\cdot\frac{\lambda t}{1 + \lambda t}\cdot\sum_{j=0}^\infty\frac{1}{1+\lambda t}\left(\frac{\lambda t}{1+\lambda t}\right)^{2j} \\ = p\cdot\frac{1}{1+\lambda t}\cdot\frac{1}{1-\left(\frac{\lambda t}{1 + \lambda t}\right)^2} + (1-p)\cdot\frac{1}{1+\lambda t}\cdot\frac{\lambda t}{1 + \lambda t}\cdot\frac{1}{1-\left(\frac{\lambda t}{1 + \lambda t}\right)^2}. $
We first combine terms:
$ P(Z(t) = 0) =p\cdot\frac{1}{1+\lambda t}\cdot\frac{1}{1-\left(\frac{\lambda t}{1 + \lambda t}\right)^2} + (1-p)\cdot\frac{1}{1+\lambda t}\cdot\frac{\lambda t}{1 + \lambda t}\cdot\frac{1}{1-\left(\frac{\lambda t}{1 + \lambda t}\right)^2} \\ = \frac{p}{(1+\lambda t)\left(1-\left(\frac{\lambda t}{1+\lambda t}\right)^2\right)} + \frac{\lambda t}{(1+\lambda t)^2\left(1-\left(\frac{\lambda t}{1+\lambda t}\right)^2\right)} - \frac{p\lambda t}{(1+\lambda t)^2\left(1-\left(\frac{\lambda t}{1+\lambda t}\right)^2\right)} \\ = \frac{p + p\lambda t}{(1+\lambda t)^2\left(1-\left(\frac{\lambda t}{1+\lambda t}\right)^2\right)} + \frac{\lambda t}{(1+\lambda t)^2\left(1-\left(\frac{\lambda t}{1+\lambda t}\right)^2\right)} - \frac{p\lambda t}{(1+\lambda t)^2\left(1-\left(\frac{\lambda t}{1+\lambda t}\right)^2\right)} \\ = \frac{p + \lambda t}{(1+\lambda t)^2\left(1-\left(\frac{\lambda t}{1+\lambda t}\right)^2\right)}. $
Then we simplify the denominator:
$ P(Z(t) = 0) = \frac{p + \lambda t}{(1+\lambda t)^2\left(1-\left(\frac{\lambda t}{1+\lambda t}\right)^2\right)} \\ = \frac{p + \lambda t}{(1+\lambda t)^2 - (\lambda t)^2} \\ = \frac{p + \lambda t}{1 + 2\lambda t}. $
Now that we have found $ P(Z(t) = 0) $, we can easily find $ P(Z(t) = 1) $ by subtracting our result from 1:
$ P(Z(t) = 1) = 1 - P(Z(t) = 0) \\ = 1 - \frac{p + \lambda t}{1 + 2\lambda t}\\ = \frac{1+2\lambda t}{1+ 2\lambda t} - \frac{p + \lambda t}{1 + 2\lambda t} \\ = \frac{1 + \lambda t-p}{1 + 2\lambda t}. $
Similar Problem
Find the mean function $ \mu(t) $ and covariance function $ C_{zz}(t_1,t_2) $ of the process $ Z(t) $.
