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| + | ==7.14 QE 2008 August== | ||
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| + | '''1''' | ||
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| + | The Weak Law of Large Numbers states that if <math>\mathbf{X}_{1},\mathbf{X}_{2},\mathbf{X}_{3},\cdots</math> is a sequence of i.i.d. random variables with finite mean <math>E\left[\mathbf{X}_{i}\right]=\mu</math> for every <math>i</math> , then the sample mean <math>\mathbf{Y}_{n}=\frac{1}{n}\sum_{i=1}^{n}\mathbf{X}_{i}</math> converges to <math>\mu</math> in probability. Suppose that instead of being i.i.d , <math>\mathbf{X}_{1},\mathbf{X}_{2},\mathbf{X}_{3},\cdots</math> each have finite mean <math>\mu</math> , and the covariance function of the sequence <math>\mathbf{X}_{n}</math> is <math>Cov\left(\mathbf{X}_{i},\mathbf{X}_{j}\right)=\sigma^{2}\rho^{\left|i-j\right|}</math> , where <math>\left|\rho\right|<1</math> and <math>\sigma^{2}>0</math> . | ||
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| + | a. (13 points) | ||
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| + | Find the mean and variance of <math>\mathbf{Y}_{n}</math> . | ||
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| + | <math>E\left[\mathbf{Y}_{n}\right]=E\left[\frac{1}{n}\sum_{i=1}^{n}\mathbf{X}_{i}\right]=\frac{1}{n}\sum_{i=1}^{n}E\left[\mathbf{X}_{i}\right]=\frac{n\mu}{n}=\mu.</math> | ||
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| + | <math>Cov\left(\mathbf{X}_{i},\mathbf{X}_{j}\right)=E\left[\left(\mathbf{X}_{i}-\mu\right)\left(\mathbf{X}_{j}-\mu\right)\right]=E\left[\mathbf{X}_{i}\mathbf{X}_{j}\right]-\mu^{2}.</math> | ||
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| + | <math>E\left[\mathbf{X}_{i}\mathbf{X}_{j}\right]=\begin{cases} | ||
| + | \begin{array}{lll} | ||
| + | \sigma^{2}+\mu^{2} & & ,\; i=j\\ | ||
| + | \sigma^{2}\rho^{\left|i-j\right|}+\mu^{2} & & ,\; i\neq j. | ||
| + | \end{array}\end{cases}</math> | ||
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| + | <math>E\left[\mathbf{Y}_{n}^{2}\right]=E\left[\frac{1}{n^{2}}\sum_{i=1}^{n}\sum_{j=1}^{n}\mathbf{X}_{i}\mathbf{X}_{j}\right]=\frac{1}{n^{2}}\left[\sum_{i=1}^{n}E\left[\mathbf{X}_{i}^{2}\right]+\underset{_{i\neq j}}{\sum_{i=1}^{n}\sum_{j=1}^{n}}E\left[\mathbf{X}_{i}\mathbf{X}_{j}\right]\right]</math><math>=\frac{1}{n^{2}}\left[n\left(\sigma^{2}+\mu^{2}\right)+2\sum_{i=1}^{n-1}\left(n-i\right)\left(\sigma^{2}\rho^{i}+\mu^{2}\right)\right]=\frac{1}{n^{2}}\left[n\sigma^{2}+2\sigma^{2}\sum_{i=1}^{n-1}\left(n-i\right)\rho^{i}\right]+\mu^{2}.</math> | ||
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| + | <math>Var\left[\mathbf{Y}_{n}\right]=E\left[\mathbf{Y}_{n}^{2}\right]-E\left[\mathbf{Y}_{n}\right]^{2}=\frac{1}{n^{2}}\left[n\sigma^{2}+2\sigma^{2}\sum_{i=1}^{n-1}\left(n-i\right)\rho^{i}\right]+\mu^{2}-\mu^{2}=\frac{1}{n^{2}}\left[n\sigma^{2}+2\sigma^{2}\sum_{i=1}^{n-1}\left(n-i\right)\rho^{i}\right].</math> | ||
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| + | Note | ||
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| + | If the element <math>m_{ij}</math> of <math>n\times n</math> matrix is defined as <math>\left|i-j\right|</math> , then | ||
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| + | <math>\left[\begin{array}{ccccc} | ||
| + | 0 & 1 & 2 & \cdots & n-1\\ | ||
| + | 1 & 0 & 1 & \ddots & \vdots\\ | ||
| + | 2 & 1 & 0 & \ddots & 2\\ | ||
| + | \vdots & \ddots & \ddots & \ddots & 1\\ | ||
| + | n-1 & \cdots & 2 & 1 & 0 | ||
| + | \end{array}\right].</math> | ||
| + | |||
| + | b. (12 points) | ||
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| + | Does the sample mean still converge to <math>\mu</math> in probability? You must justify your answer. | ||
| + | |||
| + | ref. You can see the definition of the converge in probability [CS1ConvergeInProbability]. | ||
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| + | If <math>\mathbf{Y}_{n}\rightarrow\left(p\right)\rightarrow0</math> , then for any <math>\epsilon>0</math> , <math>P\left(\left\{ \left|\mathbf{Y}_{n}-\mu\right|>\epsilon\right\} \right)\rightarrow0</math> as <math>n\rightarrow\infty</math> . According to the Chebyshev inequality [CS1ChebyshevInequality], <math>P\left(\left\{ \left|\mathbf{Y}_{n}-\mu\right|>\epsilon\right\} \right)\leq\frac{\sigma_{\mathbf{Y}}^{2}}{\epsilon^{2}}.</math> <math>\lim_{n\rightarrow\infty}\sigma_{\mathbf{Y}}^{2}</math> Therefore, we know that <math>\lim_{n\rightarrow\infty}P\left(\left\{ \left|\mathbf{Y}_{n}-\mu\right|>\epsilon\right\} \right)=0</math> . Thus <math>\mathbf{Y}_{n}\rightarrow\left(p\right)\rightarrow0</math> . | ||
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| + | 2 | ||
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| + | Let <math>\mathbf{X}_{1},\mathbf{X}_{2},\mathbf{X}_{3},\cdots</math> be a sequence of i.i.d Bernoulli random variables with <math>p=1/2</math> , and let <math>\mathbf{Y}_{n}=2^{n}\mathbf{X}_{1}\mathbf{X}_{2}\cdots\mathbf{X}_{n}</math> . | ||
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| + | a. (15 points) | ||
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| + | Does the sequence <math>\mathbf{Y}_{n}</math> converge to <math>0</math> almost everywhere? | ||
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| + | • You can see the definition of the converge almost everywhere [CS1ConvergeAlmostEverywhere]. | ||
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| + | • range of <math>\mathbf{X}_{i}=\left\{ 0,1\right\}</math> . | ||
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| + | • range of <math>\mathbf{Y}_{n}=\left\{ 0,2^{n}\right\}</math> . | ||
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| + | • the probability of <math>\mathbf{Y}_{n}=2^{n}\Longrightarrow P\left(\mathbf{Y}_{n}=2^{n}\right)=P\left(\left\{ \mathbf{X}_{1}=1\right\} \cap\cdots\cap\left\{ \mathbf{X}_{n}=1\right\} \right)=\left(\frac{1}{2}\right)^{n}.</math> | ||
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| + | • <math>\lim_{n\rightarrow\infty}P\left(\mathbf{Y}_{n}=2^{n}\right)=0</math> . <math>\lim_{n\rightarrow\infty}P\left(\mathbf{Y}_{n}=0\right)=1</math> . | ||
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| + | • Thus, <math>\left\{ \mathbf{Y}_{n}\right\} \rightarrow0</math> with probability 1. | ||
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| + | b. (15 points) | ||
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| + | Does the sequence <math>\mathbf{Y}_{n}</math> converge to 0 in the mean square sense? | ||
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| + | Note | ||
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| + | You can see the definition of the converge in mean-square [CS1ConvergeInMean-Square]. | ||
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| + | <math>E\left[\mathbf{X}^{2}\right]=\sum_{k=0}^{1}k^{2}\left(\frac{1}{2}\right)=0^{2}\cdot\frac{1}{2}+1^{2}\cdot\frac{1}{2}=\frac{1}{2}.</math> | ||
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| + | <math>E\left[\left|\mathbf{Y}_{n}-0\right|^{2}\right]=E\left[\mathbf{Y}_{n}^{2}\right]=E\left[2^{2n}\mathbf{X}_{1}^{2}\mathbf{X}_{2}^{2}\cdots\mathbf{X}_{n}^{2}\right]=2^{2n}E\left[\mathbf{X}_{1}^{2}\mathbf{X}_{2}^{2}\cdots\mathbf{X}_{n}^{2}\right]=\left(4E\left[\mathbf{X}^{2}\right]\right)^{n}=2^{n}.</math> | ||
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| + | <math>\lim_{n\rightarrow\infty}E\left[\mathbf{Y}_{n}^{2}\right]=\lim_{n\rightarrow\infty}2^{n}=\infty.</math> | ||
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| + | Thus, <math>\mathbf{Y}_{n}</math> does not converge to <math>0</math> in the mean square sense. | ||
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| + | 3 | ||
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| + | Consider a random process <math>\mathbf{X}\left(t\right)</math> that assumes values <math>\pm1</math> . Suppose that <math>\mathbf{X}\left(0\right)=\pm1</math> with probability <math>1/2</math> , and suppose that <math>\mathbf{X}\left(t\right)</math> then changes polarity with each occurrence of an event in a Poisson process of rate <math>\lambda</math> . | ||
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| + | Note: | ||
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| + | You might find the equations <math>\frac{1}{2}\left(e^{x}+e^{-x}\right)=\sum_{j=0}^{\infty}\frac{x^{2j}}{\left(2j\right)!}</math> and <math>\frac{1}{2}\left(e^{x}-e^{-x}\right)=\sum_{j=0}^{\infty}\frac{x^{2j+1}}{\left(2j+1\right)!}</math> helpful. | ||
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| + | a. (15 points) | ||
| + | |||
| + | Find the probability mass function of <math>\mathbf{X}\left(t\right)</math> . | ||
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| + | <math>P\left(\left\{ \mathbf{X}\left(t\right)=1\right\} |\left\{ \mathbf{X}\left(0\right)=1\right\} \right)=P\left(\left\{ \mathbf{N}\left(t\right)=\text{even}\right\} \right)=\sum_{n=0}^{\infty}e^{-\lambda t}\frac{\left(\lambda t\right)^{2n}}{\left(2n\right)!}=e^{-\lambda t}\sum_{n=0}^{\infty}\frac{\left(\lambda t\right)^{2n}}{\left(2n\right)!}</math><math>=e^{-\lambda t}\cdot\frac{1}{2}\left(e^{\lambda t}+e^{-\lambda t}\right)=\frac{1}{2}\left(1+e^{-2\lambda t}\right).</math> | ||
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| + | <math>P\left(\left\{ \mathbf{X}\left(t\right)=1\right\} |\left\{ \mathbf{X}\left(0\right)=-1\right\} \right)=P\left(\left\{ \mathbf{N}\left(t\right)=\text{odd}\right\} \right)=\sum_{n=0}^{\infty}e^{-\lambda t}\frac{\left(\lambda t\right)^{2n+1}}{\left(2n+1\right)!}=e^{-\lambda t}\sum_{n=0}^{\infty}\frac{\left(\lambda t\right)^{2n+1}}{\left(2n+1\right)!}</math><math>=e^{-\lambda t}\cdot\frac{1}{2}\left(e^{\lambda t}-e^{-\lambda t}\right)=\frac{1}{2}\left(1-e^{-2\lambda t}\right).</math> | ||
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| + | <math>P\left(\left\{ \mathbf{X}\left(t\right)=1\right\} \right)=P\left(\left\{ \mathbf{X}\left(t\right)=1\right\} |\left\{ \mathbf{X}\left(0\right)=1\right\} \right)P\left(\left\{ \mathbf{X}\left(0\right)=1\right\} \right)+P\left(\left\{ \mathbf{X}\left(t\right)=1\right\} |\left\{ \mathbf{X}\left(0\right)=-1\right\} \right)P\left(\left\{ \mathbf{X}\left(0\right)=-1\right\} \right)</math><math>=\frac{1}{2}\left(1+e^{-2\lambda t}\right)\cdot\frac{1}{2}+\frac{1}{2}\left(1-e^{-2\lambda t}\right)\cdot\frac{1}{2}=\frac{1}{2}.</math> | ||
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| + | Thus, <math>P\left(\left\{ \mathbf{X}\left(t\right)=1\right\} \right)=P\left(\left\{ \mathbf{X}\left(t\right)=-1\right\} \right)=\frac{1}{2}</math> . | ||
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| + | b. (15 points) | ||
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| + | Find the autocovariance function of the random process <math>\mathbf{X}\left(t\right)</math> . | ||
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| + | Note | ||
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| + | The explanation about the autocovariance function is [CS1Autocovariance]. | ||
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| + | <math>E\left[\mathbf{X}\left(t\right)\right]=1\cdot\frac{1}{2}+\left(-1\right)\cdot\frac{1}{2}=-\frac{1}{2}+\frac{1}{2}=0.</math> | ||
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| + | <math>C_{\mathbf{XX}}\left(t_{1},t_{2}\right)=E\left[\left(\mathbf{X}\left(t_{1}\right)-\mu_{\mathbf{X}}\left(t_{1}\right)\right)\left(\mathbf{X}\left(t_{2}\right)-\mu_{\mathbf{X}}\left(t_{2}\right)\right)^{*}\right]=R_{\mathbf{XX}}\left(t_{1},t_{2}\right)-\mu_{\mathbf{X}}\left(t_{1}\right)\mu_{\mathbf{X}}\left(t_{2}\right)^{*}</math><math>=R_{\mathbf{XX}}\left(t_{1},t_{2}\right)=1\cdot P\left(\left\{ \mathbf{X}\left(t_{1}=1\right)\cap\mathbf{X}\left(t_{2}=1\right)\right\} \right)-1\cdot P\left(\left\{ \mathbf{X}\left(t_{1}=-1\right)\cap\mathbf{X}\left(t_{2}=1\right)\right\} \right)</math><math>-1\cdot P\left(\left\{ \mathbf{X}\left(t_{1}=1\right)\cap\mathbf{X}\left(t_{2}=-1\right)\right\} \right)+1\cdot P\left(\left\{ \mathbf{X}\left(t_{1}=-1\right)\cap\mathbf{X}\left(t_{2}=-1\right)\right\} \right)</math><math>=1\cdot P\left(\left\{ \mathbf{X}\left(t_{1}\right)=\mathbf{X}\left(t_{2}\right)\right\} \right)-1\cdot P\left(\left\{ \mathbf{X}\left(t_{1}\right)\neq\mathbf{X}\left(t_{2}\right)\right\} \right).</math> | ||
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| + | <math>P\left(\left\{ \mathbf{X}\left(t_{1}\right)=\mathbf{X}\left(t_{2}\right)\right\} \right)=P\left(\left\{ \text{number of events occuring within }\left(t_{1},t_{2}\right)\text{is even}\right\} \right)</math> | ||
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| + | <math>=\sum_{n=0}^{\infty}e^{-\lambda\left(t_{2}-t_{1}\right)}\frac{\left[\lambda\left(t_{2}-t_{1}\right)\right]^{2n}}{\left(2n\right)!}=\frac{1}{2}e^{-\lambda\left(t_{2}-t_{1}\right)}\left(e^{\lambda\left(t_{2}-t_{1}\right)}+e^{-\lambda\left(t_{2}-t_{1}\right)}\right)</math><math>=\frac{1}{2}\left(1+e^{-2\lambda\left(t_{2}-t_{1}\right)}\right).</math> | ||
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| + | By using the same method above, we can get <math>P\left(\left\{ \mathbf{X}\left(t_{1}\right)\neq\mathbf{X}\left(t_{2}\right)\right\} \right)=\frac{1}{2}\left(1-e^{-2\lambda\left(t_{2}-t_{1}\right)}\right)</math> . Thus, | ||
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| + | <math>C_{\mathbf{XX}}\left(t_{1},t_{2}\right)=\frac{1}{2}\left(1+e^{-2\lambda\left(t_{2}-t_{1}\right)}\right)-\frac{1}{2}\left(1-e^{-2\lambda\left(t_{2}-t_{1}\right)}\right)=e^{-2\lambda\left(t_{2}-t_{1}\right)}.</math> | ||
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| + | 4 (15 points) | ||
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| + | Messages arrive at a message center according to a Poisson process of rate <math>\lambda</math> messages per hour. Every hour the messages that have arrived during the previous hour are forwarded to their destination. Find the expected value of the total time waited by all messages that arrive during the hour. | ||
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Revision as of 11:25, 23 November 2010
7.14 QE 2008 August
1
The Weak Law of Large Numbers states that if $ \mathbf{X}_{1},\mathbf{X}_{2},\mathbf{X}_{3},\cdots $ is a sequence of i.i.d. random variables with finite mean $ E\left[\mathbf{X}_{i}\right]=\mu $ for every $ i $ , then the sample mean $ \mathbf{Y}_{n}=\frac{1}{n}\sum_{i=1}^{n}\mathbf{X}_{i} $ converges to $ \mu $ in probability. Suppose that instead of being i.i.d , $ \mathbf{X}_{1},\mathbf{X}_{2},\mathbf{X}_{3},\cdots $ each have finite mean $ \mu $ , and the covariance function of the sequence $ \mathbf{X}_{n} $ is $ Cov\left(\mathbf{X}_{i},\mathbf{X}_{j}\right)=\sigma^{2}\rho^{\left|i-j\right|} $ , where $ \left|\rho\right|<1 $ and $ \sigma^{2}>0 $ .
a. (13 points)
Find the mean and variance of $ \mathbf{Y}_{n} $ .
$ E\left[\mathbf{Y}_{n}\right]=E\left[\frac{1}{n}\sum_{i=1}^{n}\mathbf{X}_{i}\right]=\frac{1}{n}\sum_{i=1}^{n}E\left[\mathbf{X}_{i}\right]=\frac{n\mu}{n}=\mu. $
$ Cov\left(\mathbf{X}_{i},\mathbf{X}_{j}\right)=E\left[\left(\mathbf{X}_{i}-\mu\right)\left(\mathbf{X}_{j}-\mu\right)\right]=E\left[\mathbf{X}_{i}\mathbf{X}_{j}\right]-\mu^{2}. $
$ E\left[\mathbf{X}_{i}\mathbf{X}_{j}\right]=\begin{cases} \begin{array}{lll} \sigma^{2}+\mu^{2} & & ,\; i=j\\ \sigma^{2}\rho^{\left|i-j\right|}+\mu^{2} & & ,\; i\neq j. \end{array}\end{cases} $
$ E\left[\mathbf{Y}_{n}^{2}\right]=E\left[\frac{1}{n^{2}}\sum_{i=1}^{n}\sum_{j=1}^{n}\mathbf{X}_{i}\mathbf{X}_{j}\right]=\frac{1}{n^{2}}\left[\sum_{i=1}^{n}E\left[\mathbf{X}_{i}^{2}\right]+\underset{_{i\neq j}}{\sum_{i=1}^{n}\sum_{j=1}^{n}}E\left[\mathbf{X}_{i}\mathbf{X}_{j}\right]\right] $$ =\frac{1}{n^{2}}\left[n\left(\sigma^{2}+\mu^{2}\right)+2\sum_{i=1}^{n-1}\left(n-i\right)\left(\sigma^{2}\rho^{i}+\mu^{2}\right)\right]=\frac{1}{n^{2}}\left[n\sigma^{2}+2\sigma^{2}\sum_{i=1}^{n-1}\left(n-i\right)\rho^{i}\right]+\mu^{2}. $
$ Var\left[\mathbf{Y}_{n}\right]=E\left[\mathbf{Y}_{n}^{2}\right]-E\left[\mathbf{Y}_{n}\right]^{2}=\frac{1}{n^{2}}\left[n\sigma^{2}+2\sigma^{2}\sum_{i=1}^{n-1}\left(n-i\right)\rho^{i}\right]+\mu^{2}-\mu^{2}=\frac{1}{n^{2}}\left[n\sigma^{2}+2\sigma^{2}\sum_{i=1}^{n-1}\left(n-i\right)\rho^{i}\right]. $
Note
If the element $ m_{ij} $ of $ n\times n $ matrix is defined as $ \left|i-j\right| $ , then
$ \left[\begin{array}{ccccc} 0 & 1 & 2 & \cdots & n-1\\ 1 & 0 & 1 & \ddots & \vdots\\ 2 & 1 & 0 & \ddots & 2\\ \vdots & \ddots & \ddots & \ddots & 1\\ n-1 & \cdots & 2 & 1 & 0 \end{array}\right]. $
b. (12 points)
Does the sample mean still converge to $ \mu $ in probability? You must justify your answer.
ref. You can see the definition of the converge in probability [CS1ConvergeInProbability].
If $ \mathbf{Y}_{n}\rightarrow\left(p\right)\rightarrow0 $ , then for any $ \epsilon>0 $ , $ P\left(\left\{ \left|\mathbf{Y}_{n}-\mu\right|>\epsilon\right\} \right)\rightarrow0 $ as $ n\rightarrow\infty $ . According to the Chebyshev inequality [CS1ChebyshevInequality], $ P\left(\left\{ \left|\mathbf{Y}_{n}-\mu\right|>\epsilon\right\} \right)\leq\frac{\sigma_{\mathbf{Y}}^{2}}{\epsilon^{2}}. $ $ \lim_{n\rightarrow\infty}\sigma_{\mathbf{Y}}^{2} $ Therefore, we know that $ \lim_{n\rightarrow\infty}P\left(\left\{ \left|\mathbf{Y}_{n}-\mu\right|>\epsilon\right\} \right)=0 $ . Thus $ \mathbf{Y}_{n}\rightarrow\left(p\right)\rightarrow0 $ .
2
Let $ \mathbf{X}_{1},\mathbf{X}_{2},\mathbf{X}_{3},\cdots $ be a sequence of i.i.d Bernoulli random variables with $ p=1/2 $ , and let $ \mathbf{Y}_{n}=2^{n}\mathbf{X}_{1}\mathbf{X}_{2}\cdots\mathbf{X}_{n} $ .
a. (15 points)
Does the sequence $ \mathbf{Y}_{n} $ converge to $ 0 $ almost everywhere?
• You can see the definition of the converge almost everywhere [CS1ConvergeAlmostEverywhere].
• range of $ \mathbf{X}_{i}=\left\{ 0,1\right\} $ .
• range of $ \mathbf{Y}_{n}=\left\{ 0,2^{n}\right\} $ .
• the probability of $ \mathbf{Y}_{n}=2^{n}\Longrightarrow P\left(\mathbf{Y}_{n}=2^{n}\right)=P\left(\left\{ \mathbf{X}_{1}=1\right\} \cap\cdots\cap\left\{ \mathbf{X}_{n}=1\right\} \right)=\left(\frac{1}{2}\right)^{n}. $
• $ \lim_{n\rightarrow\infty}P\left(\mathbf{Y}_{n}=2^{n}\right)=0 $ . $ \lim_{n\rightarrow\infty}P\left(\mathbf{Y}_{n}=0\right)=1 $ .
• Thus, $ \left\{ \mathbf{Y}_{n}\right\} \rightarrow0 $ with probability 1.
b. (15 points)
Does the sequence $ \mathbf{Y}_{n} $ converge to 0 in the mean square sense?
Note
You can see the definition of the converge in mean-square [CS1ConvergeInMean-Square].
$ E\left[\mathbf{X}^{2}\right]=\sum_{k=0}^{1}k^{2}\left(\frac{1}{2}\right)=0^{2}\cdot\frac{1}{2}+1^{2}\cdot\frac{1}{2}=\frac{1}{2}. $
$ E\left[\left|\mathbf{Y}_{n}-0\right|^{2}\right]=E\left[\mathbf{Y}_{n}^{2}\right]=E\left[2^{2n}\mathbf{X}_{1}^{2}\mathbf{X}_{2}^{2}\cdots\mathbf{X}_{n}^{2}\right]=2^{2n}E\left[\mathbf{X}_{1}^{2}\mathbf{X}_{2}^{2}\cdots\mathbf{X}_{n}^{2}\right]=\left(4E\left[\mathbf{X}^{2}\right]\right)^{n}=2^{n}. $
$ \lim_{n\rightarrow\infty}E\left[\mathbf{Y}_{n}^{2}\right]=\lim_{n\rightarrow\infty}2^{n}=\infty. $
Thus, $ \mathbf{Y}_{n} $ does not converge to $ 0 $ in the mean square sense.
3
Consider a random process $ \mathbf{X}\left(t\right) $ that assumes values $ \pm1 $ . Suppose that $ \mathbf{X}\left(0\right)=\pm1 $ with probability $ 1/2 $ , and suppose that $ \mathbf{X}\left(t\right) $ then changes polarity with each occurrence of an event in a Poisson process of rate $ \lambda $ .
Note:
You might find the equations $ \frac{1}{2}\left(e^{x}+e^{-x}\right)=\sum_{j=0}^{\infty}\frac{x^{2j}}{\left(2j\right)!} $ and $ \frac{1}{2}\left(e^{x}-e^{-x}\right)=\sum_{j=0}^{\infty}\frac{x^{2j+1}}{\left(2j+1\right)!} $ helpful.
a. (15 points)
Find the probability mass function of $ \mathbf{X}\left(t\right) $ .
$ P\left(\left\{ \mathbf{X}\left(t\right)=1\right\} |\left\{ \mathbf{X}\left(0\right)=1\right\} \right)=P\left(\left\{ \mathbf{N}\left(t\right)=\text{even}\right\} \right)=\sum_{n=0}^{\infty}e^{-\lambda t}\frac{\left(\lambda t\right)^{2n}}{\left(2n\right)!}=e^{-\lambda t}\sum_{n=0}^{\infty}\frac{\left(\lambda t\right)^{2n}}{\left(2n\right)!} $$ =e^{-\lambda t}\cdot\frac{1}{2}\left(e^{\lambda t}+e^{-\lambda t}\right)=\frac{1}{2}\left(1+e^{-2\lambda t}\right). $
$ P\left(\left\{ \mathbf{X}\left(t\right)=1\right\} |\left\{ \mathbf{X}\left(0\right)=-1\right\} \right)=P\left(\left\{ \mathbf{N}\left(t\right)=\text{odd}\right\} \right)=\sum_{n=0}^{\infty}e^{-\lambda t}\frac{\left(\lambda t\right)^{2n+1}}{\left(2n+1\right)!}=e^{-\lambda t}\sum_{n=0}^{\infty}\frac{\left(\lambda t\right)^{2n+1}}{\left(2n+1\right)!} $$ =e^{-\lambda t}\cdot\frac{1}{2}\left(e^{\lambda t}-e^{-\lambda t}\right)=\frac{1}{2}\left(1-e^{-2\lambda t}\right). $
$ P\left(\left\{ \mathbf{X}\left(t\right)=1\right\} \right)=P\left(\left\{ \mathbf{X}\left(t\right)=1\right\} |\left\{ \mathbf{X}\left(0\right)=1\right\} \right)P\left(\left\{ \mathbf{X}\left(0\right)=1\right\} \right)+P\left(\left\{ \mathbf{X}\left(t\right)=1\right\} |\left\{ \mathbf{X}\left(0\right)=-1\right\} \right)P\left(\left\{ \mathbf{X}\left(0\right)=-1\right\} \right) $$ =\frac{1}{2}\left(1+e^{-2\lambda t}\right)\cdot\frac{1}{2}+\frac{1}{2}\left(1-e^{-2\lambda t}\right)\cdot\frac{1}{2}=\frac{1}{2}. $
Thus, $ P\left(\left\{ \mathbf{X}\left(t\right)=1\right\} \right)=P\left(\left\{ \mathbf{X}\left(t\right)=-1\right\} \right)=\frac{1}{2} $ .
b. (15 points)
Find the autocovariance function of the random process $ \mathbf{X}\left(t\right) $ .
Note
The explanation about the autocovariance function is [CS1Autocovariance].
$ E\left[\mathbf{X}\left(t\right)\right]=1\cdot\frac{1}{2}+\left(-1\right)\cdot\frac{1}{2}=-\frac{1}{2}+\frac{1}{2}=0. $
$ C_{\mathbf{XX}}\left(t_{1},t_{2}\right)=E\left[\left(\mathbf{X}\left(t_{1}\right)-\mu_{\mathbf{X}}\left(t_{1}\right)\right)\left(\mathbf{X}\left(t_{2}\right)-\mu_{\mathbf{X}}\left(t_{2}\right)\right)^{*}\right]=R_{\mathbf{XX}}\left(t_{1},t_{2}\right)-\mu_{\mathbf{X}}\left(t_{1}\right)\mu_{\mathbf{X}}\left(t_{2}\right)^{*} $$ =R_{\mathbf{XX}}\left(t_{1},t_{2}\right)=1\cdot P\left(\left\{ \mathbf{X}\left(t_{1}=1\right)\cap\mathbf{X}\left(t_{2}=1\right)\right\} \right)-1\cdot P\left(\left\{ \mathbf{X}\left(t_{1}=-1\right)\cap\mathbf{X}\left(t_{2}=1\right)\right\} \right) $$ -1\cdot P\left(\left\{ \mathbf{X}\left(t_{1}=1\right)\cap\mathbf{X}\left(t_{2}=-1\right)\right\} \right)+1\cdot P\left(\left\{ \mathbf{X}\left(t_{1}=-1\right)\cap\mathbf{X}\left(t_{2}=-1\right)\right\} \right) $$ =1\cdot P\left(\left\{ \mathbf{X}\left(t_{1}\right)=\mathbf{X}\left(t_{2}\right)\right\} \right)-1\cdot P\left(\left\{ \mathbf{X}\left(t_{1}\right)\neq\mathbf{X}\left(t_{2}\right)\right\} \right). $
$ P\left(\left\{ \mathbf{X}\left(t_{1}\right)=\mathbf{X}\left(t_{2}\right)\right\} \right)=P\left(\left\{ \text{number of events occuring within }\left(t_{1},t_{2}\right)\text{is even}\right\} \right) $
$ =\sum_{n=0}^{\infty}e^{-\lambda\left(t_{2}-t_{1}\right)}\frac{\left[\lambda\left(t_{2}-t_{1}\right)\right]^{2n}}{\left(2n\right)!}=\frac{1}{2}e^{-\lambda\left(t_{2}-t_{1}\right)}\left(e^{\lambda\left(t_{2}-t_{1}\right)}+e^{-\lambda\left(t_{2}-t_{1}\right)}\right) $$ =\frac{1}{2}\left(1+e^{-2\lambda\left(t_{2}-t_{1}\right)}\right). $
By using the same method above, we can get $ P\left(\left\{ \mathbf{X}\left(t_{1}\right)\neq\mathbf{X}\left(t_{2}\right)\right\} \right)=\frac{1}{2}\left(1-e^{-2\lambda\left(t_{2}-t_{1}\right)}\right) $ . Thus,
$ C_{\mathbf{XX}}\left(t_{1},t_{2}\right)=\frac{1}{2}\left(1+e^{-2\lambda\left(t_{2}-t_{1}\right)}\right)-\frac{1}{2}\left(1-e^{-2\lambda\left(t_{2}-t_{1}\right)}\right)=e^{-2\lambda\left(t_{2}-t_{1}\right)}. $
4 (15 points)
Messages arrive at a message center according to a Poisson process of rate $ \lambda $ messages per hour. Every hour the messages that have arrived during the previous hour are forwarded to their destination. Find the expected value of the total time waited by all messages that arrive during the hour.
