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| − | =7.3 QE 2001 August= | + | =7.3 [[ECE_PhD_Qualifying_Exams|QE]] 2001 August= |
'''1. (10 Points)''' | '''1. (10 Points)''' | ||
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{| border = "1" | {| border = "1" | ||
| − | !<math>N</math>th | + | !<math class="inline">N</math>th |
| − | !<math>\left(N - 1\right)</math>th | + | !<math class="inline">\left(N - 1\right)</math>th |
| − | !<math>\left(N - 2\right)</math>th | + | !<math class="inline">\left(N - 2\right)</math>th |
| − | !<math>\left(N - 3\right)</math>th | + | !<math class="inline">\left(N - 3\right)</math>th |
| − | !<math>\cdots</math> | + | !<math class="inline">\cdots</math> |
|- | |- | ||
|- | |- | ||
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|T | |T | ||
|H | |H | ||
| − | |<math>\cdots</math> | + | |<math class="inline">\cdots</math> |
|- | |- | ||
|T | |T | ||
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|H | |H | ||
|T | |T | ||
| − | |<math>\cdots</math> | + | |<math class="inline">\cdots</math> |
|} | |} | ||
| − | <math>P\left(\left\{ N=n\right\} \right)=\frac{2}{2^{n}}=\frac{1}{2^{n-1}}\text{ for }n\geq2.</math> | + | <math class="inline">P\left(\left\{ N=n\right\} \right)=\frac{2}{2^{n}}=\frac{1}{2^{n-1}}\text{ for }n\geq2.</math> |
| − | <math>P\left(\left\{ N\leq7\right\} \right)=\sum_{k=2}^{7}\frac{1}{2^{k-1}}=\sum_{k=1}^{6}\left(\frac{1}{2}\right)^{k}=\frac{\frac{1}{2}\left(1-\left(\frac{1}{2}\right)^{6}\right)}{1-\frac{1}{2}}=1-\frac{1}{64}=\frac{63}{64}.</math> | + | <math class="inline">P\left(\left\{ N\leq7\right\} \right)=\sum_{k=2}^{7}\frac{1}{2^{k-1}}=\sum_{k=1}^{6}\left(\frac{1}{2}\right)^{k}=\frac{\frac{1}{2}\left(1-\left(\frac{1}{2}\right)^{6}\right)}{1-\frac{1}{2}}=1-\frac{1}{64}=\frac{63}{64}.</math> |
'''(b)''' | '''(b)''' | ||
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What is the probability that this experiment terminates with an even number of coin tosses? | What is the probability that this experiment terminates with an even number of coin tosses? | ||
| − | <math>P\left(\left\{ N\text{ is even}\right\} \right)=\sum_{k=1}^{\infty}\frac{1}{2^{2k-1}}=2\sum_{k=1}^{\infty}\left(\frac{1}{4}\right)^{k}=2\cdot\frac{\frac{1}{4}}{1-\frac{1}{4}}=2\cdot\frac{1}{3}=\frac{2}{3}.</math> | + | <math class="inline">P\left(\left\{ N\text{ is even}\right\} \right)=\sum_{k=1}^{\infty}\frac{1}{2^{2k-1}}=2\sum_{k=1}^{\infty}\left(\frac{1}{4}\right)^{k}=2\cdot\frac{\frac{1}{4}}{1-\frac{1}{4}}=2\cdot\frac{1}{3}=\frac{2}{3}.</math> |
'''2. (25 Points)''' | '''2. (25 Points)''' | ||
| − | Let <math>\mathbf{X}</math> and <math>\mathbf{Y}</math> be independent Poisson random variables with mean <math>\lambda</math> and <math>\mu</math> , respectively. Let <math>\mathbf{Z}</math> be a new random variable defined as <math>\mathbf{Z}=\mathbf{X}+\mathbf{Y}.</math> | + | Let <math class="inline">\mathbf{X}</math> and <math class="inline">\mathbf{Y}</math> be independent Poisson random variables with mean <math class="inline">\lambda</math> and <math class="inline">\mu</math> , respectively. Let <math class="inline">\mathbf{Z}</math> be a new random variable defined as <math class="inline">\mathbf{Z}=\mathbf{X}+\mathbf{Y}.</math> |
Note | Note | ||
| − | This problem is identical to the example [ | + | This problem is identical to the example: [[ECE 600 Exams Addition of two independent Poisson random variables|Addition of two independent Poisson random variables]]. |
'''(a)''' | '''(a)''' | ||
| − | Find the probability mass function (pmf) of <math>\mathbf{Z}</math> . | + | Find the probability mass function (pmf) of <math class="inline">\mathbf{Z}</math> . |
'''(b)''' | '''(b)''' | ||
| − | Find the conditional probability mass function (pmf) of <math>\mathbf{X}</math> conditional on the event <math>\left\{ \mathbf{Z}=n\right\}</math> . Identify the type of pmf that this is, and fully specify its parameters. | + | Find the conditional probability mass function (pmf) of <math class="inline">\mathbf{X}</math> conditional on the event <math class="inline">\left\{ \mathbf{Z}=n\right\}</math> . Identify the type of pmf that this is, and fully specify its parameters. |
'''3. (30 Points)''' | '''3. (30 Points)''' | ||
| − | Let <math>\mathbf{X}_{1},\cdots,\mathbf{X}_{n},\cdots</math> be a sequence of random variables that are not necessarily statistically independent, but that each have identical mean <math>\mu</math> and variance <math>\sigma^{2}</math> . Let <math>\mathbf{Y}_{1},\cdots,\mathbf{Y}_{n},\cdots</math> be a sequence of random variable with <math>\mathbf{Y}_{n}=\frac{1}{n}\sum_{k=1}^{n}\mathbf{X}_{k}.</math> | + | Let <math class="inline">\mathbf{X}_{1},\cdots,\mathbf{X}_{n},\cdots</math> be a sequence of random variables that are not necessarily statistically independent, but that each have identical mean <math class="inline">\mu</math> and variance <math class="inline">\sigma^{2}</math> . Let <math class="inline">\mathbf{Y}_{1},\cdots,\mathbf{Y}_{n},\cdots</math> be a sequence of random variable with <math class="inline">\mathbf{Y}_{n}=\frac{1}{n}\sum_{k=1}^{n}\mathbf{X}_{k}.</math> |
'''(a)''' | '''(a)''' | ||
| − | Given that <math>\mathbf{X}_{1},\cdots,\mathbf{X}_{n},\cdots</math> are uncorrelated, determine whether or not <math>\left\{ \mathbf{Y}_{n}\right\}</math> converges to <math>\mu</math> in the mean square sense. | + | Given that <math class="inline">\mathbf{X}_{1},\cdots,\mathbf{X}_{n},\cdots</math> are uncorrelated, determine whether or not <math class="inline">\left\{ \mathbf{Y}_{n}\right\}</math> converges to <math class="inline">\mu</math> in the mean square sense. |
| − | <math>E\left[\left|\mathbf{Y}_{n}-\mu\right|^{2}\right]=E\left[\mathbf{Y}_{n}^{2}\right]-2E\left[\mathbf{Y}_{n}\right]\mu+\mu^{2}.</math> | + | <math class="inline">E\left[\left|\mathbf{Y}_{n}-\mu\right|^{2}\right]=E\left[\mathbf{Y}_{n}^{2}\right]-2E\left[\mathbf{Y}_{n}\right]\mu+\mu^{2}.</math> |
| − | <math>E\left[\mathbf{Y}_{n}\right]=\frac{1}{n}\sum_{k=1}^{n}E\left[\mathbf{X}_{k}\right]=\mu.</math> | + | <math class="inline">E\left[\mathbf{Y}_{n}\right]=\frac{1}{n}\sum_{k=1}^{n}E\left[\mathbf{X}_{k}\right]=\mu.</math> |
| − | <math>E\left[\mathbf{Y}_{n}^{2}\right]=E\left[\frac{1}{n^{2}}\sum_{k=1}^{n}\sum_{l=1}^{n}\mathbf{X}_{k}\mathbf{X}_{l}\right]=\frac{1}{n^{2}}\sum_{k=1}^{n}\sum_{l=1}^{n}E\left[\mathbf{X}_{k}\mathbf{X}_{l}\right]</math><math>=\frac{1}{n^{2}}\sum_{k=1}^{n}E\left[\mathbf{X}_{k}^{2}\right]+\frac{1}{n^{2}}\underset{k\neq l}{\sum_{k=1}^{n}\sum_{l=1}^{n}}E\left[\mathbf{X}_{k}\right]E\left[\mathbf{X}_{l}\right]</math><math>=\frac{1}{n}\left(\mu^{2}+\sigma^{2}\right)+\frac{n\left(n-1\right)}{n^{2}}\mu^{2}=\frac{1}{n}\mu^{2}+\frac{1}{n}\sigma^{2}+\mu^{2}-\frac{1}{n}\mu^{2}</math><math>=\frac{\sigma^{2}}{n}+\mu^{2}.</math> | + | <math class="inline">E\left[\mathbf{Y}_{n}^{2}\right]=E\left[\frac{1}{n^{2}}\sum_{k=1}^{n}\sum_{l=1}^{n}\mathbf{X}_{k}\mathbf{X}_{l}\right]=\frac{1}{n^{2}}\sum_{k=1}^{n}\sum_{l=1}^{n}E\left[\mathbf{X}_{k}\mathbf{X}_{l}\right]</math><math class="inline">=\frac{1}{n^{2}}\sum_{k=1}^{n}E\left[\mathbf{X}_{k}^{2}\right]+\frac{1}{n^{2}}\underset{k\neq l}{\sum_{k=1}^{n}\sum_{l=1}^{n}}E\left[\mathbf{X}_{k}\right]E\left[\mathbf{X}_{l}\right]</math><math class="inline">=\frac{1}{n}\left(\mu^{2}+\sigma^{2}\right)+\frac{n\left(n-1\right)}{n^{2}}\mu^{2}=\frac{1}{n}\mu^{2}+\frac{1}{n}\sigma^{2}+\mu^{2}-\frac{1}{n}\mu^{2}</math><math class="inline">=\frac{\sigma^{2}}{n}+\mu^{2}.</math> |
| − | <math>E\left[\left|\mathbf{Y}_{n}-\mu\right|^{2}\right]=E\left[\mathbf{Y}_{n}^{2}\right]-2E\left[\mathbf{Y}_{n}\right]\mu+\mu^{2}=\frac{\sigma^{2}}{n}+\mu^{2}-2\mu\cdot\mu+\mu^{2}=\frac{\sigma^{2}}{n}.</math> | + | <math class="inline">E\left[\left|\mathbf{Y}_{n}-\mu\right|^{2}\right]=E\left[\mathbf{Y}_{n}^{2}\right]-2E\left[\mathbf{Y}_{n}\right]\mu+\mu^{2}=\frac{\sigma^{2}}{n}+\mu^{2}-2\mu\cdot\mu+\mu^{2}=\frac{\sigma^{2}}{n}.</math> |
| − | <math>\lim_{n\rightarrow\infty}E\left[\left|\mathbf{Y}_{n}-\mu\right|^{2}\right]=\lim_{n\rightarrow\infty}\left(\frac{\sigma^{2}}{n}\right)=0.</math> | + | <math class="inline">\lim_{n\rightarrow\infty}E\left[\left|\mathbf{Y}_{n}-\mu\right|^{2}\right]=\lim_{n\rightarrow\infty}\left(\frac{\sigma^{2}}{n}\right)=0.</math> |
Another approach | Another approach | ||
| − | <math>E\left[\left|\mathbf{Y}_{n}-\mu\right|^{2}\right]=E\left[\left|\frac{1}{n}\sum_{k=1}^{n}\left(\mathbf{X}_{k}-\mu\right)\right|^{2}\right]=\frac{1}{n^{2}}\sum_{k=1}^{n}\sum_{l=1}^{n}E\left[\left(\mathbf{X}_{k}-\mu\right)\left(\mathbf{X}_{l}-\mu\right)\right]</math><math>=\frac{1}{n^{2}}\sum_{k=1}^{n}E\left[\left(\mathbf{X}_{k}-\mu\right)^{2}\right]+\frac{1}{n^{2}}\underset{k\neq l}{\sum_{k=1}^{n}\sum_{l=1}^{n}}E\left[\mathbf{X}_{k}-\mu\right]E\left[\mathbf{X}_{l}-\mu\right]</math><math>=\frac{1}{n^{2}}\cdot n\cdot\sigma^{2}+\frac{1}{n^{2}}\cdot n\left(n-1\right)\cdot0^{2}=\frac{\sigma^{2}}{n}.</math> | + | <math class="inline">E\left[\left|\mathbf{Y}_{n}-\mu\right|^{2}\right]=E\left[\left|\frac{1}{n}\sum_{k=1}^{n}\left(\mathbf{X}_{k}-\mu\right)\right|^{2}\right]=\frac{1}{n^{2}}\sum_{k=1}^{n}\sum_{l=1}^{n}E\left[\left(\mathbf{X}_{k}-\mu\right)\left(\mathbf{X}_{l}-\mu\right)\right]</math><math class="inline">=\frac{1}{n^{2}}\sum_{k=1}^{n}E\left[\left(\mathbf{X}_{k}-\mu\right)^{2}\right]+\frac{1}{n^{2}}\underset{k\neq l}{\sum_{k=1}^{n}\sum_{l=1}^{n}}E\left[\mathbf{X}_{k}-\mu\right]E\left[\mathbf{X}_{l}-\mu\right]</math><math class="inline">=\frac{1}{n^{2}}\cdot n\cdot\sigma^{2}+\frac{1}{n^{2}}\cdot n\left(n-1\right)\cdot0^{2}=\frac{\sigma^{2}}{n}.</math> |
| − | <math>\lim_{n\rightarrow\infty}E\left[\left|\mathbf{Y}_{n}-\mu\right|^{2}\right]=\lim_{n\rightarrow\infty}\left(\frac{\sigma^{2}}{n}\right)=0.</math> | + | <math class="inline">\lim_{n\rightarrow\infty}E\left[\left|\mathbf{Y}_{n}-\mu\right|^{2}\right]=\lim_{n\rightarrow\infty}\left(\frac{\sigma^{2}}{n}\right)=0.</math> |
(b) | (b) | ||
| − | Given that the covariance between <math>\mathbf{X}_{j}</math> and <math>\mathbf{X}_{k}</math> is given by | + | Given that the covariance between <math class="inline">\mathbf{X}_{j}</math> and <math class="inline">\mathbf{X}_{k}</math> is given by |
<br> | <br> | ||
| − | <math>cov\left(\mathbf{X}_{j},\mathbf{X}_{k}\right)=\begin{cases} | + | <math class="inline">cov\left(\mathbf{X}_{j},\mathbf{X}_{k}\right)=\begin{cases} |
\begin{array}{lll} | \begin{array}{lll} | ||
\sigma^{2} \text{, for }j=k\\ | \sigma^{2} \text{, for }j=k\\ | ||
| Line 93: | Line 93: | ||
\end{array}\end{cases}</math> | \end{array}\end{cases}</math> | ||
<br> | <br> | ||
| − | where <math>-1\leq r\leq1</math> , determine whether or not <math>\left\{ \mathbf{Y}_{n}\right\}</math> converges to <math>\mu</math> in the mean square sense. | + | where <math class="inline">-1\leq r\leq1</math> , determine whether or not <math class="inline">\left\{ \mathbf{Y}_{n}\right\}</math> converges to <math class="inline">\mu</math> in the mean square sense. |
| − | <math>E\left[\left|\mathbf{Y}_{n}-\mu\right|^{2}\right]=E\left[\left|\frac{1}{n}\sum_{k=1}^{n}\left(\mathbf{X}_{k}-\mu\right)\right|^{2}\right]=\frac{1}{n^{2}}\sum_{k=1}^{n}\sum_{l=1}^{n}E\left[\left(\mathbf{X}_{k}-\mu\right)\left(\mathbf{X}_{l}-\mu\right)\right]</math><math>=\frac{1}{n^{2}}\sum_{k=1}^{n}E\left[\left(\mathbf{X}_{k}-\mu\right)^{2}\right]+\frac{1}{n^{2}}\underset{k\neq l}{\sum_{k=1}^{n}\sum_{l=1}^{n}}E\left[\left(\mathbf{X}_{k}-\mu\right)\left(\mathbf{X}_{l}-\mu\right)\right]</math><math>=\frac{1}{n}\sigma^{2}+\frac{2\left(n-1\right)}{n^{2}}r\sigma^{2}.</math> | + | <math class="inline">E\left[\left|\mathbf{Y}_{n}-\mu\right|^{2}\right]=E\left[\left|\frac{1}{n}\sum_{k=1}^{n}\left(\mathbf{X}_{k}-\mu\right)\right|^{2}\right]=\frac{1}{n^{2}}\sum_{k=1}^{n}\sum_{l=1}^{n}E\left[\left(\mathbf{X}_{k}-\mu\right)\left(\mathbf{X}_{l}-\mu\right)\right]</math><math class="inline">=\frac{1}{n^{2}}\sum_{k=1}^{n}E\left[\left(\mathbf{X}_{k}-\mu\right)^{2}\right]+\frac{1}{n^{2}}\underset{k\neq l}{\sum_{k=1}^{n}\sum_{l=1}^{n}}E\left[\left(\mathbf{X}_{k}-\mu\right)\left(\mathbf{X}_{l}-\mu\right)\right]</math><math class="inline">=\frac{1}{n}\sigma^{2}+\frac{2\left(n-1\right)}{n^{2}}r\sigma^{2}.</math> |
| − | <math>\lim_{n\rightarrow\infty}E\left[\left|\mathbf{Y}_{n}-\mu\right|^{2}\right]=\lim_{n\rightarrow\infty}\left(\frac{1}{n}\sigma^{2}+\frac{2\left(n-1\right)}{n^{2}}r\sigma^{2}\right)=0.</math> | + | <math class="inline">\lim_{n\rightarrow\infty}E\left[\left|\mathbf{Y}_{n}-\mu\right|^{2}\right]=\lim_{n\rightarrow\infty}\left(\frac{1}{n}\sigma^{2}+\frac{2\left(n-1\right)}{n^{2}}r\sigma^{2}\right)=0.</math> |
| − | Thus, <math>\mathbf{Y}_{n}</math> converges in the mean square sense to <math>\mu</math> . | + | Thus, <math class="inline">\mathbf{Y}_{n}</math> converges in the mean square sense to <math class="inline">\mu</math> . |
4. (35 Points) | 4. (35 Points) | ||
| − | Let <math>\left\{ t_{k}\right\}</math> be the set of Poisson points corresponding to a homogeneous Poisson process with parameters <math>\lambda</math> on the real line such that if <math>\mathbf{N}\left(t_{1},t_{2}\right)</math> is defined as the number of points in the interval <math>\left[t_{1},t_{2}\right)</math> , then <math>P\left(\left\{ N\left(t_{1},t_{2}\right)=k\right\} \right)=\frac{\left[\lambda\left(t_{2}-t_{1}\right)\right]^{k}e^{-\lambda\left(t_{2}-t_{1}\right)}}{k!}\;,\qquad k=0,1,2,\cdots,\; t_{2}>t_{1}\geq0. Let \mathbf{X}\left(t\right)=\mathbf{N}\left(0,t\right)</math> be the Poisson counting process for <math>t>0</math> (note that <math>\mathbf{X}\left(0\right)=0</math> ). | + | Let <math class="inline">\left\{ t_{k}\right\}</math> be the set of Poisson points corresponding to a homogeneous Poisson process with parameters <math class="inline">\lambda</math> on the real line such that if <math class="inline">\mathbf{N}\left(t_{1},t_{2}\right)</math> is defined as the number of points in the interval <math class="inline">\left[t_{1},t_{2}\right)</math> , then <math class="inline">P\left(\left\{ N\left(t_{1},t_{2}\right)=k\right\} \right)=\frac{\left[\lambda\left(t_{2}-t_{1}\right)\right]^{k}e^{-\lambda\left(t_{2}-t_{1}\right)}}{k!}\;,\qquad k=0,1,2,\cdots,\; t_{2}>t_{1}\geq0. Let \mathbf{X}\left(t\right)=\mathbf{N}\left(0,t\right)</math> be the Poisson counting process for <math class="inline">t>0</math> (note that <math class="inline">\mathbf{X}\left(0\right)=0</math> ). |
(a) | (a) | ||
| − | Find the (first order) characteristic function of <math>\mathbf{X}\left(t\right)</math> . | + | Find the (first order) characteristic function of <math class="inline">\mathbf{X}\left(t\right)</math> . |
| − | <math>\Phi_{\mathbf{X}}\left(\omega\right)=E\left[e^{i\omega\mathbf{X}}\right]=\sum_{k=0}^{\infty}e^{i\omega k}\frac{\left(\lambda t\right)^{k}e^{-\lambda t}}{k!}=e^{-\lambda t}\sum_{k=0}^{\infty}\frac{\left(\lambda te^{i\omega}\right)^{k}}{k!}=e^{-\lambda t}e^{\lambda te^{i\omega}}=e^{-\lambda t\left(1-e^{i\omega}\right)}.</math> | + | <math class="inline">\Phi_{\mathbf{X}}\left(\omega\right)=E\left[e^{i\omega\mathbf{X}}\right]=\sum_{k=0}^{\infty}e^{i\omega k}\frac{\left(\lambda t\right)^{k}e^{-\lambda t}}{k!}=e^{-\lambda t}\sum_{k=0}^{\infty}\frac{\left(\lambda te^{i\omega}\right)^{k}}{k!}=e^{-\lambda t}e^{\lambda te^{i\omega}}=e^{-\lambda t\left(1-e^{i\omega}\right)}.</math> |
(b) | (b) | ||
| − | Find the mean and variance of <math>\mathbf{X}\left(t\right)</math> . | + | Find the mean and variance of <math class="inline">\mathbf{X}\left(t\right)</math> . |
| − | <math>E\left[\mathbf{X}\left(t\right)\right]=\frac{d}{di\omega}\Phi_{\mathbf{X}}\left(\omega\right)\biggl|_{i\omega=0}=\frac{d}{di\omega}e^{-\lambda t}e^{\lambda te^{i\omega}}\biggl|_{i\omega=0}=e^{-\lambda t}\cdot\frac{d}{di\omega}e^{\lambda te^{i\omega}}\biggl|_{i\omega=0}</math><math>=e^{-\lambda t}\cdot e^{\lambda te^{i\omega}}\cdot\lambda te^{i\omega}\biggl|_{i\omega=0}=e^{-\lambda t}\cdot e^{\lambda t}\cdot\lambda t=\lambda t.</math> | + | <math class="inline">E\left[\mathbf{X}\left(t\right)\right]=\frac{d}{di\omega}\Phi_{\mathbf{X}}\left(\omega\right)\biggl|_{i\omega=0}=\frac{d}{di\omega}e^{-\lambda t}e^{\lambda te^{i\omega}}\biggl|_{i\omega=0}=e^{-\lambda t}\cdot\frac{d}{di\omega}e^{\lambda te^{i\omega}}\biggl|_{i\omega=0}</math><math class="inline">=e^{-\lambda t}\cdot e^{\lambda te^{i\omega}}\cdot\lambda te^{i\omega}\biggl|_{i\omega=0}=e^{-\lambda t}\cdot e^{\lambda t}\cdot\lambda t=\lambda t.</math> |
| − | <math>E\left[\mathbf{X}^{2}\left(t\right)\right]=\frac{d}{d\left(i\omega\right)^{2}}\Phi_{\mathbf{X}}\left(\omega\right)\biggl|_{i\omega=0}=\frac{d}{di\omega}\lambda te^{-\lambda t}e^{\lambda te^{i\omega}}e^{i\omega}\biggl|_{i\omega=0}</math><math>=\lambda te^{-\lambda t}\cdot\frac{d}{di\omega}e^{\lambda te^{i\omega}}e^{i\omega}\biggl|_{i\omega=0}</math><math>=\lambda te^{-\lambda t}\left(e^{\lambda te^{i\omega}}\lambda te^{i\omega}e^{i\omega}+e^{\lambda te^{i\omega}}e^{i\omega}\right)\biggl|_{i\omega=0}</math><math>=\lambda te^{-\lambda t}\left(\lambda te^{\lambda te^{i\omega}}e^{2i\omega}+e^{\lambda te^{i\omega}}e^{i\omega}\right)\biggl|_{i\omega=0}=\lambda te^{-\lambda t}\left(\lambda te^{\lambda t}+e^{\lambda t}\right)</math><math>=\lambda t\left(\lambda t+1\right)=\left(\lambda t\right)^{2}+\lambda t.</math> | + | <math class="inline">E\left[\mathbf{X}^{2}\left(t\right)\right]=\frac{d}{d\left(i\omega\right)^{2}}\Phi_{\mathbf{X}}\left(\omega\right)\biggl|_{i\omega=0}=\frac{d}{di\omega}\lambda te^{-\lambda t}e^{\lambda te^{i\omega}}e^{i\omega}\biggl|_{i\omega=0}</math><math class="inline">=\lambda te^{-\lambda t}\cdot\frac{d}{di\omega}e^{\lambda te^{i\omega}}e^{i\omega}\biggl|_{i\omega=0}</math><math class="inline">=\lambda te^{-\lambda t}\left(e^{\lambda te^{i\omega}}\lambda te^{i\omega}e^{i\omega}+e^{\lambda te^{i\omega}}e^{i\omega}\right)\biggl|_{i\omega=0}</math><math class="inline">=\lambda te^{-\lambda t}\left(\lambda te^{\lambda te^{i\omega}}e^{2i\omega}+e^{\lambda te^{i\omega}}e^{i\omega}\right)\biggl|_{i\omega=0}=\lambda te^{-\lambda t}\left(\lambda te^{\lambda t}+e^{\lambda t}\right)</math><math class="inline">=\lambda t\left(\lambda t+1\right)=\left(\lambda t\right)^{2}+\lambda t.</math> |
| − | <math>Var\left[\mathbf{X}\left(t\right)\right]=E\left[\mathbf{X}^{2}\left(t\right)\right]-\left(E\left[\mathbf{X}\left(t\right)\right]\right)^{2}=\left(\lambda t\right)^{2}+\lambda t-\left(\lambda t\right)^{2}=\lambda t.</math> | + | <math class="inline">Var\left[\mathbf{X}\left(t\right)\right]=E\left[\mathbf{X}^{2}\left(t\right)\right]-\left(E\left[\mathbf{X}\left(t\right)\right]\right)^{2}=\left(\lambda t\right)^{2}+\lambda t-\left(\lambda t\right)^{2}=\lambda t.</math> |
(c) | (c) | ||
| − | Deriven an expression for the autocorrelation function of <math>\mathbf{X}\left(t\right)</math> . | + | Deriven an expression for the autocorrelation function of <math class="inline">\mathbf{X}\left(t\right)</math> . |
| − | <math>R_{\mathbf{XX}}\left(t_{1},t_{2}\right)</math> | + | <math class="inline">R_{\mathbf{XX}}\left(t_{1},t_{2}\right)</math> |
(d) | (d) | ||
| − | Assuming that <math>t_{2}>t_{1}</math> , find an expression for <math>P\left(\left\{ \mathbf{X}\left(t_{1}\right)=m\right\} \cap\left\{ \mathbf{X}\left(t_{2}\right)=n\right\} \right)</math> , for all <math>m=0,1,2,\cdots</math> and <math>n=0,1,2,\cdots</math> . | + | Assuming that <math class="inline">t_{2}>t_{1}</math> , find an expression for <math class="inline">P\left(\left\{ \mathbf{X}\left(t_{1}\right)=m\right\} \cap\left\{ \mathbf{X}\left(t_{2}\right)=n\right\} \right)</math> , for all <math class="inline">m=0,1,2,\cdots</math> and <math class="inline">n=0,1,2,\cdots</math> . |
| − | <math>P\left(\left\{ \mathbf{X}\left(t_{1}\right)=m\right\} \cap\left\{ \mathbf{X}\left(t_{2}\right)=n\right\} \right)</math> | + | <math class="inline">P\left(\left\{ \mathbf{X}\left(t_{1}\right)=m\right\} \cap\left\{ \mathbf{X}\left(t_{2}\right)=n\right\} \right)</math> |
---- | ---- | ||
[[ECE600|Back to ECE600]] | [[ECE600|Back to ECE600]] | ||
| − | [[ECE 600 QE|Back to ECE 600 QE]] | + | [[ECE 600 QE|Back to my ECE 600 QE page]] |
| + | |||
| + | [[ECE_PhD_Qualifying_Exams|Back to the general ECE PHD QE page]] (for problem discussion) | ||
Latest revision as of 08:33, 27 June 2012
7.3 QE 2001 August
1. (10 Points)
Consider the following random experiment: A fair coin is repeatedly tossed until the same outcome (H or T) appears twice in a row.
(a)
What is the probability that this experiment terminates on or before the seventh coin toss?
Let N be the number of toss until the same outcome appears twice in a row.
| $ N $th | $ \left(N - 1\right) $th | $ \left(N - 2\right) $th | $ \left(N - 3\right) $th | $ \cdots $ |
|---|---|---|---|---|
| H | H | T | H | $ \cdots $ |
| T | T | H | T | $ \cdots $ |
$ P\left(\left\{ N=n\right\} \right)=\frac{2}{2^{n}}=\frac{1}{2^{n-1}}\text{ for }n\geq2. $
$ P\left(\left\{ N\leq7\right\} \right)=\sum_{k=2}^{7}\frac{1}{2^{k-1}}=\sum_{k=1}^{6}\left(\frac{1}{2}\right)^{k}=\frac{\frac{1}{2}\left(1-\left(\frac{1}{2}\right)^{6}\right)}{1-\frac{1}{2}}=1-\frac{1}{64}=\frac{63}{64}. $
(b)
What is the probability that this experiment terminates with an even number of coin tosses?
$ P\left(\left\{ N\text{ is even}\right\} \right)=\sum_{k=1}^{\infty}\frac{1}{2^{2k-1}}=2\sum_{k=1}^{\infty}\left(\frac{1}{4}\right)^{k}=2\cdot\frac{\frac{1}{4}}{1-\frac{1}{4}}=2\cdot\frac{1}{3}=\frac{2}{3}. $
2. (25 Points)
Let $ \mathbf{X} $ and $ \mathbf{Y} $ be independent Poisson random variables with mean $ \lambda $ and $ \mu $ , respectively. Let $ \mathbf{Z} $ be a new random variable defined as $ \mathbf{Z}=\mathbf{X}+\mathbf{Y}. $
Note
This problem is identical to the example: Addition of two independent Poisson random variables.
(a)
Find the probability mass function (pmf) of $ \mathbf{Z} $ .
(b)
Find the conditional probability mass function (pmf) of $ \mathbf{X} $ conditional on the event $ \left\{ \mathbf{Z}=n\right\} $ . Identify the type of pmf that this is, and fully specify its parameters.
3. (30 Points)
Let $ \mathbf{X}_{1},\cdots,\mathbf{X}_{n},\cdots $ be a sequence of random variables that are not necessarily statistically independent, but that each have identical mean $ \mu $ and variance $ \sigma^{2} $ . Let $ \mathbf{Y}_{1},\cdots,\mathbf{Y}_{n},\cdots $ be a sequence of random variable with $ \mathbf{Y}_{n}=\frac{1}{n}\sum_{k=1}^{n}\mathbf{X}_{k}. $
(a)
Given that $ \mathbf{X}_{1},\cdots,\mathbf{X}_{n},\cdots $ are uncorrelated, determine whether or not $ \left\{ \mathbf{Y}_{n}\right\} $ converges to $ \mu $ in the mean square sense.
$ E\left[\left|\mathbf{Y}_{n}-\mu\right|^{2}\right]=E\left[\mathbf{Y}_{n}^{2}\right]-2E\left[\mathbf{Y}_{n}\right]\mu+\mu^{2}. $
$ E\left[\mathbf{Y}_{n}\right]=\frac{1}{n}\sum_{k=1}^{n}E\left[\mathbf{X}_{k}\right]=\mu. $
$ E\left[\mathbf{Y}_{n}^{2}\right]=E\left[\frac{1}{n^{2}}\sum_{k=1}^{n}\sum_{l=1}^{n}\mathbf{X}_{k}\mathbf{X}_{l}\right]=\frac{1}{n^{2}}\sum_{k=1}^{n}\sum_{l=1}^{n}E\left[\mathbf{X}_{k}\mathbf{X}_{l}\right] $$ =\frac{1}{n^{2}}\sum_{k=1}^{n}E\left[\mathbf{X}_{k}^{2}\right]+\frac{1}{n^{2}}\underset{k\neq l}{\sum_{k=1}^{n}\sum_{l=1}^{n}}E\left[\mathbf{X}_{k}\right]E\left[\mathbf{X}_{l}\right] $$ =\frac{1}{n}\left(\mu^{2}+\sigma^{2}\right)+\frac{n\left(n-1\right)}{n^{2}}\mu^{2}=\frac{1}{n}\mu^{2}+\frac{1}{n}\sigma^{2}+\mu^{2}-\frac{1}{n}\mu^{2} $$ =\frac{\sigma^{2}}{n}+\mu^{2}. $
$ E\left[\left|\mathbf{Y}_{n}-\mu\right|^{2}\right]=E\left[\mathbf{Y}_{n}^{2}\right]-2E\left[\mathbf{Y}_{n}\right]\mu+\mu^{2}=\frac{\sigma^{2}}{n}+\mu^{2}-2\mu\cdot\mu+\mu^{2}=\frac{\sigma^{2}}{n}. $ $ \lim_{n\rightarrow\infty}E\left[\left|\mathbf{Y}_{n}-\mu\right|^{2}\right]=\lim_{n\rightarrow\infty}\left(\frac{\sigma^{2}}{n}\right)=0. $
Another approach
$ E\left[\left|\mathbf{Y}_{n}-\mu\right|^{2}\right]=E\left[\left|\frac{1}{n}\sum_{k=1}^{n}\left(\mathbf{X}_{k}-\mu\right)\right|^{2}\right]=\frac{1}{n^{2}}\sum_{k=1}^{n}\sum_{l=1}^{n}E\left[\left(\mathbf{X}_{k}-\mu\right)\left(\mathbf{X}_{l}-\mu\right)\right] $$ =\frac{1}{n^{2}}\sum_{k=1}^{n}E\left[\left(\mathbf{X}_{k}-\mu\right)^{2}\right]+\frac{1}{n^{2}}\underset{k\neq l}{\sum_{k=1}^{n}\sum_{l=1}^{n}}E\left[\mathbf{X}_{k}-\mu\right]E\left[\mathbf{X}_{l}-\mu\right] $$ =\frac{1}{n^{2}}\cdot n\cdot\sigma^{2}+\frac{1}{n^{2}}\cdot n\left(n-1\right)\cdot0^{2}=\frac{\sigma^{2}}{n}. $
$ \lim_{n\rightarrow\infty}E\left[\left|\mathbf{Y}_{n}-\mu\right|^{2}\right]=\lim_{n\rightarrow\infty}\left(\frac{\sigma^{2}}{n}\right)=0. $
(b)
Given that the covariance between $ \mathbf{X}_{j} $ and $ \mathbf{X}_{k} $ is given by
$ cov\left(\mathbf{X}_{j},\mathbf{X}_{k}\right)=\begin{cases} \begin{array}{lll} \sigma^{2} \text{, for }j=k\\ r\sigma^{2} \text{, for }\left|j-k\right|=1\\ 0 \text{, elsewhere, } \end{array}\end{cases} $
where $ -1\leq r\leq1 $ , determine whether or not $ \left\{ \mathbf{Y}_{n}\right\} $ converges to $ \mu $ in the mean square sense.
$ E\left[\left|\mathbf{Y}_{n}-\mu\right|^{2}\right]=E\left[\left|\frac{1}{n}\sum_{k=1}^{n}\left(\mathbf{X}_{k}-\mu\right)\right|^{2}\right]=\frac{1}{n^{2}}\sum_{k=1}^{n}\sum_{l=1}^{n}E\left[\left(\mathbf{X}_{k}-\mu\right)\left(\mathbf{X}_{l}-\mu\right)\right] $$ =\frac{1}{n^{2}}\sum_{k=1}^{n}E\left[\left(\mathbf{X}_{k}-\mu\right)^{2}\right]+\frac{1}{n^{2}}\underset{k\neq l}{\sum_{k=1}^{n}\sum_{l=1}^{n}}E\left[\left(\mathbf{X}_{k}-\mu\right)\left(\mathbf{X}_{l}-\mu\right)\right] $$ =\frac{1}{n}\sigma^{2}+\frac{2\left(n-1\right)}{n^{2}}r\sigma^{2}. $
$ \lim_{n\rightarrow\infty}E\left[\left|\mathbf{Y}_{n}-\mu\right|^{2}\right]=\lim_{n\rightarrow\infty}\left(\frac{1}{n}\sigma^{2}+\frac{2\left(n-1\right)}{n^{2}}r\sigma^{2}\right)=0. $
Thus, $ \mathbf{Y}_{n} $ converges in the mean square sense to $ \mu $ .
4. (35 Points)
Let $ \left\{ t_{k}\right\} $ be the set of Poisson points corresponding to a homogeneous Poisson process with parameters $ \lambda $ on the real line such that if $ \mathbf{N}\left(t_{1},t_{2}\right) $ is defined as the number of points in the interval $ \left[t_{1},t_{2}\right) $ , then $ P\left(\left\{ N\left(t_{1},t_{2}\right)=k\right\} \right)=\frac{\left[\lambda\left(t_{2}-t_{1}\right)\right]^{k}e^{-\lambda\left(t_{2}-t_{1}\right)}}{k!}\;,\qquad k=0,1,2,\cdots,\; t_{2}>t_{1}\geq0. Let \mathbf{X}\left(t\right)=\mathbf{N}\left(0,t\right) $ be the Poisson counting process for $ t>0 $ (note that $ \mathbf{X}\left(0\right)=0 $ ).
(a)
Find the (first order) characteristic function of $ \mathbf{X}\left(t\right) $ .
$ \Phi_{\mathbf{X}}\left(\omega\right)=E\left[e^{i\omega\mathbf{X}}\right]=\sum_{k=0}^{\infty}e^{i\omega k}\frac{\left(\lambda t\right)^{k}e^{-\lambda t}}{k!}=e^{-\lambda t}\sum_{k=0}^{\infty}\frac{\left(\lambda te^{i\omega}\right)^{k}}{k!}=e^{-\lambda t}e^{\lambda te^{i\omega}}=e^{-\lambda t\left(1-e^{i\omega}\right)}. $
(b)
Find the mean and variance of $ \mathbf{X}\left(t\right) $ .
$ E\left[\mathbf{X}\left(t\right)\right]=\frac{d}{di\omega}\Phi_{\mathbf{X}}\left(\omega\right)\biggl|_{i\omega=0}=\frac{d}{di\omega}e^{-\lambda t}e^{\lambda te^{i\omega}}\biggl|_{i\omega=0}=e^{-\lambda t}\cdot\frac{d}{di\omega}e^{\lambda te^{i\omega}}\biggl|_{i\omega=0} $$ =e^{-\lambda t}\cdot e^{\lambda te^{i\omega}}\cdot\lambda te^{i\omega}\biggl|_{i\omega=0}=e^{-\lambda t}\cdot e^{\lambda t}\cdot\lambda t=\lambda t. $
$ E\left[\mathbf{X}^{2}\left(t\right)\right]=\frac{d}{d\left(i\omega\right)^{2}}\Phi_{\mathbf{X}}\left(\omega\right)\biggl|_{i\omega=0}=\frac{d}{di\omega}\lambda te^{-\lambda t}e^{\lambda te^{i\omega}}e^{i\omega}\biggl|_{i\omega=0} $$ =\lambda te^{-\lambda t}\cdot\frac{d}{di\omega}e^{\lambda te^{i\omega}}e^{i\omega}\biggl|_{i\omega=0} $$ =\lambda te^{-\lambda t}\left(e^{\lambda te^{i\omega}}\lambda te^{i\omega}e^{i\omega}+e^{\lambda te^{i\omega}}e^{i\omega}\right)\biggl|_{i\omega=0} $$ =\lambda te^{-\lambda t}\left(\lambda te^{\lambda te^{i\omega}}e^{2i\omega}+e^{\lambda te^{i\omega}}e^{i\omega}\right)\biggl|_{i\omega=0}=\lambda te^{-\lambda t}\left(\lambda te^{\lambda t}+e^{\lambda t}\right) $$ =\lambda t\left(\lambda t+1\right)=\left(\lambda t\right)^{2}+\lambda t. $
$ Var\left[\mathbf{X}\left(t\right)\right]=E\left[\mathbf{X}^{2}\left(t\right)\right]-\left(E\left[\mathbf{X}\left(t\right)\right]\right)^{2}=\left(\lambda t\right)^{2}+\lambda t-\left(\lambda t\right)^{2}=\lambda t. $
(c)
Deriven an expression for the autocorrelation function of $ \mathbf{X}\left(t\right) $ .
$ R_{\mathbf{XX}}\left(t_{1},t_{2}\right) $
(d)
Assuming that $ t_{2}>t_{1} $ , find an expression for $ P\left(\left\{ \mathbf{X}\left(t_{1}\right)=m\right\} \cap\left\{ \mathbf{X}\left(t_{2}\right)=n\right\} \right) $ , for all $ m=0,1,2,\cdots $ and $ n=0,1,2,\cdots $ .
$ P\left(\left\{ \mathbf{X}\left(t_{1}\right)=m\right\} \cap\left\{ \mathbf{X}\left(t_{2}\right)=n\right\} \right) $
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