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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> ). | 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 class="inline">\mathbf{X}\left(t\right)</math> . |
| − | Find the ( | + | (b) Find the mean and variance of <math class="inline">\mathbf{X}\left(t\right)</math> . |
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| + | (c) Derive an expression for the autocorrelation function of <math class="inline">\mathbf{X}\left(t\right)</math> . | ||
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| + | (d) 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> . | ||
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| + | ---- | ||
| + | ==Share and discuss your solutions below.== | ||
| + | ---- | ||
| + | ==Solution 1== | ||
| + | (a) | ||
<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> | <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) | ||
| − | |||
| − | |||
<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 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> | ||
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<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 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 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> | + | <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) | ||
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<math class="inline">R_{\mathbf{XX}}\left(t_{1},t_{2}\right)</math> | <math class="inline">R_{\mathbf{XX}}\left(t_{1},t_{2}\right)</math> | ||
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(d) | (d) | ||
| − | |||
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<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> | <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> | ||
| + | |||
| + | ---- | ||
Latest revision as of 16:48, 13 March 2015
Communication, Networking, Signal and Image Processing (CS)
Question 1: Probability and Random Processes
August 2001
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) $ .
(b) Find the mean and variance of $ \mathbf{X}\left(t\right) $ .
(c) Derive an expression for the autocorrelation function of $ \mathbf{X}\left(t\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 $ .
Solution 1
(a)
$ \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)
$ 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)
$ R_{\mathbf{XX}}\left(t_{1},t_{2}\right) $
(d)
$ P\left(\left\{ \mathbf{X}\left(t_{1}\right)=m\right\} \cap\left\{ \mathbf{X}\left(t_{2}\right)=n\right\} \right) $
