Difference between revisions of "ECE600 QE 2007 August" - Rhea

Latest revision as of 07:29, 27 June 2012

7.13 QE 2007 August

1. (25 Points)

Let $ \mathbf{X} $ and $ \mathbf{Y} $ be two independent identically distributed random variables taking on values in $ \mathbf{N} $ (the natural numbers) with $ P\left(\left\{ \mathbf{X}=i\right\} \right)=P\left(\left\{ \mathbf{Y}=i\right\} \right)=\frac{1}{2^{i}}\;,\qquad i=1,2,3,\cdots. $

(a)

Find $ P\left(\left\{ \min\left(\mathbf{X},\mathbf{Y}\right)=k\right\} \right) $ , for $ k\in\mathbf{N} $ .

Note

This problem is different from $ P\left(\left\{ \min\left(\mathbf{X},\mathbf{Y}\right)>k\right\} \right) $ .

$ P\left(\left\{ \mathbf{Y}>k\right\} \right)=1-P\left(\left\{ \mathbf{Y}\leq k\right\} \right)=1-\sum_{i=1}^{k}\frac{1}{2^{i}}=1-\frac{\frac{1}{2}\left(1-\left(\frac{1}{2}\right)^{k}\right)}{1-\frac{1}{2}}=1-\left(1-\left(\frac{1}{2}\right)^{k}\right)=\left(\frac{1}{2}\right)^{k}. $

$ P\left(\left\{ \min\left(\mathbf{X},\mathbf{Y}\right)=k\right\} \right)=P\left(\left\{ \mathbf{X}=k\right\} \cap\left\{ \mathbf{Y}>k\right\} \right)+P\left(\left\{ \mathbf{X}>k\right\} \cap\left\{ \mathbf{Y}=k\right\} \right)+P\left(\left\{ \mathbf{X}=k\right\} \cap\left\{ \mathbf{Y}=k\right\} \right) $$ =2\cdot P\left(\left\{ \mathbf{X}=k\right\} \right)\cdot P\left(\left\{ \mathbf{Y}>k\right\} \right)+P\left(\left\{ \mathbf{X}=k\right\} \right)\cdot P\left(\left\{ \mathbf{Y}=k\right\} \right) $$ =2\cdot\left(\frac{1}{2}\right)^{k}\cdot\left(\frac{1}{2}\right)^{k}+\left(\frac{1}{2}\right)^{k}\cdot\left(\frac{1}{2}\right)^{k}=3\cdot\left(\frac{1}{2}\right)^{2k}=\frac{3}{4^{k}}. $

(b)

Find $ P\left(\left\{ \mathbf{X}=\mathbf{Y}\right\} \right) $ .

$ P\left(\left\{ \mathbf{X}=\mathbf{Y}\right\} \right)=\sum_{k=1}^{\infty}P\left(\left\{ \mathbf{X}=k\right\} \cap\left\{ \mathbf{Y}=k\right\} \right)=\sum_{k=1}^{\infty}P\left(\left\{ \mathbf{X}=k\right\} \right)\cdot P\left(\left\{ \mathbf{Y}=k\right\} \right) $$ =\sum_{k=1}^{\infty}\left(\frac{1}{2}\right)^{k}\left(\frac{1}{2}\right)^{k}=\sum_{k=1}^{\infty}\left(\frac{1}{4}\right)^{k}=\frac{\frac{1}{4}}{1-\frac{1}{4}}=\frac{1}{3}. $

(c)

Find $ P\left(\left\{ \mathbf{Y}>\mathbf{X}\right\} \right) $ .

$ P\left(\left\{ \mathbf{Y}>\mathbf{X}\right\} \right)=\sum_{k=1}^{\infty}P\left(\left\{ \mathbf{Y}>k\right\} \cap\left\{ \mathbf{X}=k\right\} \right)=\sum_{k=1}^{\infty}P\left(\left\{ \mathbf{Y}>k\right\} \right)\cdot P\left(\left\{ \mathbf{X}=k\right\} \right) $$ =\sum_{k=1}^{\infty}\left(\frac{1}{2}\right)^{k}\left(\frac{1}{2}\right)^{k}=\sum_{k=1}^{\infty}\left(\frac{1}{4}\right)^{k}=\frac{\frac{1}{4}}{1-\frac{1}{4}}=\frac{1}{3}. $

(d)

Find $ P\left(\left\{ \mathbf{Y}=k\mathbf{X}\right\} \right) $ for a given natural number $ k $ .

$ P\left(\left\{ \mathbf{Y}=k\mathbf{X}\right\} \right)=\sum_{i=1}^{\infty}P\left(\left\{ \mathbf{Y}=i\right\} \cap\left\{ \mathbf{X}=ki\right\} \right)=\sum_{i=1}^{\infty}P\left(\left\{ \mathbf{Y}=i\right\} \right)\cdot P\left(\left\{ \mathbf{X}=ki\right\} \right) $$ =\sum_{i=1}^{\infty}\frac{1}{2^{i}}\cdot\frac{1}{2^{ki}}=\sum_{i=1}^{\infty}\left(\frac{1}{2^{k+1}}\right)^{i}=\frac{\frac{1}{2^{k+1}}}{1-\frac{1}{2^{k+1}}}=\frac{1}{2^{k+1}-1}. $

2. (25 Points)

Let $ \left\{ \mathbf{X}_{n}\right\} _{n\geq1} $ be a sequence of binomially distributed random variables, with the $ n $ -th random variable $ \mathbf{X}_{n} $ having pmf $ p_{\mathbf{X}_{n}}\left(k\right)=P\left(\left\{ \mathbf{X}_{n}=k\right\} \right)=\left(\begin{array}{c} n\\ k \end{array}\right)p_{n}^{k}\left(1-p_{n}\right)^{n-k}\;,\qquad k=0,\cdots,n,\quad p_{n}\in\left(0,1\right). $

Show that, if the $ p_{n} $ have the property that $ np_{n}\rightarrow\lambda $ as $ n\rightarrow\infty $ , where $ \lambda $ is a positive constant, then the sequence $ \left\{ \mathbf{X}_{n}\right\} _{n\geq1} $ converges in distribution to a Poisson random variable $ \mathbf{X} $ with mean $ \lambda $ .

Hint:

You may find the following fact useful:

$ \lim_{n\rightarrow\infty}\left(1+\frac{x}{n}\right)^{n}=e^{x}. $

Answer

Please see the example that is identical to this problem.

3. (25 Points)

Let $ \mathbf{X}\left(t\right) $ be a real Gaussian random process with mean function $ \mu\left(t\right) $ and autocovariance function $ C_{\mathbf{XX}}\left(t_{1},t_{2}\right) $ .

(a)

Write the expression for the $ n $ -th order characteristic function of $ \mathbf{X}\left(t\right) $ in terms of $ \mu\left(t\right) $ and $ C_{\mathbf{XX}}\left(t_{1},t_{2}\right) $ .

ref.

There are the note about the n-th order characteristic function of Gaussians random process . The only difference between the note and this problem is that this problem use the $ \mu\left(t\right) $ rather than $ \eta_{\mathbf{X}}\left(t\right)=E\left[\mathbf{X}\left(t\right)\right] $ .

Solution

$ \Phi_{\mathbf{X}\left(t_{1}\right)\cdots\mathbf{X}\left(t_{n}\right)}\left(\omega_{1},\cdots,\omega_{n}\right)=\exp\left\{ i\sum_{k=1}^{n}\mu_{\mathbf{X}}\left(t_{k}\right)\omega_{k}-\frac{1}{2}\sum_{j=1}^{n}\sum_{k=1}^{n}C_{\mathbf{XX}}\left(t_{j},t_{k}\right)\omega_{j}\omega_{k}\right\} $ .

(b)

Show that the probabilistic description of $ \mathbf{X}\left(t\right) $ is completely characterized by $ \mu\left(t\right) $ and autocovariance function $ C_{\mathbf{XX}}\left(t_{1},t_{2}\right) $ .

Solution

From (a), the characteristic function of $ \mathbf{X}\left(t\right) $ is specified completely in terms of $ \mu_{\mathbf{X}}\left(t\right) $ and $ C_{\mathbf{XX}}\left(t_{1},t_{2}\right) $ . Thus, probabilistic description of $ \mathbf{X}\left(t\right) $ is completely characterized by the characteristic function.

Note

$ f_{\mathbf{X}}\left(x\right)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-i\omega x}\Phi_{\mathbf{X}}\left(\omega\right)d\omega. $

(c)

Show that if $ \mathbf{X}\left(t\right) $ is wide-sense stationary then it is also strict-sense stationary.

Note

You can use the theorem and its proof for solving this problem.

4. (25 Points)

Let $ \mathbf{X}_{1},\mathbf{X}_{2},\mathbf{X}_{3},\cdots $ be a sequence of independent, identically distributed random variables, each having Cauchy pdf $ f\left(x\right)=\frac{1}{\pi\left(1+x^{2}\right)}\;,\qquad-\infty<x<\infty. Let \mathbf{Y}_{n}=\frac{1}{n}\sum_{i=1}^{n}\mathbf{X}_{i}. $ Find the pdf of $ \mathbf{Y}_{n} $ . Describe how the pdf of $ \mathbf{Y}_{n} $ depends on $ n $ . Does the sequence $ \mathbf{Y}_{1},\mathbf{Y}_{2},\mathbf{Y}_{3},\cdots $ converge in distribution? If yes, what is the distribution of the random variable it converges to?

Note

You can see the definition of the converge in distribution. Furthermore, you have to know the characteristic function of Cauchy distributed random varaible.

Solution

According to the characteristic function of Cauchy distributed random variable,

$ \Phi_{\mathbf{X}}\left(\omega\right)=e^{-\left|\omega\right|}. $

$ \Phi_{\mathbf{Y}_{n}}\left(\omega\right)=E\left[\exp\left\{ i\omega\mathbf{Y}_{n}\right\} \right]=E\left[\exp\left\{ i\frac{\omega}{n}\sum_{k=1}^{n}\mathbf{X}_{k}\right\} \right]=E\left[\prod_{k=1}^{n}\exp\left\{ i\frac{\omega}{n}\mathbf{X}_{k}\right\} \right] $$ =E\left[\exp\left\{ i\frac{\omega}{n}\mathbf{X}\right\} \right]^{n}=\Phi_{\mathbf{X}}\left(\frac{\omega}{n}\right)^{n}=\left[e^{-\left|\omega/n\right|}\right]^{n}=e^{-\left|\omega\right|}. $

$ f_{\mathbf{Y}_{n}}\left(\omega\right)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-i\omega y}e^{-\left|\omega\right|}d\omega=\frac{1}{2\pi}\left[\int_{-\infty}^{0}e^{-i\omega y}e^{\omega}d\omega+\int_{0}^{\infty}e^{-i\omega y}e^{-\omega}d\omega\right] $$ =\frac{1}{2\pi}\left[\int_{-\infty}^{C}e^{\omega\left(1-iy\right)}+\int_{C}^{\infty}e^{-\omega\left(1+iy\right)}d\omega\right]=\frac{1}{2\pi}\left[\frac{1}{1-iy}e^{\omega\left(1-iy\right)}\biggl|_{-\infty}^{C}+\frac{-1}{1+iy}e^{-\omega\left(1+iy\right)}\biggl|_{C}^{\infty}\right] $$ =\frac{1}{2\pi}\left[\frac{1}{1-iy}+\frac{1}{1+iy}\right]=\frac{1}{2\pi}\left[\frac{1+iy+1-iy}{1+y^{2}}\right]=\frac{1}{2\pi}\cdot\frac{2}{1+y^{2}}=\frac{1}{\pi\left(1+y^{2}\right)}. $

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