Communication, Networking, Signal and Image Processing (CS)

Question 1: Probability and Random Processes

August 2007

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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