Question

3. Let N = R and P be the probability distribution on (R, B(R)) with density 1 -131 XER. Put X(w):=w2, WEN. (a) Describe o(X) (the o-algebra on 2 generated by X). Justify your answer. (b) Derive the distribution function Fx of the random variable X. (c) Compute the mean EX and moment generating function y(t) == EetX 500, t> 0, for the random variable (RV) X. (3+3+3 = 9]

Answer 1

a)

The $\sigma-algebra \hspace{0.1 cm} \sigma(X)$ is set of subsets of $\Omega =\mathbb{R}^{2}$ such that for $w \in \sigma(X)=>w^{c}\in \sigma(X)$ .

b)

Let Y follow a double exponential distribution with parameter 1(that is Y has the pdf given in the question).

Here $X=Y^{2}$

Jacobian:$J=\pm \frac{1}{2\sqrt{x}}$

PDF of X:

$f_{X}(x)=\frac{1}{4\sqrt{x}}e^{-\sqrt{x}},x\geq 0, and , 0,otherwise$

CDF of X:

$F_{X}(x)=0,x <0,and ,\int_{0}^{x}f_{X}(t)dt,x\geq 0\newline =0,x <0,and ,\int_{0}^{x}\frac{1}{4\sqrt{t}}e^{-\sqrt{t}}dt,x\geq 0\newline =0,x <0,and ,\int_{0}^{\sqrt{x}}\frac{1}{2}e^{-z}dt,x\geq 0, substituting \hspace{0.1 cm}z=\sqrt{x}\newline =0,x <0,and ,\frac{1}{2}(1-e^{-\sqrt{x}}),x\geq 0$

c)

$E(X)=\int_{0}^{\infty}\frac{1}{4\sqrt{t}}te^{-\sqrt{t}}dt=\int_{0}^{\infty}\frac{1}{4}\sqrt{t}e^{-\sqrt{t}}dt=\int_{0}^{\infty}\frac{1}{2}z^{2}e^{-z}dt,\newline subtituting\hspace{0.1 cm}z=\sqrt{t},\newline =\frac{\Gamma(3)}{2}=1$

$E(e^{tX})=\int_{0}^{\infty}\frac{1}{4\sqrt{x}}e^{tx}e^{-\sqrt{x}}dx=\int_{0}^{\infty}\frac{1}{2}e^{tz^{2}-z}dz,\newline subtituting\hspace{0.1 cm}z=\sqrt{t}\newline =\int_{0}^{\infty}\frac{1}{2}e^{-(\sqrt{t}z-\frac{1}{2\sqrt{t}})^{2}}dz=\int_{0}^{\infty}\frac{1}{2}e^{-t(z-\frac{1}{2t})^{2}}dz\newline =\int_{0}^{\infty}\frac{1}{2}e^{-t(z-\frac{1}{2t})^{2}}dz\newline =\frac{1}{2}e^{-\frac{1}{4t}}\frac{1}{\sqrt{2t}}\int_{0}^{\infty}\frac{1}{\sqrt{2\pi \frac{1}{2t}}}e^{-t(z-\frac{1}{2t})^{2}}dz,\frac{1}{\sqrt{2\pi \frac{1}{2t}}}e^{-t(z-\frac{1}{2t})^{2}}\equiv N(\frac{1}{2t},\frac{1}{2t})\newline =\frac{1}{2}e^{-\frac{1}{4t}}\frac{1}{\sqrt{2t}}(\int_{0}^{\frac{1}{2t}}\frac{1}{\sqrt{2\pi \frac{1}{2t}}}e^{-t(z-\frac{1}{2t})^{2}}dz+\int_{\frac{1}{2t}}^{\infty}\frac{1}{\sqrt{2\pi \frac{1}{2t}}}e^{-t(z-\frac{1}{2t})^{2}}dz)\newline$

$=\frac{1}{2}e^{-\frac{1}{4t}}\frac{1}{\sqrt{2t}}(\Phi(0)-\Phi(-\frac{1}{\sqrt{2t}})+\frac{1}{2})\newline =\frac{1}{2}e^{-\frac{1}{4t}}\frac{1}{\sqrt{2t}}(1-\Phi(-\frac{1}{\sqrt{2t}}))$