Question

11-5 (20) Assume X1.X2Xn are IID random variables, each with density fx (x)-6x-(0+1)n(x-1) , where θ > 0 and u(x) İs the unit step function. Use the CLT to find the approximate pdf of the random variable Z- In((X,X2X]. What happens as n Hint: make a change of variable x -ey and integration by parts, or results at the end of this problem set.

Answer 1

Given are IID random variables with PDF,

$f_X\left ( x \right )=\theta x^{-\theta -1};x>1,\theta >0$

Now the PDF of the random variable $Y=g\left ( X \right )=\ln X$ is

$f_Y\left ( y \right )=f_X\left ( g^{-1}\left ( y \right ) \right )\left | \frac{\mathrm{d} g^{-1}\left ( y \right ) }{\mathrm{d} y} \right |$

Here $X=e^Y\Rightarrow g^{-1}\left ( y \right )=e^y$ . Thus the PDF of $Y=g\left ( X \right )=\ln X$ is

$f_Y\left ( y \right )=\theta \left (e^y \right )^{-\theta -1}\left | \frac{\mathrm{d} e^y }{\mathrm{d} y} \right |\\ f_Y\left ( y \right )=\theta e^{-\theta y};y>0$

Thst is the PDF of $Y=g\left ( X \right )=\ln X$ is exponential.

Consider the random variable

$Z=\ln\left ( X_1X_2... X_n \right )^{1/n}\\ Z=\frac{\ln X_1+\ln X_2+...+\ln X_n}{n}\\ Z=\frac{Y_1+Y_2+...+Y_n}{n}$

According to CLT, the distribution of $Z=\frac{Y_1+Y_2+...+Y_n}{n}$ is approximately normal with

$\mu =E\left ( Y \right ),\sigma ^2=Var\left ( Y \right )/n$

We know the mean of exponential random variable $E\left ( Y \right )=1/\theta ,Var\left ( Y \right )=1/\theta ^2$ .

Thus the approximate PDF of

${\color{Blue} Z=\ln\left ( X_1X_2... X_n \right )^{1/n}\sim N\left ( 1/\theta ,1/\left (n\theta ^2 \right )\right )}$

When $n\rightarrow \infty$ , the above PDF becomes,

$Z=\ln\left ( X_1X_2... X_n \right )^{1/n}\sim \lim_{n\rightarrow \infty}N\left ( 1/\theta ,1/\left (n\theta ^2 \right )\right )$

So the distribution is $f_Z\left ( z \right )= \delta \left ( z-1/\theta \right );-\infty <z<\infty$

The PDF becomes Dirac's delta function.

Kindly upvote.