Answer to Question #330807 in Statistics and Probability for Tarie

Question

The random variable Y = LogX has N(12,6) distribution. Find (a) The pdf of X (b) Mean and variance of X (c) P ( X ≤ 1000

Answer 1

a). The pdf of $X$ is a density of $X$. The variable $Y=log\,X$ has a normal distribution with parameters $\mu=12$ and $\sigma^2=6$. Variable $X>0$, since it appears under the function $log$. Therefore, $P(X\leq0)=0$. The distribution of $X$ is similar to the log-normal distribution.

We have: $P(log\,X\leq z)=P(Y\leq z)=\int_{-\infty}^zg(y)dy,$ where $g(y)=\frac{1}{\sqrt{12\pi}}e^{-\frac12\left(\frac{y-12}{\sqrt{6}}\right)^2}$. $P(X\leq s)=P(log\,X\leq log\,s)=\int_{0}^{log(s)}g(y)dy$. The derivative(due to the Leibniz’s Theorem) is: $f(s)=(P(X\leq s))'=g(log(s))\frac{1}{s\,\,ln(10)}=\frac{1}{(\sqrt{12\pi})\,\,(ln(10))s}e^{-\frac12\left(\frac{log(s)-12}{\sqrt{6}}\right)^2}$.

Thus, the pdf of $X$ is: $f(s)=\frac{1}{(\sqrt{12\pi})\,\,(ln(10))s}e^{-\frac12\left(\frac{log(s)-12}{\sqrt{6}}\right)^2}$

b). $E[X]=\int_{0}^{+\infty}\frac{1}{(\sqrt{12\pi})\,\,(ln(10))}e^{-\frac12\left(\frac{log(s)-12}{\sqrt{6}}\right)^2}ds.$ Make the change of variables: $x=ln(s)$. Then, we receive: $s=e^x$. $E[X]=\int_{-\infty}^{+\infty}\frac{e^x}{(\sqrt{12\pi})\,\,(ln(10))}e^{-\frac12\left(\frac{\frac{x}{ln(10)}-12}{\sqrt{6}}\right)^2}dx=e^{\frac1{12}(a^2-144)}\int_{-\infty}^{+\infty}\frac{1}{\,ln\,(10)\sqrt{6\pi}\,\,}e^{-\frac12\left({\frac{x-a\,ln(10)}{\sqrt{6}\,ln(10)}}\right)^2}dx,$

where $a=6\,ln(10)+3$. The last integral is equal to $1$ (it becomes the Gaussian integral after simple changes). Thus, we get: $E[X]=e^{\frac1{12}((6\,ln(10)+3)^2-144)}$.

$Var[X]=E[X^2]-(E[X])^2$.

$E[X^2]=\int_{0}^{+\infty}\frac{s}{(\sqrt{12\pi})\,\,(ln(10))}e^{-\frac12\left(\frac{log(s)-12}{\sqrt{6}}\right)^2}ds.$ Make the change $s=e^x$. We get: $E[X^2]=\int_{-\infty}^{+\infty}\frac{e^{2x}}{(\sqrt{12\pi})\,\,(ln(10))}e^{-\frac12\left(\frac{\frac{x}{ln(10)}-12}{\sqrt{6}}\right)^2}dx=e^{\frac1{12}(b^2-144)}\int_{-\infty}^{+\infty}\frac{1}{\,ln\,(10)\sqrt{12\pi}\,\,}e^{-\frac12\left({\frac{x-b\,ln(10)}{\sqrt{6},ln(10)}}\right)^2}dx,$

where $b=12\,ln(10)+3.$ Thus, $Var[X]=e^{\frac1{12}((12\,ln(10)+3)^2-144)}-e^{\frac1{6}((6\,ln(10)+3)^2-144)}$.

We received: $E[X]=e^{\frac1{12}((6\,ln(10)+3)^2-144)}$, $Var[X]=e^{\frac1{12}((12\,ln(10)+3)^2-144)}-e^{\frac1{6}((6\,ln(10)+3)^2-144)}$

c). $P(X\leq1000)=\int_{0}^{1000}\frac{1}{(\sqrt{12\pi})\,\,(ln(10))s}e^{-\frac12\left(\frac{log(s)-12}{\sqrt{6}}\right)^2}ds\approx0.0001$ (it is rounded to 4 decimal places)