Answer to Question #265866 in Statistics and Probability for talie

Question

The random variable Y = LogX has N(10,4) distribution. Find (a) The pdf of X (b) Mean and variance of X (c) P ( X ≤ 1000

Answer 1

for distribution N(10,4):

$\mu=10, \sigma^2=4$

a)

for pdf of Y:

$f_Y(y)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{1}{2}(\frac{y-\mu}{\sigma})^2}=\frac{1}{2\sqrt{2\pi}}e^{-\frac{1}{2}(\frac{y-10}{2})^2}$

$\int^{\infin}_{-\infin}f_Y(y)dy=1$

$dy=e^{-y}dx$

then:

$f_Y(y)=e^yf_{X>0}(x)=2e^yf_{X}(x)$

$f_{X}(x)=f_Y(y)/(2e^y)=\frac{1}{4x\sqrt{2\pi}}e^{-\frac{1}{2}(\frac{logx-10}{2})^2}$

b)

mean of X:

$E(X)=\int^{\infin}_0xf_X(x)dx=\frac{1}{4\sqrt{2\pi}}\int^{\infin}_0e^{-\frac{1}{2}(\frac{logx-10}{2})^2}dx=$

$=\frac{e^{12}}{4}erf(\frac{lnx-14}{2\sqrt2})|^{\infin}_0=e^{12}/4$

where erf is the error function.

variance of X:

$Var(X)=E(X^2)-(E(X))^2$

$E(X^2)=\int^{\infin}_0x^2f(x)dx=\frac{1}{4\sqrt{2\pi}}\int^{\infin}_0xe^{-\frac{1}{2}(\frac{logx-10}{2})^2}dx=$

$=\frac{e^{28}}{4}erf(\frac{lnx-18}{2\sqrt2})|^{\infin}_0=e^{28}/4$

$Var(X)=e^{28}/4-e^{24}/4$

c)

$P ( X ≤ 1000)=\int^{1000}_0f(x)dx=\frac{1}{4\sqrt{2\pi}}\int^{1000}_0\frac{e^{-\frac{1}{2}(\frac{logx-10}{2})^2}}{x}dx=$

$=\frac{1}{4}erf(\frac{lnx-10}{2\sqrt2})|^{1000}_0=\frac{1}{4}(erf(\frac{ln1000-10}{2\sqrt2})+1)=0.0765$