Standard Normal Distribution Transformation Z=lnY

Question

I'm not sure if my approach to this problem is correct and I need help

I need to apply $Z=\ln{Y}$ to the following standard normal distribution and then find the distribution of $Y$

$f(z)=\frac1{\sqrt{2\pi}}e^{\frac{-z^2}{2}}~$

my approach was like this

$z=\ln{y}$

$y=e^z$

$dz=\frac1{y}dy$

then the distribution of $Y$ is

$f(y)=\frac1{y\sqrt{2\pi}}e^{\frac{-(\ln{y})^2}{2}}$

is my approach correct?

I am a beginner at statistics and I want to get better so any help is welcomed

Answer 1

If $Z \text{~}N(0,1)$ variable with $Y=e^Z$.

Then $ F_Y(y) = Pr(Y<y) = Pr(e^Z<y) = Pr(Z<ln(y)) = F_Z(ln(y)) $ ) ,

Now note that $ S_Z = (-\infty,\infty) $ , so $ S_Y = (0,\infty) $. Hence,

$ F_Y(y) = \begin{cases} F_Z(ln(y)), & \text{if }{y \in (0,\infty)} \\ 0, & \text{otherwise.} \end{cases} $

Then observe that the pdf of Y is $ f_Y(y) = { dF_Y(y) \over dy } = \begin{cases} { dF_Z(ln(y)) \over dy } , & \text{if }{y \in (0,\infty)}, \\ 0, & \text{otherwise.} \end{cases}= \begin{cases} { { 1 \over y }{ f_Z(ln(y)) } } , & \text{if }{y \in (0,\infty)}, \\ 0, & \text{otherwise.} \end{cases}= \begin{cases} { 1 \over { y\sqrt { 2\pi } } }{ e^{-(ln(y))^2 \over 2 } }, & \text{if }{y \in (0,\infty)}, \\ 0, & \text{otherwise.} \end{cases},$
(what you got).

So yes the approach is correct-ish, what I meant before is that you didn't have the correct structure. What I said before about the distribution, ignore... I had a derp moment.. >.<"