If $X \sim N(0,1)$ and $Y = e^X$, find the PDF of $Y$ using the two methods:
(i) Find the CDF of of $Y$ and then differentiate. Use the notation $\Phi(x)$ and $\phi(x)$ for the CDF and PDF of $X$ respectively. You may use the fact that $\phi(x) = \Phi'(x)$.
So I'm not sure how to differentiate $\Phi\big(\dfrac{\ln x-\mu}{\sigma} \bigg)$ to get $\dfrac{1}{x\sigma\sqrt{2\pi}}e^-\frac{(\ln x-\mu)^2}{2\sigma^2}$
(ii) Use the transformation formula.
I'm not sure where to even begin with this one.
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If $Y=e^X$, then $\varphi^{-1}(Y)= \log Y$. Hence, $f_X (y) =\frac{1}{\sqrt{2 \pi}}e^{-\frac{\log^2 y}{2}}$ and differentiate $|\frac{d \varphi^{-1}(Y)}{dy} | = |\frac{1}{Y}|$. Hence, the pdf of $Y$ is $$ h_{Y}(y) = \frac{1}{\sqrt{2 \pi}}e^{-\frac{\log^2 y}{2}}\frac{1}{|y|} $$
Alex
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This is called function of the single random variable, that can probably be referred to as transformation. – Alex Apr 22 '13 at 20:48
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http://en.wikipedia.org/wiki/Random_variable#Functions_of_random_variables – Alex Apr 22 '13 at 20:48
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2Could you possibly go into a little more detail? I.e. how did you get from $log Y$ to the following expression? – Mathlete Apr 22 '13 at 20:48
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I still don't understand - it's the transition to the expression for $f_X(y)$ that I'm having trouble with. – Mathlete Apr 22 '13 at 20:51
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You appear to have combined (i) and (ii) into one, so I'm getting confused... – Mathlete Apr 22 '13 at 20:57
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1No, I didn't use CDF of $X$ anywhere and I didn't differentiate it either. I differentiated $X= \log Y$, which is necessary for the transformation formula. – Alex Apr 22 '13 at 21:02
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\mathrm{where}\phi(y)=\frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}}.$$ – Dilip Sarwate Apr 23 '13 at 03:12