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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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  • $\begingroup$ For (i), just use the chain rule for differentiation, remembering that $$\frac{d}{dx}\Phi(g(x))=\phi(g(x))g'(x) ~~\mathrm{where}~~ \phi(y)=\frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}}.$$ $\endgroup$ – Dilip Sarwate Apr 23 '13 at 3:12
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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 $$ h_{Y}(y) = \frac{1}{\sqrt{2 \pi}}e^{-\frac{\log^2 y}{2}}\frac{1}{|y|} $$

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  • $\begingroup$ This is called function of the single random variable, that can probably be referred to as transformation. $\endgroup$ – Alex Apr 22 '13 at 20:48
  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$ – Alex Apr 22 '13 at 20:48
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    $\begingroup$ Could you possibly go into a little more detail? I.e. how did you get from $log Y$ to the following expression? $\endgroup$ – Mathlete Apr 22 '13 at 20:48
  • $\begingroup$ Just log both sides. Then take a derivative wrt to $y$ $\endgroup$ – Alex Apr 22 '13 at 20:49
  • $\begingroup$ I still don't understand - it's the transition to the expression for $f_X(y)$ that I'm having trouble with. $\endgroup$ – Mathlete Apr 22 '13 at 20:51

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