Pdf of a function of a random variable

**Jame** · May 6th 2011, 06:55 PM

I am supposed to fin the pdf f $Y = e^X$ where $X$ is normal r.v. with parameters $\mu, \sigma^2$

The pdf of X is

$\frac{1}{ \sqrt{2\pi\sigma^2}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$

(Sorry I can't get this to appear in tex for some reason)

If I integrate this function from negative infinity to ln(y), then differeniate with respect to y, I should get the function I seek.

However I am not sure how to integrate the normal distribution function, could someone please explain to me some way I way go about doing that?

Thank you for your help.

**Sambit** · May 6th 2011, 10:54 PM

Let $F_Y(.)$ be cdf of $Y$ and $F_X(.)$ be cdf of $X$ ; $f_Y(.)$ and $f_X(.)$ are pdf of $Y$ and $X$ respectively.

$F_Y(y) = P(Y\leq y) = P(e^X\leq y) = P(X\leq \ln y) = F_X(\ln y)$

Hence,differentiating, $f_Y(y) = \frac{d}{dx}F_X(\ln y) = f_X(\ln y)\frac{d}{dx}\ln y = f_X(\ln y)\frac{1}{y}$
$= \frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(\ln y-\mu)^2}{2\sigma^2}}\frac{1}{y} = \frac{1}{y\sqrt{2\pi\sigma^2}}e^{-\frac{(\ln y-\mu)^2}{2\sigma^2}$

This is called Log-Normal distribution.

**matheagle** · May 7th 2011, 09:58 PM

minor correction, the derivative of F(y) wrt y is f(y)
you have d/dx, you mean d/dy.
Otherwise you would need the chain rule

This is what you meant....

Hence,differentiating, $f_Y(y) = \frac{d}{dy}F_X(\ln y) = f_X(\ln y)\frac{d}{dy}\ln y = f_X(\ln y)\frac{1}{y}$

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