# Density function ln(x)

• April 29th 2009, 05:57 AM
memanuelson
D
If ln(x) has a normal distribution N(0,1) how do I find and prove the distribution for x?

And if S(t)=S(0)exp((mu-(sigma^2)/2)t+sigma*sqrt(tao)*G), where G N(0,1) what is the density function for S(t)?
• April 30th 2009, 12:02 AM
memanuelson
How do you guys do to make those nice equations?
• April 30th 2009, 09:40 AM
Moo
Hello,
Quote:

Originally Posted by memanuelson
And if S(t)=S(0)exp((mu-(sigma^2)/2)t+sigma*sqrt(tao)*G), where G N(0,1) what is the density function for S(t)?

Click on the "picture" to see the code ;)
Anyway, there's a latex sub-forum, where you may be able to learn some things about writing equations.

$S(t)=S(0) \exp\left(\frac{\mu-\sigma^2}{2} \cdot t+\sigma \sqrt{\tau} G\right)$

But what is "tao" ?

Quote:

If ln(x) has a normal distribution N(0,1) how do I find and prove the distribution for x?
The pdf of $T=\ln(x)$ is $\frac{1}{\sqrt{2\pi}} \cdot e^{-\frac{t^2}{2}}$

Now, I can never remember the method you guys use... But mine is similar, using the Jacobian of the transformation.

For any bounded and continuous function f, we have :
$\mathbb{E}(f(X))=\mathbb{E}(f(e^T))=\int_{-\infty}^\infty f(e^t) \cdot \frac{1}{\sqrt{2\pi}} \cdot e^{-\frac{t^2}{2}} ~dt$

Make the substitution $x=e^t$ :
$dx=e^t ~dt=x dt$

So $\mathbb{E}(f(X))=\int_0^\infty f(x) \cdot \frac{1}{\sqrt{2\pi}} \cdot e^{-\frac{(\ln(x))^2}{2}} \cdot \frac 1x ~dx$

Thus the pdf of $X$ is $\frac{1}{\sqrt{2\pi}} \cdot e^{-\frac{(\ln(x))^2}{2}} \cdot \frac 1x$, for x>0
Which is the pdf of the log-normal distribution, with parameters (0,1).
• April 30th 2009, 11:59 PM
memanuelson
Thank you so much!