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)?
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)?
Hello,
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Anyway, there's a latex sub-forum, where you may be able to learn some things about writing equations.
$\displaystyle S(t)=S(0) \exp\left(\frac{\mu-\sigma^2}{2} \cdot t+\sigma \sqrt{\tau} G\right)$
But what is "tao" ?
The pdf of $\displaystyle T=\ln(x)$ is $\displaystyle \frac{1}{\sqrt{2\pi}} \cdot e^{-\frac{t^2}{2}}$If ln(x) has a normal distribution N(0,1) how do I find and prove the distribution for x?
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 :
$\displaystyle \mathbb{E}(f(X))=\mathbb{E}(f(e^T))=\int_{-\infty}^\infty f(e^t) \cdot \frac{1}{\sqrt{2\pi}} \cdot e^{-\frac{t^2}{2}} ~dt$
Make the substitution $\displaystyle x=e^t$ :
$\displaystyle dx=e^t ~dt=x dt$
So $\displaystyle \mathbb{E}(f(X))=\int_0^\infty f(x) \cdot \frac{1}{\sqrt{2\pi}} \cdot e^{-\frac{(\ln(x))^2}{2}} \cdot \frac 1x ~dx$
Thus the pdf of $\displaystyle X$ is $\displaystyle \frac{1}{\sqrt{2\pi}} \cdot e^{-\frac{(\ln(x))^2}{2}} \cdot \frac 1x$, for x>0
Which is the pdf of the log-normal distribution, with parameters (0,1).