Thread: Normally Distributed Random Variable-Plzzzz Help

A Normally distributed random variable with mean [mu] has a probability density function given by

[gamma]/2 Pi *sigma exp(-gamma^2/sigma*x-[mu]^2/2).

Its standard deviation is given by:



*Sorry I'm still learning how to use Latex

Originally Posted by KayPee

A Normally distributed random variable with mean [mu] has a probability density function given by

[gamma]/2 Pi *sigma exp(-gamma^2/sigma*x-[mu]^2/2).

Its standard deviation is given by:



*Sorry I'm still learning how to use Latex

Sorry but I can't make sense of the expression you have posted for the pdf.

If it's a normal distribution then it has the usual form found here: Normal distribution - Wikipedia, the free encyclopedia

So unless you can re-post a clearer expression I suggest you read the link, fit your pdf to the normal pdf given in the link and then read off what the variance and hence standard deviation is.

Originally Posted by mr fantastic

Sorry but I can't make sense of the expression you have posted for the pdf.

If it's a normal distribution then it has the usual form found here: Normal distribution - Wikipedia, the free encyclopedia

So unless you can re-post a clearer expression I suggest you read the link, fit your pdf to the normal pdf given in the link and then read off what the variance and hence standard deviation is.

γ/Sqrt(2πσ)exp(-γ/σ*(x-μ)^2/2))

Originally Posted by KayPee

γ/Sqrt(2πσ)exp(-γ/σ*(x-μ)^2/2))

Is it $\displaystyle \frac{y}{\sqrt{2 \pi \sigma}} e^{-\frac{y (x - \mu)^2}{2 \sigma}}$ ?

That can't be right. It has to be either $\displaystyle \frac{y}{\sqrt{2 \pi \sigma}} e^{-\frac{y^2 (x - \mu)^2}{2 \sigma}}$ or $\displaystyle \frac{\sqrt{y}}{\sqrt{2 \pi \sigma}} e^{-\frac{y (x - \mu)^2}{2 \sigma}}$

Assuming the former expression, then I would have thought that the comparison with the pdf in the link I gave you is straightforward:


$\displaystyle \frac{y}{\sqrt{2 \pi \sigma}} = \frac{1}{\sigma_X \sqrt{2 \pi}}$ therefore $\displaystyle \sigma_X = \, ....$

If it's the latter expression, then you do a similar thing.