Thread: Density Functions
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Old May 15th, 2008, 07:23 AM
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Originally Posted by janvdl View Post
I've asked this before in a previous post, however, while not quite necessary then, the answer did not include full workings, and I think I sort of need that just now to make sure I understand this.
I'm writing another semester test tomorrow, so I would appreciate if someone could show me how this is done.



(Not sure whether it would be classified as "bumping" to bring up the old topic again, I decided to just as well post it in a new thread. Please show full workings.)

Thanks in advance guys
Let F(x) = \Pr \left( \frac{1}{2} M V^2 < x \right)

\Rightarrow F(x) = \Pr \left( -\sqrt{\frac{2x}{M}} < V < \sqrt{\frac{2x}{M}} \right)

= \frac{1}{\sigma \sqrt{2 \pi}} \, \int_{-\sqrt{\frac{2x}{M}} }^{\sqrt{\frac{2x}{M}} } e^{-u^2/(2\sigma^2)} \, du

= \frac{2}{\sigma \sqrt{2 \pi}} \, \int_{0}^{\sqrt{\frac{2x}{M}} } e^{-u^2/(2 \sigma^2)} \, du.

Therefore f(x) = \frac{dF}{dx} = \frac{2}{\sigma \sqrt{2 \pi}} \, e^{-x/(\sigma^2 M)} \, \left (\frac{1}{\sqrt{2Mx}} \right) \, , x > 0


where the Fundamental Theorem of Calculus and the chain rule have been used to get the derivative


= \frac{e^{-x/(\sigma^2 M)}}{\sigma \sqrt{\pi M x}} \, , x > 0

and f(x) = 0 for x < 0.


\int_{-\infty}^{+\infty} f(x) \, dx = 1 so I don't think I've slipped up anywhere here .......
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Last edited by mr fantastic; May 15th, 2008 at 07:44 AM. Reason: Had sd = 1 instead of sigma.
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