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Thread: Density Functions

  1. #1
    Bar0n janvdl's Avatar
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    Density Functions

    A particle of mass $\displaystyle M$ has a random velocity, $\displaystyle V$, which is normally distributed with parameters $\displaystyle \mu = 0$ and $\displaystyle \sigma$. Find the density function of the kinetic energy.
    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
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  2. #2
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    Quote 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 $\displaystyle F(x) = \Pr \left( \frac{1}{2} M V^2 < x \right)$

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

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

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

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


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


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

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


    $\displaystyle \int_{-\infty}^{+\infty} f(x) \, dx = 1$ so I don't think I've slipped up anywhere here .......
    Last edited by mr fantastic; May 15th 2008 at 05:44 AM. Reason: Had sd = 1 instead of sigma.
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  3. #3
    Bar0n janvdl's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Let $\displaystyle F(x) = \Pr \left( \frac{1}{2} M V^2 < x \right)$

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

    $\displaystyle = \frac{1}{\sqrt{2 \pi}} \, \int_{-\sqrt{\frac{2x}{M}} }^{\sqrt{\frac{2x}{M}} } e^{-u^2/2} \, du$

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

    Therefore $\displaystyle f(x) = \frac{dF}{dx} = \frac{2}{\sqrt{2 \pi}} \, e^{-x/M} \, \left (\frac{1}{\sqrt{2Mx}} \right) \, $ , $\displaystyle x > 0$
    where the Fundamental Theorem of Calculus and the chain rule have been used to get the derivative

    $\displaystyle = e^{-x/M} \, \frac{1}{\sqrt{\pi M x}} \, $ , $\displaystyle x > 0$

    and f(x) = 0 for x < 0.
    Thank you Mr F, I will work through this now and try to understand
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