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Thread: Probability density function

  1. #1
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    Probability density function

    Hi, I have this question that I've been struggling with.

    Given $\displaystyle D$ is a random distance with probability density function $\displaystyle f_d$ and expected value $\displaystyle \mu$, then the sampled distance distribution has probability density function $\displaystyle g$, where

    $\displaystyle g(x)=\frac{xf_d(x)}{\mu}, x \geq 0$

    Find the form of the distance probability density function $\displaystyle f_d$ and its expected value $\displaystyle \mu$ if the sample distances are fitted by the probability distance function $\displaystyle g(x)=\frac{x^2 e^{-\frac{x}{\alpha}}}{2 \alpha^3}, x \geq 0$ and $\displaystyle \alpha = 8.1$.

    Anyone help will be appreciated. thanks.

    I have tried equating the two g(x) but I keep getting something ridiculous like showing $\displaystyle E_d(X) = \mu$
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  2. #2
    Moo
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    Hello,

    Do you agree that we have $\displaystyle \frac{f_d(x)}{\mu}=\frac{xe^{-x/\alpha}}{2\alpha^3}$

    Then $\displaystyle f_d(x)=\frac{\mu}{2\alpha^3}\cdot xe^{-x/\alpha}$, for any positive x.

    But we want $\displaystyle f_d(x)$ to be a pdf. So its integral over 0 to infinity should be 1. And this computation will give you $\displaystyle \mu$

    Any further question ?
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