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Thread: need help variance of poisson dist.

  1. #1
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    need help variance of poisson dist.

    Can anyone provide a proof for the variance of poisson distribution?
    Show that it is Var(N(t)) = lamda*t without using the Bernoulli distribution and independence way..( which is the typical way of summations or expectations)
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  2. #2
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    Quote Originally Posted by yoavperel View Post
    Can anyone provide a proof for the variance of poisson distribution?
    Show that it is Var(N(t)) = lamda*t without using the Bernoulli distribution and independence way..( which is the typical way of summations or expectations)
    Actually the variance = mean = $\displaystyle \lambda$.

    To find the variance, all you need to do is calculate $\displaystyle E(X^2)$.

    Then $\displaystyle Var(X) = E(X^2) - [E(X)]^2 = E(X^2) - \lambda^2$.


    $\displaystyle E(X^2) = \sum_{x=0}^{\infty} x^2 \frac{e^{-\lambda} \lambda^x}{x!} = \lambda \, e^{-\lambda} \sum_{x=1}^{\infty} \frac{x \, \lambda^{x-1}}{(x-1)!}$


    $\displaystyle = \lambda \, e^{-\lambda} \left( 1 + \frac{2 \lambda}{1} + \frac{3 \lambda^2}{2!} + \frac{4 \lambda^3}{3!} + ... \right)$


    $\displaystyle = \lambda \, e^{-\lambda} \left( \left[ 1 + \frac{\lambda}{1} + \frac{\lambda^2}{2!} + \frac{\lambda^3}{3!} + ... \right] + \left[ \frac{\lambda}{1} + \frac{2 \lambda^2}{2!} + \frac{3 \lambda^3}{3!} + ... \right] \right)$


    $\displaystyle = \lambda \, e^{-\lambda} \left( \left[ e^{\lambda} \right] + \lambda \left[ 1 + \frac{2 \lambda}{2!} + \frac{3 \lambda^2}{3!} + ... \right] \right)$


    $\displaystyle = \lambda \, e^{-\lambda} \left( e^{\lambda} + \lambda \left[ 1 + \lambda + \frac{\lambda^2}{2!} + ... \right] \right)$


    $\displaystyle = \lambda \, e^{-\lambda} \left( e^{\lambda} + \lambda \left[ e^{\lambda} \right] \right)$


    $\displaystyle = \lambda \, e^{-\lambda} \left( e^{\lambda} + \lambda\, e^{\lambda}\right) = \lambda (1 + \lambda) = \lambda + \lambda^2$.
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  3. #3
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    There is another way I've learnt, using generating function (not sure you know it...)

    $\displaystyle \chi=\mathbb{N}$

    $\displaystyle f(t)=\mathbb{E}(t^X)=\sum_{x \in \chi} t^x \mathbb{P}(X=x)$

    $\displaystyle =\sum_{x \in \chi} t^x \cdot e^{-\lambda} \cdot \frac{\lambda^x}{x!}$

    $\displaystyle =e^{-\lambda} \sum_{x \in \chi} \frac{(\lambda t)^x}{x!}$

    $\displaystyle =e^{-\lambda} \cdot e^{\lambda t}$

    $\displaystyle \boxed{f(t)=e^{\lambda(t-1)}}$

    ------------------------

    Now, we know that $\displaystyle \mathbb{E}(X)=f'(1)$

    and $\displaystyle \mathbb{E}(X(X-1))=f''(1)$

    Linearity of mean : $\displaystyle \mathbb{E}(X^2)=\mathbb{E}(X(X-1))+\mathbb{E}(X)=f''(1)+f'(1)$

    Therefore $\displaystyle \text{var}(X)=\mathbb{E}(X^2)-(\mathbb{E}(X))^2=f''(1)+f'(1)-[f'(1)]^2$
    Last edited by Moo; Jul 5th 2008 at 06:10 AM.
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    Quote Originally Posted by Moo View Post
    There is another way I've learnt, using generating function (not sure you know it...)

    $\displaystyle \chi=\mathbb{N}$

    $\displaystyle f(t)=\mathbb{E}(t^X)=\sum_{x \in \chi} t^x \mathbb{P}(X=x)$

    $\displaystyle =\sum_{x \in \chi} t^x \cdot e^{-\lambda} \cdot \frac{\lambda^x}{x!}$

    $\displaystyle =e^{-\lambda} \sum_{x \in \chi} \frac{(\lambda t)^x}{x!}$

    $\displaystyle =e^{-\lambda} \cdot e^{\lambda t}$

    $\displaystyle \boxed{f(t)=e^{\lambda(t-1)}}$

    ------------------------

    Now, we know that $\displaystyle \mathbb{E}(X)=f'(1)$

    and $\displaystyle \mathbb{E}(X(X-1))=f''(1)$

    Linearity of mean : $\displaystyle \mathbb{E}(X^2)=\mathbb{E}(X(X-1))+\mathbb{E}(X)=f''(1)+f'(1)$

    Therefore $\displaystyle \text{var}(X)=\mathbb{E}(X^2)-(\mathbb{E}(X))^2=f''(1)+f'(1)-[f'(1)]^2$
    I think I would have used the moment generating function $\displaystyle M(t)=e^{-\lambda}e^{\lambda e^t}$, which is just as easy to find as the generating function.

    Then the n-th moment is $\displaystyle E(X^n)=\left. D^{(n)}M(t) \right|_{t=0}$ etc ...

    RonL
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    Thank you all , it's very helpfull ..

    ..
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  6. #6
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    x^2 question

    Quote Originally Posted by mr fantastic View Post
    Actually the variance = mean = $\displaystyle \lambda$.

    To find the variance, all you need to do is calculate $\displaystyle E(X^2)$.

    Then $\displaystyle Var(X) = E(X^2) - [E(X)]^2 = E(X^2) - \lambda^2$.


    $\displaystyle E(X^2) = \sum_{x=0}^{\infty} x^2 \frac{e^{-\lambda} \lambda^x}{x!} = \lambda \, e^{-\lambda} \sum_{x=1}^{\infty} \frac{x \, \lambda^{x-1}}{(x-1)!}$


    $\displaystyle = \lambda \, e^{-\lambda} \left( 1 + \frac{2 \lambda}{1} + \frac{3 \lambda^2}{2!} + \frac{4 \lambda^3}{3!} + ... \right)$


    $\displaystyle = \lambda \, e^{-\lambda} \left( \left[ 1 + \frac{\lambda}{1} + \frac{\lambda^2}{2!} + \frac{\lambda^3}{3!} + ... \right] + \left[ \frac{\lambda}{1} + \frac{2 \lambda^2}{2!} + \frac{3 \lambda^3}{3!} + ... \right] \right)$


    $\displaystyle = \lambda \, e^{-\lambda} \left( \left[ e^{\lambda} \right] + \lambda \left[ 1 + \frac{2 \lambda}{2!} + \frac{3 \lambda^2}{3!} + ... \right] \right)$


    $\displaystyle = \lambda \, e^{-\lambda} \left( e^{\lambda} + \lambda \left[ 1 + \lambda + \frac{\lambda^2}{2!} + ... \right] \right)$


    $\displaystyle = \lambda \, e^{-\lambda} \left( e^{\lambda} + \lambda \left[ e^{\lambda} \right] \right)$


    $\displaystyle = \lambda \, e^{-\lambda} \left( e^{\lambda} + \lambda\, e^{\lambda}\right) = \lambda (1 + \lambda) = \lambda + \lambda^2$.
    Mr. Fantastic,
    Will you tell me why the x^2 doesn't affect the f(x) in the formula for the expected value? I have been working on this problem for almost a week and have just discovered your solution.

    Thanks,
    Yvonne
    Last edited by yvonnehr; Sep 28th 2009 at 09:13 PM.
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  7. #7
    Moo
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    Quote Originally Posted by yvonnehr View Post
    Mr. Fantastic,
    Will you tell me why the x^2 doesn't affect the f(x) in the formula for the expected value? I have been working on this problem for almost a week and have just discovered your solution.

    Thanks,
    Yvonne
    Hello,

    Because $\displaystyle \text{Var}(X)=\mathbb{E}(X^2)-(\mathbb{E}(X))^2=\mathbb{E}(X^2)-\lambda^2$

    This is why in the end you don't have $\displaystyle \lambda^2$ anymore.
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  8. #8
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    To Mr. Fantastic

    Thanks a lot, i hv been thinking of this proof in other methods and cannot figure it out until i found this simple one.
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