# need help variance of poisson dist.

• Jul 5th 2008, 03:37 AM
yoavperel
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)
• Jul 5th 2008, 05:42 AM
mr fantastic
Quote:

Originally Posted by yoavperel
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 $\displaystyle Var(X) = E(X^2) - [E(X)]^2 = E(X^2) - \lambda^2$.

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

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

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

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

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

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

$\displaystyle = \lambda \, e^{-\lambda} \left( e^{\lambda} + \lambda\, e^{\lambda}\right) = \lambda (1 + \lambda) = \lambda + \lambda^2$.
• Jul 5th 2008, 05:56 AM
Moo
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 $\displaystyle \text{var}(X)=\mathbb{E}(X^2)-(\mathbb{E}(X))^2=f''(1)+f'(1)-[f'(1)]^2$
• Jul 5th 2008, 12:58 PM
CaptainBlack
Quote:

Originally Posted by Moo
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
• Jul 5th 2008, 08:29 PM
yoavperel
Thank you all , it's very helpfull ..
..
• Sep 28th 2009, 09:00 PM
yvonnehr
x^2 question
Quote:

Originally Posted by mr fantastic
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
• Oct 1st 2009, 09:09 AM
Moo
Quote:

Originally Posted by yvonnehr
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.
• Jul 26th 2010, 02:48 AM
undine99
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.(Clapping)