# Thread: need help variance of poisson dist.

1. ## 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)

2. 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 = $\lambda$.

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

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

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

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

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

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

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

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

$= \lambda \, e^{-\lambda} \left( e^{\lambda} + \lambda\, e^{\lambda}\right) = \lambda (1 + \lambda) = \lambda + \lambda^2$.

3. There is another way I've learnt, using generating function (not sure you know it...)

$\chi=\mathbb{N}$

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

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

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

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

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

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

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

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

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

Therefore $\text{var}(X)=\mathbb{E}(X^2)-(\mathbb{E}(X))^2=f''(1)+f'(1)-[f'(1)]^2$

4. Originally Posted by Moo
There is another way I've learnt, using generating function (not sure you know it...)

$\chi=\mathbb{N}$

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

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

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

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

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

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

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

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

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

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

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

RonL

..

6. ## x^2 question

Originally Posted by mr fantastic
Actually the variance = mean = $\lambda$.

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

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

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

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

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

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

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

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

$= \lambda \, e^{-\lambda} \left( e^{\lambda} + \lambda\, e^{\lambda}\right) = \lambda (1 + \lambda) = \lambda + \lambda^2$.
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

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

This is why in the end you don't have $\lambda^2$ anymore.

8. ## 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.