# show that V(X) = lambda

• Jan 11th 2009, 03:30 AM
Animalxxv
show that V(X) = lambda
I have a problem and I don't really want the solution just a push in the right direction.

given that $\displaystyle \Pr(X=r) = \frac{e^{\lambda}\cdot{\lambda ^{r}}}{r!}$ r = 0,1,.....

show that Varience of X equals $\displaystyle \lambda$.

• Jan 11th 2009, 03:40 AM
Moo
Hello,
Quote:

Originally Posted by Animalxxv
I have a problem and I don't really want the solution just a push in the right direction.

given that P(X=r) = $\displaystyle \frac{e^{lambda}\cdot{lambda ^{r}}}{r!}$ r = 0,1,.....

show that Varience of X equals lambda.

Note that $\displaystyle \mathbb{E}(f(X))=\sum_{r=0}^\infty \mathbb{P}(X=r) f(r)$
where $\displaystyle \mathbb{E}$ is the expectation.

We know that the variance V is given by this formula :
$\displaystyle V(X)=\mathbb{E}(X^2)-\mathbb{E}(X)^2$

$\displaystyle V(X)=\sum_{r=0}^\infty \mathbb{P}(X=r) r^2-\left(\sum_{r=0}^\infty \mathbb{P}(X=r)r\right)^2$

Can you do it ?

Edit : there are other ways such as using generating functions. This was the most basic one that came in my mind ><
• Jan 11th 2009, 03:42 AM
Animalxxv
Yep thank you very much for your help(Rofl) (Hi)(Hi)(Hi)
• Jan 11th 2009, 03:50 AM
mr fantastic
Quote:

Originally Posted by Animalxxv
I have a problem and I don't really want the solution just a push in the right direction.

given that $\displaystyle \Pr(X=r) = \frac{e^{\lambda}\cdot{\lambda ^{r}}}{r!}$ r = 0,1,.....

show that Varience of X equals $\displaystyle \lambda$.