# Poisson Distribution homework

1. Feb 17, 2009

### buzzmath

1. The problem statement, all variables and given/known data
An actuary has discovered that policyholders are three times as likely to file two claims as to file four claims. If the number of claims filed has a Poisson distribution, what is the variance of the number of claims filed?

2. Relevant equations[/]
P(x)=e^(-lamda)*(lamda^x)/x! Var(x)=lamda

3. The attempt at a solution
I know that P(2)=3P(4) and that P(2)=e^(-lamda)*(lamda^2)/2 and that P(4)=e^(-lamda)*(lamda^4)/4! So I set these equal to find lamda and that is the variance. The solution given to me is P(2)=e^(lamda)*(lamda^2)/2=3*e^(-lamda)*(lamda^4)/4!=3*P(4)*24*lamda^2=6*lamda^4 I don't understand the last two equalities. Any help? where do the 24*lamda^2 comefrom and how do they get 6*lamda^4?
Thanks

2. Feb 19, 2009

### buzzmath

Is this a typo? if so how would I go about solving the problem using the method I've started? or any other way?
Thanks

3. Feb 19, 2009

### Staff: Mentor

From the given information,
$$P(2) = \frac{e^{-\lambda}\lambda^2}{2!} = 3P(4) = 3\frac{e^{-\lambda}\lambda^4}{4!}$$
From this, we get
$$\frac{e^{-\lambda}\lambda^2}{2} = 3\frac{e^{-\lambda}\lambda^4}{24}$$
so
$$\frac{e^{-\lambda}\lambda^2}{2} = \frac{e^{-\lambda}\lambda^4}{8}$$
or
$$4e^{-\lambda}\lambda^2} = e^{-\lambda}\lambda^4}$$
That works out to $4\lambda^2 = \lambda^4$
Can you take it from there?

BTW, the name of this Greek letter is lambda.