The following question is what I was working on.
a% of the population has a risk of incurring damage that has a Poisson distribution with mean 1. Similarly, b% has a distribution with mean 2 and the rest (c%) has a distribution of with mean 3. A random person is chosen from this population and has a mean damage of 2.59 and a variance of 2.1. Find the probability of no claims occurring.
I understand that the weighed mean can be figured out as $$a(1)+b(2)+c(3)=2.1$$
However, I am not quite understanding why my book tells me that the second moment is found by $$a(2)+b(6)+c(12)=2.59+2.1^2$$
I understand the right hand side because it's simply
$$Var[x]+(E[X])^2=E[X]$$
But I though that the second moment should be $$a(1^2)+b(2^2)+c(3^2)$$
Can someone explain to me what is going on?
If you are measuring the second moment about zero using your $E[X^2] = Var[X]+(E[X])^2$ then you want
$$E[X^2]= aE[X_1^2]+bE[X_2^2]+cE[X_3^2] = a(1+1^2)+b(2+2^2)+c(3+3^2)$$ on the left hand side.