Example (3.47), Actex Manual, Mod.3, Pag 28

[Actuario]

Hi,

Using the direct calculation method, I am not seeing how E(N²) is calculated as:

0.5(5² + 5) + 0.5(1² + 1) = 16

In my mind, this would've been:

0.5(5²) + 0.5(1²) = 13

I do see how the double expectation theorem works though.

What am I not understanding?

[Kazodev]

Originally Posted by Actuario

Hi,

Using the direct calculation method, I am not seeing how E(N²) is calculated as:

0.5(5² + 5) + 0.5(1² + 1) = 16

In my mind, this would've been:

0.5(5²) + 0.5(1²) = 13

I do see how the double expectation theorem works though.

What am I not understanding?

Care to post the actual question?

[Abraham Weishaus]

Let me guess (I don't have the Actex manual):

You are given an equally weighted mixture of two Poisson distributions, with the first having a mean of 5 and the second having a mean of 1. Calculate the variance of the mixture.

How did I do?

[Qisoneminusp]

Yes, that is exactly what kind of problem this is.

Given 50% Good drivers, 50% Bad drivers.
LambdaG = 1, LambdaB = 5
Find E(N^2).

It then shows:
E(N^2) = .5(Nbad^2) + .5(Ngood^2)
E(N^2) = .5(5^2 + 5) + .5(1^2 + 1)

What formula does this come from, or what is the logic behind this?

[Gandalf]

Line one is just the standard conditional expectation applied to N^2:

E(anything) = E(anything | A) * prob(A) = E(anything | ~A) * prob(~ A)

Line 2 is just substituting in E(N^2|good) and E(N^2|bad)

[Actuario]

Originally Posted by Gandalf

Line one is just the standard conditional expectation applied to N^2:

E(anything) = E(anything | A) * prob(A) = E(anything | ~A) * prob(~ A)

Line 2 is just substituting in E(N^2|good) and E(N^2|bad)

Line 1 should read:

E(anything) = E(anything | A) * prob(A) + E(anything | ~A) * prob(~ A)

I still don't understand how line 2 works, I'm feeling

How do you get to 0.5(5² + 5) + 0.5(1² + 1) = 16 ?

[Gandalf]

If the distribution of Nbad is Poisson with mean 5, then
Var(Nbad) = E(Nbad^2) - (E(Nbad))^2; definition of variance
5 = E(Nbad^2) - 5^2 since Poisson with mean = variance =5
E(Nbad)^2 = 5^2 + 5
Similarly for Ngood.
.5's are P(good) and P(not good) = P(bad)

[Qisoneminusp]

:: slaps forehead ::

duh! Thanks for clearing that up!

[Actuario]

Awesome, thank you very much!!!