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# Thread: Problem 135 in SOA-137

1. ## Problem 135 in SOA-137

I need an explanation of problem number 135 in the SOA-137.

N is Poisson distributed with mean lamda. Lamba is uniformly distributed on [0,3].
Find Var[N].

a) lamda
b) 2*lamda[/FONT]
c) 0.75
d) 1.5
e) 2.25

The answer given in SOA-137 is 2.25. I understand the formula used to get 2.25. My question is: if N is Poisson distributed, what is E[N]? and, if different from Var[N], how then can N be Poisson distributed since the mean and variance should be the same? I feel that the answer should be 1.5 but this contradicts the formula used in the SOA-137 solution.

A similar problem is worked in the Guo manual with Lamda uniformly distributed on [0,6] and the variance is calculated at 3 with only part of the variance formula short-cut sans explanation.

izzy:

2. Originally Posted by Apollon
I need an explanation of problem number 135 in the SOA-137.

N is Poisson distributed with mean lamda. Lamba is uniformly distributed on [0,3].
Find Var[N].

a) lamda
b) 2*lamda[/FONT]
c) 0.75
d) 1.5
e) 2.25

The answer given in SOA-137 is 2.25. I understand the formula used to get 2.25. My question is: if N is Poisson distributed, what is E[N]? and, if different from Var[N], how then can N be Poisson distributed since the mean and variance should be the same? I feel that the answer should be 1.5 but this contradicts the formula used in the SOA-137 solution.

A similar problem is worked in the Guo manual with Lamda uniformly distributed on [0,6] and the variance is calculated at 3 with only part of the variance formula short-cut sans explanation.

izzy:
E(N) = V(N) if your lambda is fixed, here your lambda is varying according to a distribution.

3. ## Hmmm

What properties do they use here to show that V(N|λ) = λ and that E(N|λ) = λ?

4. Originally Posted by edwinjaxfl
What properties do they use here to show that V(N|λ) = λ and that E(N|λ) = λ?
That's coming from the fact that N ~ Poisson(lambda). E(N|lambda) means what's the expected value of N given that N~Poisson(lambda), well that's lambda. Similarly with V(N|lambda). What they ARE asking is for V(N) which is V(E(N|lambda)) + E(V(N|lambda)) = V(lambda) + E(lambda). Now since lambda ~ U[0,3], we know that V(lambda) = 9/12 and E(lambda) = 3/2.

5. Originally Posted by NoMoreExams
That's coming from the fact that N ~ Poisson(lambda). E(N|lambda) means what's the expected value of N given that N~Poisson(lambda), well that's lambda. Similarly with V(N|lambda). What they ARE asking is for V(N) which is V(E(N|lambda)) + E(V(N|lambda)) = V(lambda) + E(lambda). Now since lambda ~ U[0,3], we know that V(lambda) = 9/12 and E(lambda) = 3/2.
After the V(N|λ) = λ and the E(N|λ) = λ, it's a piece of cake. It is given that N is Poisson with µ = λ. I'm not convinced that this automatically means that the conditional mean and conditional variance of N given λ is λ. I'm looking for a mathematical proof.

6. Originally Posted by edwinjaxfl
After the V(N|λ) = λ and the E(N|λ) = λ, it's a pieace of cake. It is given that N is Poisson with µ = λ. I'm not convinced that this automatically means that the conditional mean and conditional variance of N given λ is λ. I'm looking for a mathematical proof.
I'd like to find that too.
For this line really trips me.

7. Originally Posted by NoMoreExams
E(N) = V(N) if your lambda is fixed, here your lambda is varying according to a distribution.
I think I get it now. Guo mentions this but then appears to contradict himself in the example. Hence, my confusion. Thank you.

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