2005 13 & 19

[jennyzhang]

P13
I do not agree with the solution. Mixture of two poisson distributions are still a poisson distribution. It's mean and variance should be the same, isn't it?


P19
My 2005 spring version ASM talked about the K-S test about grouped data. Why B is wrong here?


Thanks a lot in advance!

[Abraham Weishaus]

P13: No it isn't. Mixing distributions tends to increase the variance. The mixture of two Poisson distributions has a variance higher than its mean.

You are mixing (pun intended) sums of random variables with mixtures. If X is a Poisson process mean lambda1 and Y is another one mean lambda2, then X+Y is Poisson mean lambda1+lambda2. But that's not a mixture. In a mixture, you have a single random variable Z with some proportion of it Poisson mean lambda1 and another proportion of it Poisson mean lambda2. Since the sub-distribution from which Z is picked is also random (it could be the lambda1 one or the lambda2 one), this adds variance.

Perhaps an example will clarify the difference. An example of a sum would be:
I have a medical policy with 1 claim/year and a dental policy with 2 claims/year, both Poisson. How many claims do I expect next year on these 2 policies combined, and what's the variance? Answer: 3 and 3.

An example of a mixture would be:
Half my policyholders have medical and half of them have dental. The medical policyholders submit 1 claim/yr and the dental ones 2/year, both Poisson. What is the total number of claims I expect next year from 2 randomly selected policyholders, and what's the variance? Answer: 3 and 3.5.

P.19 The K-S test cannot be used for grouped data, but bounds for the statistic may be calculated.

If you feel uncomfortable with this, you're in good company. In another thread, Mr. Mahler expresses his discomfort with this question.

[Sparky5]

P13:
An example of a mixture would be:
Half my policyholders have medical and half of them have dental. The medical policyholders submit 1 claim/yr and the dental ones 2/year, both Poisson. What is the total number of claims I expect next year from 2 randomly selected policyholders, and what's the variance? Answer: 3 and 3.5.


What does the distribution of one of these random samples look like?

Y = .5(L1) + .5(L2),

L1 is distribution of medical claims
L2 is distribution of dental claims

The variance of two samples:

Var(2Y) = 2^2(.5^2 Var(L1) + .5^2 Var(L2)) = 3

[Abraham Weishaus]

Well, that's half a sum of random variables, and the constant "0.5" could've just as well been sqrt(2) (There's no need for the constants to add up to 1 -- they're not weights). Multiplying a Poisson variable by a constant yields a Poisson.

[Sparky5]

Well, that's half a sum of random variables, and the constant "0.5" could've just as well been sqrt(2) (There's no need for the constants to add up to 1 -- they're not weights). Multiplying a Poisson variable by a constant yields a Poisson.

Is it that the mixing means the parameter has a distribution on it? In the example you gave the random sample would look like:

Y ~ ?(x,th|th = t)
Where pdf of th is f(1) = .5 and f(2) = .5

The variance is the "predictive variance" not Var(.5L1 + .5L2)

Here Var(Y) = E(Var(x|th)) + Var(E(x|th)) = 1.5 + (1-2)^2(.5)^2 = 1.75

Two random Samples would be:

Var(Y + Y) = 3.5

Am I thinking of this correctly?

[Abraham Weishaus]

I think you have it. Of course, Y's have different subscripts, so Y+Y is not 2Y (double, don't quadruple, the variance).

[Sparky5]

Now that I think about it some more I don't see how your example is a mixed distribution.

In the expression I wrote down

Var(Y) = E(Var(Y|th)) + Var(E(Y|th)) = 1.5 + (1-2)^2(.5)^2 = 1.75

How can we say E(Var(Y|th)) = 1.5. All we know is that th is distributed as two poissons, but we know nothing about the distribution of Y. In other words what is the prior distribution of Y?

Isn't it easier to see Jenny's confusion if we look at the variance of a Poisson mixed distribution?

For a Mixed Distribution Poisson/Poisson = Y~pois|th~pois,

E(Y) = E(E(x|th)) = E(th)

Var(Y) = E(var(x|th)) + var(E(x|th)) =

= E(th) + var(E(x|th))

>= E(th)

Sorry for all of the rambling, but I just got it, unless Dr.W says this is all trash.

[Sparky5]

Now that I look at is some more the same result is obtained no matter how th is distributed. You could have Poisson mixed with any other distribution.

I suppose that is why the call is a poisson mixed instead of a poisson/poisson mixed.