PDA

View Full Version : Poisson process, payment, standard deviation

Ego7894612
08-31-2010, 08:59 AM
We have an insurance company that pays out claims at times of a Poisson process, with rate 4 per week. Suppose that the mean payment is 10K dollar and the standard deviation is 6K dollar (K=1000). Find the standard deviation of the total payment for 4 weeks.

This is how I reasoned:

Let P_i be the i-th payment (in K dollars). Note that all these P_i have the same distribution (that is, I think we are allowed to assume this), and denote this distribution by P.)

Then we have

(\mbox{Standaarddeviation of the total payment for 4 weeks})^2

= ( \mbox{Standarddeviation}( P_1 + ... + P_{N(4)} ) )^2

= Var( P_1 + ... + P_{N(4)} )

= Var( N(4) \cdot P )

= (E[P])^2 Var(N(4)) + (E[N(4)])^2 Var(P) + Var(N(4)) Var(P)

= 10^2 . 16 + 16^2 . 36 + 16 . 36 = 11392

Hence the standarddeviation of the total payment for 4 weeks =

\sqrt{11392} = 106.73 K.

Here's the problem: in the numerical solution manual it says the correct answer is 46.65 K.

So what's wrong with my solution, and how do you get the right solution, namely 46.65 K ?

Ego7894612
08-31-2010, 10:44 AM
I've found a solution which gives the correct result.

However, who can tell me what is exactly wrong with my previous, apparently wrong, calculation?

EDIT: I guess the "wrong step" HAS to be

= Var( P_1 + ... + P_{N(4)} )

= Var( N(4) \cdot P )

But what's wrong with this exactly?

Dason
08-31-2010, 11:09 AM
= Var( N(4) \cdot P )

= (E[P])^2 Var(N(4)) + (E[N(4)])^2 Var(P) + Var(N(4)) Var(P)

So what's wrong with my solution, and how do you get the right solution, namely 46.65 K ?

How do you make the jump between those two steps? Both N(4) and P are random variables so you can't just pull them out like constants. You said you eventually got it and really the better approach is to model this as a hierarchial model and use the fact that if you have something like X|Y ~ something, Y ~ somethingelse, then Var(X) = E[Var(X|Y)]+Var[E(X|Y)]

BGM
08-31-2010, 12:42 PM
Let N(t) be the Poisson Process in t weeks
Then N(t) \sim Poisson(4t)
\Rightarrow N(4) \sim Poisson(16)
Note Var[N(4)] = E[N(4)] = 16

Now given E[P_i] = 10K, Var[P_i] = (6K)^2, i = 1, 2, 3, ...
and they are a sequence of independent random variables,

Var\left[\sum_{i=1}^{N(4)} P_i\right]
= E\left[Var\left[\left.\sum_{i=1}^{N(4)} P_i\right|N(4)\right]\right]
+ Var\left[E\left[\left.\sum_{i=1}^{N(4)} P_i\right|N(4)\right]\right]

= E[N(4)Var[P_1]] + Var[E[P_1]N(4)]

= E[N(4)]Var[P_1] + E[P_1]^2Var[N(4)]

= E[N(4)]\left(Var[P_1] + E[P_1]^2\right)

= 16\left[(6K)^2 + (10K)^2\right]

= 2176K^2

\Rightarrow SD\left[\sum_{i=1}^{N(4)} P_i\right]
= \sqrt{2176}K \approx 46.64762K

Ego7894612
08-31-2010, 01:38 PM
How do you make the jump between those two steps? Both N(4) and P are random variables so you can't just pull them out like constants. You said you eventually got it and really the better approach is to model this as a hierarchial model and use the fact that if you have something like X|Y ~ something, Y ~ somethingelse, then Var(X) = E[Var(X|Y)]+Var[E(X|Y)]

This is not true. I'm not sure what you're saying, but I'm definitely not 'pulling them out as constants'. Of course N(4) and P are random variables. But if X and Y are *independent* r.v.'s, I think the following relationship holds between them:

Var(X Y) = (EY)^2 Var(X) + (EX)^2 Var(Y) + Var(X) Var(Y).

I do not think this step is the problem. Like I said before, I'm convinced the problem is in the step

= Var( P_1 + ... + P_{N(4)} )

= Var( N(4) \cdot P )

I guess the r.v.'s P_1 + ... + P_{N(4)} and N(4) \cdot P do not have the same distribution ?

Any further ideas or comments ?

@ BGM: That's indeed the solution I found after my first attempt.

Dason
08-31-2010, 02:15 PM
This is not true. I'm not sure what you're saying, but I'm definitely not 'pulling them out as constants'. Of course N(4) and P are random variables. But if X and Y are *independent* r.v.'s, I think the following relationship holds between them:

Var(X Y) = (EY)^2 Var(X) + (EX)^2 Var(Y) + Var(X) Var(Y).

Yup, you're right. My mistake. I skimmed over that and forgot about that little fact. I just saw something that looked like Var(XY) = Y^2Var(x) ... and wasn't paying as much attention as I should.

= Var( P_1 + ... + P_{N(4)} )

= Var( N(4) \cdot P )

I guess the r.v.'s P_1 + ... + P_{N(4)} and N(4) \cdot P do not have the same distribution ?
I'm thinking you're right. It reminds me of a brain teaser in calculus:

d/dx(x^2) = 2x. We all know this.

Now consider:
d/dx(x + x + x ... + x) <-- x number of x's
= d/dx(x) + d/dx(x) + ... + d/dx(x) <-- x number of derivatives
= 1 + 1 + ... + 1
= x
Now something is clearly wrong because (x + x + x ... + x) with x number of x's is equal to x^2 but the derivative isn't the same and that's because the "x number of x's" is part of the function and needs to be accounted for in the derivative.

Ego7894612
09-01-2010, 09:06 PM
Yup, you're right. My mistake. I skimmed over that and forgot about that little fact. I just saw something that looked like Var(XY) = Y^2Var(x) ... and wasn't paying as much attention as I should.

No problem man! Thanks for your help :)