I wanted to generate a bivariate Poisson distribution (D1,D2) and to do so, I did according to the indication found on wikipedia (bottom of the page, http://en.wikipedia.org/wiki/Poisson_distribution), therefore, I have generated three independent random variables which have a Poisson distribution (Y1,Y2,Y3), with means λ1 = 1, λ2 = 4, λ3 = 1.
Provided I have set:
D1= Y1 + Y3
D2= Y2 + Y3
Then,
D1 ~ Poisson(λ1+λ3) = Poisson(2)
D2 ~ Poisson(λ2+λ3) = Poisson(5)
In order to compute the bivariate probability I use the formula found on wikipedia (bottom of the page, http://en.wikipedia.org/wiki/Poisson_distribution), which was really straight forward to calculate.
My problem occurs due to the fact that at a certain point in my problem, I have to compute P(D1 >= k1, (D1 + D2) >= k2) and I am unsure how the formula presented on wikipedia (bottom of the page, http://en.wikipedia.org/wiki/Poisson_distribution) changes.
Provided D1 ~ Poisson(λ1 + λ3) and D2 ~ Poisson(λ2 + λ3), if we denote D = D1 + D2, I would say D ~ Poisson (λ1 + λ2 + 2*λ3) = Poisson(7). However, I was contradicted on the validity of this statement. Given that I must calculate P(D1 >= k1, (D1 + D2) >= k2), I feel I am reaching a dead end.
If we consider:
D1 = demand for product 1 (which can be satisfied from stock of product 1 (k1) OR in case stock 1 is not enough, it will be satisfied from stock of product 2 (k2), after the demand for product 2 was satisfied)
D2 = demand for product 2 (which can only be satisfied from stock of product 2 (k2)).
Basically, P(D1 >= k1, (D1 + D2) >= k2) is associated with inability of satisfying demand for product 1 (D1) from the stock of product 1 (k1) (thus, D1 >=k1) together with the inability of satisfying both demands from both stocks (D1+D2 >= k1+k2). Therefore, that means either that both demands are greater than the existing stock (aka D1 >= k1 and D2 >= k2) OR that demand 2 can be satisfied from stock 2, but demand 1 cannot be fully satisfied from stock 1 and 2 (D1 >= (k1 + m), D2 = k2 - m, where m takes values from 1 to k2). To be more specific, if we have P(D1 >= 3, (D1 + D2) >= 7) I have tried to modify it into something I can compute, hence I got, according to the previously presented reasoning:
P(D1 >= 3, (D1 + D2) >= 7) = P(D1 >= 3, D2 >= 4) + P(D1 >= 4, D2 = 3) + P(D1 >= 5, D2 = 2) + P(D1 >= 6, D2 = 1) + P(D1 >= 7, D2 = 0).
However, I am aware it is not fully correct, since I get the value of the sum as being greater than 1, which obviously indicates I do something wrong with my reasoning. My question is, provided the information I have presented before:
- How can I compute P(D1 >= k1, (D1 + D2) >= k2) aka how does the formula ((bottom of the page, http://en.wikipedia.org/wiki/Poisson_distribution)) changes?
OR
- How can I modify P(D1 >= k1, (D1 + D2) >= k2) into something that can be computed?
$$P\left(D_{1}\geq k_{1}\wedge D_{1}+D_{2}\geq k_{2}\right)=\sum_{k=k_{1}}^{\infty}p\left(k,k_{2}\right)$$ where: $$p\left(k,k_{2}\right):=P\left(Y_{1}+Y_{3}=k\wedge Y_{2}+Y_{3}\geq k_{2}-k\right)$$
If $k_{1}\leq k<k_{2}$ then $$p\left(k,k_{2}\right)=\sum_{i=0}^{k}P\left(Y_{1}=k-i\right)P\left(Y_{2}\geq k_{2}-k-i\right)P\left(Y_{3}=i\right)$$
If $k_{2}\leq k$ then $$p\left(k,k_{2}\right)=P\left(Y_{1}+Y_{3}=k\right)$$