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Product of two binomials 
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#1
Dec1410, 05:32 PM

P: 3

Hi guys,
Here's a description of the problem: Suppose we have x apples and y oranges where each apple has a probability pa to rot and each orange has a probability po to rot, then what's the pdf of the total number of fruit z being rotten. This pdf looks like a product of two binomial distributions on the surface. However, since z = x+y, x >= 0, y >= 0, then it is actually necessary to sum over all combinations of scenarios (i.e. 6 fruit rotten = 3 apples rotten + 3 oranges rotten or 6 fruits rotten = 1 apple rotten + 5 oranges rotten). By intuition, I worked out the expectation to be x*pa + y *po and it appears to be correct when I manually tested my problem on a spreadsheet. However, I am not quite sure how that expectation can be derived from this messy pdf. Any suggestions? Thanks. 


#2
Dec1510, 01:11 AM

P: 38

The probability that z fruits will rot, is P(A)^0感(O)^z+P(A)^1感(O)^(z1)+P(A)^2感(O)^(z2)+...+P(A)^z感(O)^0
You could also write that as: [tex]\sum^{z}_{k=0}P(A)^kP(O)^{zk}[/tex] 


#3
Dec1510, 04:26 AM

P: 3

This is close to the derivation I had, but instead of P(A)^[something] and P(O)^[something], I had two different binomial distributions. I have no problem getting the pdf, but I have trouble simplifying the expectation.



#4
Dec1510, 07:48 AM

P: 38

Product of two binomials
Are you familiar with this division?
[tex](a^nb^n):(ab)=a^{n1}+a^{n2}b+a^{n3}b^2+\ldots+b^{n1}[/tex] 


#5
Dec1510, 01:51 PM

P: 3

I saw this sum in the derivation of the binomial expectation. Anyhow, this is what I have for the probability of having 2 rotten fruits:
prob of having 2 rotten apples * prob of having 0 rotten oranges + prob of having 1 rotten apple * prob of having 1 rotten orange + prob of having 0 rotten apples * prob of having 2 rotten oranges This seems to be the product of two binomial distributions and then summed over all possible combinations. 


#6
Dec1610, 12:16 AM

P: 38

(Ignore my first post, there are errors in it.)
Let's say you have 2 fruits. Then: Both fruits can be apples. The probability for those being rotten is P(A)^{2} Fruit 1 can be apple, fruit 2 can be orange. The probability for those being rotten is P(A)感(O) Fruit 1 can be orange, fruit 2 can be apple. The probability for those being rotten is P(A)感(O) Both fruits can be oranges. The probability for those being rotten is P(O)^{2} Since we don't know how many fruits we have, the probability that both the fruits will rot, is the average of the probabilities above, which is: (P(A)^{2}+2 P(A)感(O)+P(O)^{2})/4 The probability that z fruits will be rotten is: [tex]\frac{\binom{z}{0}P(A)^zP(O)^0+\binom{z}{1}P(A)^{z1}P(O)^1+\binom{z}{2}P(A)^{z2}P(O)^2+\ldots+\binom{z}{z}P(A)^0P(O)^z}{2^z}[/tex] This is exactly the same, but said in a shorter way: [tex]\frac{\sum^{z}_{k=0}\binom{z}{k}P(A)^{zk}P(O)^k}{2^z}[/tex] By using this equation: [tex](a+b)^n=\binom{n}{0}a^nb^0+\binom{n}{1}a^{n1}b^1+\binom{n}{2}a^{n2}b^2+\ldots+\binom{n}{n}a^0b^n[/tex] We get that: [tex]\frac{\sum^{z}_{k=0}\binom{z}{k}P(A)^{zk}P(O)^k}{2^z}=\frac{(P(A)+P(O))^z}{2^z}=\left(\frac{P(A)+P(O)}{2}\righ t)^z[/tex] 


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