# Product of two binomials

1. Dec 14, 2010

### ouiouiwewe

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. Dec 15, 2010

### Yayness

The probability that z fruits will rot, is P(A)^0·P(O)^z+P(A)^1·P(O)^(z-1)+P(A)^2·P(O)^(z-2)+...+P(A)^z·P(O)^0

You could also write that as:
$$\sum^{z}_{k=0}P(A)^kP(O)^{z-k}$$

Last edited: Dec 15, 2010
3. Dec 15, 2010

### ouiouiwewe

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. Dec 15, 2010

### Yayness

Are you familiar with this division?
$$(a^n-b^n):(a-b)=a^{n-1}+a^{n-2}b+a^{n-3}b^2+\ldots+b^{n-1}$$

5. Dec 15, 2010

### ouiouiwewe

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. Dec 16, 2010

### Yayness

(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)·P(O)
-Fruit 1 can be orange, fruit 2 can be apple. The probability for those being rotten is P(A)·P(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)·P(O)+P(O)2)/4

The probability that z fruits will be rotten is:
$$\frac{\binom{z}{0}P(A)^zP(O)^0+\binom{z}{1}P(A)^{z-1}P(O)^1+\binom{z}{2}P(A)^{z-2}P(O)^2+\ldots+\binom{z}{z}P(A)^0P(O)^z}{2^z}$$

This is exactly the same, but said in a shorter way:
$$\frac{\sum^{z}_{k=0}\binom{z}{k}P(A)^{z-k}P(O)^k}{2^z}$$

By using this equation: $$(a+b)^n=\binom{n}{0}a^nb^0+\binom{n}{1}a^{n-1}b^1+\binom{n}{2}a^{n-2}b^2+\ldots+\binom{n}{n}a^0b^n$$

We get that:
$$\frac{\sum^{z}_{k=0}\binom{z}{k}P(A)^{z-k}P(O)^k}{2^z}=\frac{(P(A)+P(O))^z}{2^z}=\left(\frac{P(A)+P(O)}{2}\right)^z$$