Estimate p from sample of two Binomially Distributions

[Scootertaj]

1. Suppose X~B(5,p) and Y~(7,p) independent of X. Sampling once from each population gives x=3,y=5. What is the best (minimum-variance unbiased) estimate of p?

Homework Equations

P(X=x)=\binom{n}{x}p^x(1-p)^{n-x}

The Attempt at a Solution

My idea is that Maximum Likelihood estimators are unbiased, and have asymptotic variance = Cramer-Rao lower bound. Also, C-R lower bound = minimum variance of unbiased estimators.
So, since X and Y independent, X+Y \sim B(5+7,p)=B(12,p)
Thus, can we just compute the likelihood function and take the derivative?
L = \binom{12}{8}p^8(1-p)^4
\frac{dL}{dp} = 8p^7(1-p)^4 - 4p^8(1-p)^3
Thus, p=8/12=2/3

Is that legit?

 

[haruspex]

The method is OK as far as I know (not an expert on unbiased estimators), but I have a problem with this step:

\frac{dL}{dp} = 8p^7(1-p)^4 - 4p^8(1-p)^3
Thus, p=8/12=3/4

That's not what I get by putting dL/dp =0.

 

[Scootertaj]

\frac{dL}{dp} = 8p^7(1-p)^4 - 4p^8(1-p)^3=0
p^7(1-p)^3[8(1-p)]-4p]=0
8-8p-4p=0 ignoring p=0,p=1
8=12p \Rightarrow p=8/12=2/3

Did I miss something?
Note: I screwed up my fraction simplification previously if that's what you meant.

 

[haruspex]

I screwed up my fraction simplification previously if that's what you meant.

Yes.

 

[Ray Vickson]

1. Suppose X~B(5,p) and Y~(7,p) independent of X. Sampling once from each population gives x=3,y=5. What is the best (minimum-variance unbiased) estimate of p?

Homework Equations

P(X=x)=\binom{n}{x}p^x(1-p)^{n-x}

The Attempt at a Solution

My idea is that Maximum Likelihood estimators are unbiased, and have asymptotic variance = Cramer-Rao lower bound. Also, C-R lower bound = minimum variance of unbiased estimators.
So, since X and Y independent, X+Y \sim B(5+7,p)=B(12,p)
Thus, can we just compute the likelihood function and take the derivative?
L = \binom{12}{8}p^8(1-p)^4
\frac{dL}{dp} = 8p^7(1-p)^4 - 4p^8(1-p)^3
Thus, p=8/12=2/3

Is that legit?

I disagree with your expression for L, but agree with the final answer you get later. You only get B(n,p)+B(m,p) = B(n+m,p) when you perform a convolution (that is, sum over all the outcomes for each term). In your case you just have one outcome from each binomial, so you should use
L = {5 \choose 3}p^3(1-p)^2 {7 \choose 5} p^5 (1-p)^2. Of course, this has the form $c p^8(1-p)^4,$ so will give the same result you got.

 

[Scootertaj]

Ahh yes, I thought about that also. It's nice to see I wasn't off-base by considering that way.
Why does it not make statistical sense to use the L I suggested (which is based off the fact that X+Y~B(12,p) ?

 

[Ray Vickson]

Ahh yes, I thought about that also. It's nice to see I wasn't off-base by considering that way.
Why does it not make statistical sense to use the L I suggested (which is based off the fact that X+Y~B(12,p) ?

I already gave the explanation, although it was very brief. I mean that P(X+Y=8) = P(X=1,Y=7) + P(X=2,Y=6) + P(X=3,Y=5) + P(X=4,Y=4)+ P(X=5,Y=3), a sum of 5 terms, but you have only the term P(X=3,Y=5). Each separate term has p^8 * (1-p)^4, but with different coefficients.