# Maximum likelihood estimate

1. Feb 13, 2008

### icedsake

in need of help for how to do this question
given probability mass function:
x 1 2 3 4
p(x) 1/4(θ+2) 1/4(θ) 1/4(1-θ) 1/4(1-θ)

Marbles
1=green
2=blue
3=red
4=white

For 3839 randomly picked marbles
green=1997
blue=32
red=906
white=904

what is the max likelihood of θ using this data?

2. Feb 13, 2008

### EnumaElish

What is the likelihood function in this case?

3. Feb 13, 2008

### icedsake

oops i left out that x=1,2,3,4 are of binomial distributions...
would the likelihood function be the pmf of binomial dist.?
= (nCx) p^x (1-p)^(n-x)

and the loglikelihood function be:
L(p)= log(nCx) + xlog(p) + (n-x)log(1-p) ??

4. Feb 13, 2008

### EnumaElish

Is it a binomial, or a multinomial distribution? Binomial has two possible outcomes; here you have four.

5. Feb 13, 2008

### icedsake

i'm a little lost at this point, in the above section it says that for example green marbles is modelled by a r.v. N1 with a binomial (n, 1/4(θ+2)) distribution and blue is modelled by r.v. N2 with a binomial (n,1/4(θ)) dist. where n in both cases is total # of marbles (3839 in this case)

so i'm assuming red and white have similar binomial dist.

6. Feb 13, 2008

### EnumaElish

It is possible to look at multinomial r.v.'s as a vector of binomial r.v.'s.

The likelihood function (nCx) p^x (1-p)^(n-x) represents just one of the 4 variables, though (e.g., green vs. not green). To capture all individual colors you need to think in terms of a multinomial distribution with multiple (> 2) outcomes.

7. Feb 13, 2008

### icedsake

hmm..so in this case i should use the multinomial prob. mass function to get the likelihood function.. then take the natural log of it correct?
Do I differentiate now and how do I arrive at the estimate for theta?

8. Feb 13, 2008

### EnumaElish

You should set up the log likelihood function L, then differentiate it with respect to theta, set it to zero, and solve for theta: L'(θ) = 0 so θ* = L'-1(0). Then check L"(θ*) < 0 to make sure it's a maximum and not a minimum.

9. Feb 13, 2008

### icedsake

thanks for the clarifications =)