MLE of Binomial

[shrutigupta17]

When I am finding the MLE of Binomial distribution with parameters n and p:

I am confused at this step:

L(p) = ∏ (n xi) ( p^xi ) ( (1-p)^(n-xi) )

then it changes to :

ln L(p) = constant x [∑xi x ln (p)] x [∑(n-xi) x ln (1-p)]

In this equation ln (1-p) is multiplied by ∑(n-xi) = n² -∑ xi, but all books say n-∑xi

Can anyone please explain me where am I going wrong?

Cheers
S

 

[DevonMatthews]

Hello Shruti,

The problem is we n samples, from a Bi(>>M<<,p) distribution, so the likelihood function you have written down is incorrect, it should be L(p) = ∏(i=1->n) (m xi) ( p^xi ) ( (1-p)^(m-xi) . I wont do any more because you obviously know what you're doing, you just made a minor slip up.

Best of luck for the forthcoming exam.

 

[shrutigupta17]

Hello Shruti,

The problem is we n samples, from a Bi(>>M<<,p) distribution, so the likelihood function you have written down is incorrect, it should be L(p) = ∏(i=1->n) (m xi) ( p^xi ) (1-p)^(m-xi) . I wont do any more because you obviously know what you're doing, you just made a minor slip up.

Best of luck for the forthcoming exam.

Hey, Thanks for the help. However when I tried evaluating the ∏(1-p)^(m-xi) part I had a slight problem:

L(p) = ∏(i=1->n) (m xi) ( p^xi ) (1-p)^(m-xi)

the last part is: (1-p)^(m-x1) * (1-p)^(m-x2) * .............* (1-p)^(m-xn)

Now when we add up the coefficients we should get nm - ∑(i=1->n) xi ,
however the books say m - ∑(i=1->n) xi, which I didnt understand.

Thanks in advance

Cheers

 

[DevonMatthews]

Hey, Thanks for the help. However when I tried evaluating the ∏(1-p)^(m-xi) part I had a slight problem:

L(p) = ∏(i=1->n) (m xi) ( p^xi ) (1-p)^(m-xi)

the last part is: (1-p)^(m-x1) * (1-p)^(m-x2) * .............* (1-p)^(m-xn)

Now when we add up the coefficients we should get nm - ∑(i=1->n) xi ,
however the books say m - ∑(i=1->n) xi, which I didnt understand.

Thanks in advance

Cheers

Why are you expanding the product literally? It's alot easier if you maximise the logL(p) and use the fact that the log of a product is equal to the sum of the logs. You should have Sum(i=1->n) [Log (m xi) + xiLogp + (m-xi)log(1-p)]. Differentiating that with respect to p gives the required answer

 

[shrutigupta17]

Why are you expanding the product literally? It's alot easier if you maximise the logL(p) and use the fact that the log of a product is equal to the sum of the logs. You should have Sum(i=1->n) [Log (m xi) + xiLogp + (m-xi)log(1-p)]. Differentiating that with respect to p gives the required answer

Thanks Devon, it really helped me....all the best for the exams