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Topic: Algebra Question
Replies: 5   Last Post: Nov 24, 2012 3:29 AM

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R

Posts: 31
Registered: 10/8/10
Re: Algebra Question
Posted: Nov 23, 2012 9:39 AM
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In fact, there is more. To avoid any confusion:

If we assume the underlying probability of a discrete random variable y is binomial, we have:

pr(y;theta)=(nCy*theta^y)(1-theta)^(n-y)

where

---the possible values of y are the (n+1) integer values 0,1,2,...,n

---(nCy*theta^y) = n! / [y!(n-y)!]

---
Notes:

1. That statistical estimation problem concerns how to use n and y to obtain an estimator of theta, "theta_hat", which is a random variable since it is a function of the random variable, y.

2. The likelihood function gives the probability of the observed data (i.e., y) as a mathematical function of the unknown parameter, theta.

3. The mathematical problem addressed by maximum likelihood estimation is to determine the value of theta, "theta_hat", which maximizes L(theta)

--The maximum liklihood estimator of theta is a numerical value that agrees most closely with the observed data in a sense of providing the largest possible value for the probability L(theta).

Using calculus to maximize the function,(nCy*theta^y)(1-theta)^(n-y), by setting the derivative of L(theta) with respect to theta equal to zero and then solving the resulting equation for theta to obtain theta_hat:

(d/d_theta)[L(theta)] = nCy([y*theta^(y-1)]-[n-y]theta^y[1-theta]^[n-y-1])

I can see how the chain rule has been applied above. However, the textbook goes on to simplify above (which confuses me) as follows:

nCy[(y*theta^[y-1])*([1-theta]^[n-y-1])*(y-n*theta)]

Any tips would be appreciated.

Thanks!

R

p.s. apologies if I've made mathematical notation errors.



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