Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
Drexel University or The Math Forum.


Nick
Posts:
20
Registered:
4/12/08


Re: Algebra question
Posted:
Nov 23, 2012 5:33 PM


On 23/11/2012 14:37, R wrote: > Apologies. Let me give the full context: > > If we assume the underlying probability of a discrete random variable y is binomial, we have: > > pr(y;theta)=(nCy*theta^y)(1theta)^(ny) > > where > > the possible values of y are the (n+1) integer values 0,1,2,...,n > > (nCy*theta^y) = n! / [y!(ny)!] > >  > 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)(1theta)^(ny), 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^(y1)][ny]theta^y[1theta]^[ny1]) > > I can see how the chain rule has been applied above. However, the textbook goes on to simply above (which confuses me) as follows: > > nCy[(y*theta^[y1])*([1theta]^[ny1])*(yn*theta)] > > Hope this clarifies my question. > > R > > p.s. apologies if I've made mathematical notation errors. >
Its not errors in notations it is just plain errors.
I can only guess but your equations both look to be wrong.
Firstly your result from the application of the chain rule is wrong and then you also seem to have added an extra y multiplier to the second equation. e.g. I think it should be
nCy[(theta^[y1])*([1theta]^[ny1])*(yn*theta)]
Starting from the wrong place and going to the wrong place do make the problem more difficult. So I would advise you to spend more time checking what you do know how to do.


Date

Subject

Author

11/23/12


R

11/23/12


Nick

11/23/12


R

11/23/12


Nick

11/23/12


R


