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Likelihood distribution

by aymer
Tags: distribution, likelihood
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aymer
#1
Oct18-10, 12:35 PM
P: 16
Hey everyone,

i am new to using statistics and have come across a problem. I am trying to regenerate the result of a paper in which a theoretical parameter is constrained.
I have calculated the chi squared and plotted the likelihood function (exp[-chisquared])to see what value of the parameter has the highest likelihood.
I am getting almost the same value as the paper but my graph is not scaled like it. I would like the maximum value of likelihood to be 1. What normalization constants am i missing?

Also in the likelihood function do we use the reduced chisquared?
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Chalnoth
#2
Oct19-10, 01:05 AM
Sci Advisor
P: 4,782
Well, first of all, there may be a factor of two error. The way the Chi square is usually defined, you should be dividing it by two in the exponent to the probability distribution.

That said, the key point here is that for most problems, the normalization is simply not known when the Chi square is generated, so you have to manually integrate the probability distribution to determine the normalization. If your distribution is Gaussian, you can do this analytically, though for real distributions we typically make use of Markov-Chain Monte Carlo to perform the integration numerically.

At any rate, you can read up on the multivariate normal distribution here:
http://en.wikipedia.org/wiki/Multiva...l_distribution
aymer
#3
Oct19-10, 03:48 AM
P: 16
Thanx,
I thought of integrating the likelihood and equating it to find the normalization constant but the problem is my chi-squared itself is a function of an unknown parameter,the value of which I want to find. So we have two unknowns-the parameter and the normalization constant and only one equation!!!!

Chalnoth
#4
Oct19-10, 04:24 AM
Sci Advisor
P: 4,782
Likelihood distribution

Quote Quote by aymer View Post
Thanx,
I thought of integrating the likelihood and equating it to find the normalization constant but the problem is my chi-squared itself is a function of an unknown parameter,the value of which I want to find. So we have two unknowns-the parameter and the normalization constant and only one equation!!!!
Well, no, the unknown parameter is left variable like so:

[tex]\int P(x) dx = 1[/tex]

Where [itex]x[/itex] here is the set of all parameters in the chi square, and the integral is over the entire available parameter space. If your chi square is Gaussian-distributed, this normalization is relatively straightforward. If not, you have to actually perform the integral numerically.

Basically, the unknown input parameters to the chi square are never solved for, we merely find their probability distributions.
aymer
#5
Oct20-10, 08:36 AM
P: 16
hii,
i tried doing the normalization as you suggested by doing

[A*integral(exp[-chisquared/2])] over the parameter(-inf,inf) and equated it to 1 to calculate A.
then i plotted A*exp[-chisquared/2] vs the parameter. but still it is not normalized to 1. the maximum value the function takes is nearly 2.
what am i doing wrong?
aymer
#6
Oct20-10, 08:45 AM
P: 16
i also tried normalizing each event(i have 17 data points) to 1 and find corresponding normalization constants. Then i plotted (product of constants)*exp[-chisquared/2],but even that is not scaled to 1.
Chalnoth
#7
Oct20-10, 09:08 AM
Sci Advisor
P: 4,782
Quote Quote by aymer View Post
hii,
i tried doing the normalization as you suggested by doing

[A*integral(exp[-chisquared/2])] over the parameter(-inf,inf) and equated it to 1 to calculate A.
then i plotted A*exp[-chisquared/2] vs the parameter. but still it is not normalized to 1. the maximum value the function takes is nearly 2.
what am i doing wrong?
What do you mean it isn't normalized to one? If the integral of the probability distribution is equal to one, then it is normalized to one. Depending upon the width of the distribution, it could easily take on values at any given point that are much larger than one. This isn't a problem.

Normalizing the individual events changes the problem significantly, and should not be done.
aymer
#8
Oct20-10, 10:55 AM
P: 16
Quote Quote by Chalnoth View Post
What do you mean it isn't normalized to one? If the integral of the probability distribution is equal to one, then it is normalized to one. Depending upon the width of the distribution, it could easily take on values at any given point that are much larger than one. This isn't a problem.

Normalizing the individual events changes the problem significantly, and should not be done.
Thanx,

So I guess what i have done is correct and the authors of the paper must have just scaled the graph to one somehow.I was worried because I was getting the peak at the same point and my graph intersects the x-axis at the same coordinates as the paper.
Chalnoth
#9
Oct20-10, 12:59 PM
Sci Advisor
P: 4,782
Quote Quote by aymer View Post
Thanx,

So I guess what i have done is correct and the authors of the paper must have just scaled the graph to one somehow.I was worried because I was getting the peak at the same point and my graph intersects the x-axis at the same coordinates as the paper.
Right, so, a lot of the time people don't care to normalize probability distributions. So if they simply set the peak equal to one, that would explain their graph.


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