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Thread: maximum likelihood of 1+(a/b)cos (theta)

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    maximum likelihood of 1+(a/b)cos (theta)



    Hello everyone. I am stuck in a problem which is the following. I will explain my problem, then my attempt.

    The idea is to evaluate an estimator for (a/b) by calculing the maximum likelihood with respect to cos theta. Experimentaly, a series of theta is given in the exercise, theta.1,theta.2...theta.n

    what i first do is to set up the product of all the [1+(a/b)cos theta.i]. I the take the log of it , and then i differentiate with respect to theta.i. I the get something like

    Sum [cos theta.i]/(1+(a/b)cos theta.i).

    I don't know how to go from there. i have tried a taylor expansion but afte having equalled this expression to zero, i get the wrong answer. Does anyone have an idea?

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    Re: maximum likelihood of 1+(a/b)cos (theta)

    It's not clear to me what you're doing.

    What distribution are you using? Please provide more details. Also note that a lot of times there isn't a closed form solution for the MLEs in a given problem.
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    Re: maximum likelihood of 1+(a/b)cos (theta)

    I don't know the name of distribution as it is actually a distribution used in physics (muon decay). basically, we are given a series of experimental values for "cos theta" (and not theta, as i said above, sorry). based on this, we actually want to estimate "a/b", coefficient of the cos in this equation. So I first construct the product mentioned, then differentiated with respect to a/b, and finally obtained Sum [cos theta.i]/(1+(a/b)cos theta.i). . From there I am stuck

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    Re: maximum likelihood of 1+(a/b)cos (theta)

    What is the range of theta? It seems that it depends on the unknown parameter as well such that the given pdf is valid.

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    Re: maximum likelihood of 1+(a/b)cos (theta)

    the variable is cos theta rather than theta itself, and from what it seems, it thus ranges from -1 to 1. I can ensure that the pdf is valid and normalisable using a constant multiplicative factor easy to determine.

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    Re: maximum likelihood of 1+(a/b)cos (theta)

    Sorry misread your information previously.

    If you do some reparametrizations, the pdf under consideration is

    f(x;\eta) = \frac {1} {2} (1 + \eta x), -1 < x < 1, -1 < \eta < 1

    So the product of it will be a high degree polynomial of \eta which has no closed form solution for its maximum as mentioned by Dason, although it should behave nicely numerically.

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    Re: maximum likelihood of 1+(a/b)cos (theta)


    Thanks for this BGM, but then, do you mean that te only way to actually maximize the likelihood function is to check one of the available experimental value of cos theta gives us the higher value for the likelihood function? I was thinking to this, even though it looks kind of brute force thing.

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