(New page: I think you start by working the maximum likelihood estimation formula of a binomial RV. The number of photons captured is (1,000,000) and the probability of the camera catching a photon i...)
 
 
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I think you start by working the maximum likelihood estimation formula of a binomial RV. The number of photons captured is (1,000,000) and the probability of the camera catching a photon is p, n (the number of photons total) is what we are looking for.
 
I think you start by working the maximum likelihood estimation formula of a binomial RV. The number of photons captured is (1,000,000) and the probability of the camera catching a photon is p, n (the number of photons total) is what we are looking for.
  
<math>\hat n_{ML} = \text{max}_a ( f_{X}(x_i;a) ) = |x_i|</math>
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<math>\hat n_{ML} = \text{max}_n ( \binom{n}{k} p^{k} (1-p)^{n-k} )</math>
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<math>\hat n_{ML} = \text{max}_n ( \binom{n}{1000000} p^{1000000} (1-p)^{n-1000000} )</math>
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But to find the maximum I think you have to take the derivative of an n!... Does anyone know how to do this? Or am I going about the problem completely wrong?
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//Comment Anand Gautam - I am also stuck on taking the derivative. anyone know how to do this?

Latest revision as of 16:04, 10 November 2008

I think you start by working the maximum likelihood estimation formula of a binomial RV. The number of photons captured is (1,000,000) and the probability of the camera catching a photon is p, n (the number of photons total) is what we are looking for.

$ \hat n_{ML} = \text{max}_n ( \binom{n}{k} p^{k} (1-p)^{n-k} ) $

$ \hat n_{ML} = \text{max}_n ( \binom{n}{1000000} p^{1000000} (1-p)^{n-1000000} ) $

But to find the maximum I think you have to take the derivative of an n!... Does anyone know how to do this? Or am I going about the problem completely wrong?

//Comment Anand Gautam - I am also stuck on taking the derivative. anyone know how to do this?

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Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang