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<math>\hat n_{ML} = \text{max}_n ( \binom{n}{k} p^{k} (1-p)^{n-k} )</math> | <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?
