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N coins have probability $p_n = e^{-t_n/s}$ of heads, $t_n$ being specific for each coin. Coins 1 to m came up heads and m+1 to N came up tails. Now I'm trying to estimate $s$ using the Maximum Likelihood Method.

$L(s) = p_1 p_2 \dots p_m (1-p_{m+1})\dots(1-p_N)$

But this function is difficult to maximize. Do I have to resort to numerical methods?

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it may be important, how do t_n grow and how small is m/N – Fedor Petrov Dec 2 '10 at 13:12
Fedor: Both are unknown. This is actually a model of memory, where $t_n$ is the interval between successive reminders and $m$ is the number of tims the item was remembered. – user7946 Dec 2 '10 at 13:48
seems like $s\to \infty$ and that the likelihood might not take on its maximum. – Suvrit Jan 2 '11 at 21:32

As @Suvrit wrote in a comment, there's not always a solution. We have $$\frac{L'(s)}{L(s)} = \sum_{i\le m} \frac{t_i}{s^2} - \sum_{i>m} \frac{t_i}{s^2}\frac{p_i}{1-p_i} $$ $$=\sum_{i\le m} \frac{t_i}{s^2} - \sum_{i>m} \frac{t_i}{s^2}\left(\frac{1}{1-p_i}-1\right)=\sum_{i\le N} \frac{t_i}{s^2} - \sum_{i>m} \frac{t_i}{s^2}\left(\frac{1}{1-p_i}\right)=0$$ if \begin{equation} \sum_{i\le N} t_i = \sum_{i>m} t_i\frac{1}{1-p_i}. \end{equation} Let's consider the case $m=N-2=0$: $$(1-e^{-t_2/s})(1-e^{-t_1/s})(t_1+t_2)=t_2(1-e^{-t_1/s})+t_1(1-e^{-t_2/s})$$ Let $(t_1,t_2)=(1,1)$ and $u=1/s$, then $$(1-e^{-u})(1-e^{-u})(2)=(1-e^{-u})+(1-e^{-u})$$ $$(1-e^{-u})=1$$ which has no solution.

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I think there's a mistake in the derivative... – Ori Gurel-Gurevich Feb 8 '15 at 9:57
Thanks, fixed now – Bjørn Kjos-Hanssen Feb 9 '15 at 1:55

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