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$$f_T(t;B,C) = \frac{\exp(-t/C)-\exp(-t/B)}{C-B}$$

where our mean is $C+B$ and $t>0$.

so far i have found my log likelihood functions and differentiated them as follows:

$$dl/dB = \sum[t\exp(t/C) / (B^2(\exp(t/c)-\exp(t/B)))] +n/(C-B) = 0$$

i have also found a similar $dl/dC$.

I have now been asked to comment what you can find in the way of sufficient statistics for estimating these parameters and why there is no simple way of using Maximum Likelihood for estimation in the problem. I am simply unsure as to what to comment upon. Any help would be appreciated. Thanks, Rachel

Editor's Note: Given here is the probability density function $$ f_T (t;B,C) = \frac{{e^{ - t/C} - e^{ - t/B} }}{{C - B}}, \;\; t > 0, $$ where $B$ and $C$ are positive constants such that $C > B$. The mean is $C+B$. For the log likelihood function, see the last equation in my answer to this related question, and differentiate accordingly (with respect to $B$ and $C$).

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It is worth noting for you that there is also – Shai Covo Jan 25 '11 at 13:14
Of course, you meant to write $(\exp ( - t/C) - \exp ( - t/B))/(C - B)$. – Shai Covo Jan 25 '11 at 14:01
yes, and the sum is concerning t. i know it is very confusing to read, but i do not know how else to write it. – R.M Jan 25 '11 at 14:17
The first moment (that is, the mean) is $C+B$, and you should be able to show that the second moment is $$ \int_0^\infty {t^2 \frac{{e^{ - t/C} - e^{ - t/B} }}{{C - B}}\,{\rm d}t} = 2(B^2 + BC + C^2 ). $$ – Shai Covo Jan 25 '11 at 20:55
The reasons exponential families are great for log likelihood is because go from a product to a sum. However, with linear combinations you can not make use of the logs to turn it into a sum. – Chinny84 Sep 13 '14 at 23:34

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