Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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 $C$).

share|improve this question
It is worth noting for you that there is also stats.stackexchange.com –  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
Since posting i think i have answered my own question, however i am now unsure as to how to show that the method of moments estimator is B + C. any ideas?? –  R.M Jan 25 '11 at 15:16
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
add comment

1 Answer

I was just looking at this question, as I completed the same problem myself. However, I am unsure as to why you found the second moment for the final part?

share|improve this answer
Using the notation in en.wikipedia.org/wiki/Method_of_moments_(statistics), $m_1$ is trivially the method of moments estimator for $C+B$ (since $C+B$ is the mean). However, using the second moment you might be able to estimate $B$ and $C$ (separately) in terms of $m_1$ and $m_2$, by solving a pair of equations. –  Shai Covo Jan 25 '11 at 23:03
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.