Maximum likelihood and sufficient statistics $$f_T(t;B,C) = \frac{\exp(-t/C)-\exp(-t/B)}{C-B}$$
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 $C$).

Edit: Made a mistake in the MLE part
There is no MLE because the likelihood is maximized at at a parameter specification outside of the parameter range.
\begin{equation}
f_T (t;B,C) = \frac{{e^{ - \frac{t}{C}}  - e^{ -\frac{t}{B}} }}{{C - B}}
\end{equation}
Taking the gradient we get: 
\begin{equation}
\nabla f_T (t;B,C) = \{\frac{e^{\frac{-1}{x}}}{x^2(x-y)}-\frac{e^{\frac{-1}{x}}-e^{\frac{-1}{y}}}{(x-y)^2},\frac{e^{\frac{-1}{x}}-e^{\frac{-1}{y}}}{(x-y)^2}-\frac{e^{\frac{-1}{y}}}{y^2(x-y)}\}
\end{equation}
Setting this equal to 0 and re-arranging some terms gives us the following system of equations:
\begin{array}{lcl} \frac{e^{\frac{-1}{C}}}{C^2} & = & \frac{e^{\frac{-1}{C}}-e^{\frac{-1}{B}}}{(C-B)}\\ 
\frac{e^{\frac{-1}{B}}}{B^2} & = & \frac{e^{\frac{-1}{C}}-e^{\frac{-1}{B}}}{(C-B)}
\end{array}
\begin{array}{lcl} \frac{(C-B)}{C^2} & = & 1-e^{\frac{-1}{B}+\frac{1}{C}}\\ 
\frac{(C-B)}{B^2} & = & 1-e^{\frac{-1}{C}+\frac{1}{B}}
\end{array}
Which is maximized when $C = B$, which is not a valid parameterization.
If you have that $E[X] = C+B$ and $E[X^2] = 2(B^2 + BC + C^2)$ then:
\begin{equation}
Var[X] = 2B^2 + 2BC + 2C^2 - B^2 -2BC - C^2 = B^2 + C^2  
\end{equation}
Using method of moments we get that $C = \bar{X} - B \Rightarrow \hat{\sigma^2} = B^2 + B^2 + \bar{X}^2 -2\bar{X}B \Rightarrow 0 = 2B^2 -2\bar{X}B + \bar{X}^2 - \hat{\sigma^2}$.
The roots of that quadratic get you $B$ (one is probably negative so take the positive one), and plugging that estimate for $B$ into $C = \bar{X} - B$ gives you the moment estimator for C.
Edit: Examining the methods of moment estimators
The root for the polynomial in $B$ is $\hat{B} = \frac{1}{2}(\sqrt{2\hat{\sigma}^2 - \bar{X}^2}+\bar{X})$. Plugging that into the expression for $C$ gives $\hat{C} = \frac{1}{2}(\bar{X} - \sqrt{2\hat{\sigma}^2 - \bar{X}^2})$. 
Note that with this estimate $\hat{C} < \hat{B}$, but because in the estimating equations we didn't imply an ordering on $B$ or $C$, and because the formula for the mean and variance are symmetric in the parameters, we can just switch the estimators to make $\hat{B} = \frac{1}{2}(\bar{X} - \sqrt{2\hat{\sigma}^2 - \bar{X}^2})$ and $\hat{C} = \frac{1}{2}(\sqrt{2\hat{\sigma}^2 - \bar{X}^2}+\bar{X})$.
Noting that the MLE didn't converge to anything it might be interesting to examine whether these estimates are unbiased for the mean and variance and have asymptotic consistency.
\begin{equation}
E[\hat{C} + \hat{B}] = E[\frac{\bar{X} + \bar{X}}{2}] = E[\bar{X}] = \mu
\end{equation}
\begin{equation}
\begin{split}
E[\hat{C}^2 + \hat{B}^2]& = E[\frac{1}{4}((\bar{X}^2 +2\bar{X}\sqrt{2\hat{\sigma}^2 -\bar{X}^2} + 2\hat{\sigma}^2 - \bar{X}^2)) + (\bar{X}^2 -2\bar{X}\sqrt{2\hat{\sigma}^2 -\bar{X}^2} + 2\hat{\sigma}^2 - \bar{X}^2)] \\
& = E[\frac{2\hat{\sigma}^2 + 2\hat{\sigma}^2}{4}] \\
& = E[\hat{\sigma}^2] = \sigma^2
\end{split}
\end{equation}
To show convergence, note that by the law of large numbers convergence in probability to a constant $\bar{X}\overset{P}{\rightarrow}\mu$. Applying Slutsky's theorem to $2\hat{\sigma^2} - \bar{X}^2 \overset{P}{\rightarrow} 2\sigma^{2^*} - \mu^2$, where $\sigma^{2^*}$ is the limiting distribution of $\hat{\sigma}^2$.
Applying the continuous mapping theorem to the square root, then Slutsky's theorem again to the additional $\bar{X}$, we have that:
\begin{equation}
\hat{B} \overset{P}{\rightarrow} \frac{1}{2}(\mu- \sqrt{2\sigma^{2^*} - \mu^2})
\end{equation}
\begin{equation}
\hat{C} \overset{P}{\rightarrow} \frac{1}{2}(\sqrt{2\sigma^{2^*} - \mu^2} + \mu)
\end{equation}
By the law of large numbers $\sigma^{2^*} \overset{P}{\rightarrow} \sigma^2$. Plugging $\sigma^2 = B^2 + C^2$ and $\mu = B + C$ into these expressions gives us:
\begin{equation}
\begin{split}
\hat{B} & \overset{P}{\rightarrow} \frac{1}{2}(\mu- \sqrt{2\sigma^{2^*} - \mu^2})\\
& = \frac{1}{2}(B + C- \sqrt{B^2 -2BC + C^2})\\
& = \frac{1}{2}(B + C- \sqrt{(C-B)^2})\\
& = \frac{1}{2}(B + C- C + B)\\
& = B
\end{split}
\end{equation}
\begin{equation}
\begin{split}
\hat{C} & \overset{P}{\rightarrow} \frac{1}{2}( \sqrt{2\sigma^{2^*} - \mu^2} + \mu)\\
& = \frac{1}{2}(B + C + \sqrt{B^2 -2BC + C^2})\\
& = \frac{1}{2}(B + C + \sqrt{(C-B)^2})\\
& = \frac{1}{2}(B + C + C - B)\\
& = C
\end{split}
\end{equation}
