# Show that the minimal sufficient statistics for the shifted exponential is complete for $n = 2$

If we had $$Y_i$$, $$i = 1, 2, ..., n$$ are $$iid$$ and have the density $$f(y) = \lambda e^{-\lambda(y - \mu)} I_{y > \mu} , ~~~y >0$$ where $$\lambda >0,~ \mu >0$$ are unknown parameters.

I was able to show through factorization theorem that the minimum order statistic $$T_1 = Y_{(1)}$$ and $$T_2 = \sum_i^n y_i$$ are sufficient statistics. How would I go about showing that these are complete for $$n = 2$$?

Is the idea to find the distributions of $$T_1$$ and $$T_2$$, and then showing that $$E(g(T_1)) = 0 \implies g(T_1) = 0$$ and $$E(g(T_2)) = 0 \implies g(T_2) = 0?$$ It seems it should be simpler than this considering $$n = 2$$.

• Hi. Please note that these are self-study questions. I have added the tag. In future, please do the same. Commented Jan 2, 2023 at 4:41
• No, you have to show that $$\mathbb E_{\lambda,\mu}[g(T_1,T_2)]=0$$ for all $(\lambda,\mu)$ implies that $g(\cdot,\cdot)=0$. Commented Jan 2, 2023 at 8:38

Since this is question, I will not provide any solution but rather jot down the general approach to attack the problem.

It has been shown $$(T_1, T_2):= \left(Y_{(1)}, \sum Y_i\right)\equiv \left(Y_{(1)}, \sum \left(Y_i-Y_{(1)}\right)\right)$$ are jointly sufficient for $$(\mu, \lambda).$$

What should be the approach to show $$(T_1, T_2)$$ is jointly complete?

$$\bullet$$ Stick to the definition (unlike what OP attempted): Show that $$\mathbb E_{\mu,\lambda}[f(T_1, T_2)] = 0 \implies f\equiv 0 ~~\textrm{a.s.}\tag 1\label 1$$

$$\bullet$$ Show that $$T_1, ~T_2$$ are independently distributed. Find out their respective distributions.

$$\bullet$$ For fixed $$\lambda,$$ apply Fubini's theorem in $$\eqref 1$$ to get $$\int\underbrace{\left[\int f(t_1, t_2)f_{T_2}(t_2)~\mathrm dt_2\right]}_{=~\mathbb E_\lambda[f(t_1, T_2)]}f_{T_1}(t_1)~\mathrm dt_1 = 0\tag 2\label 2$$ for all $$\mu.$$

$$\bullet$$ Conclude $$\mathbb E_\lambda f(t_1, T_2) = 0~~\textrm{a.s.}$$ (How?)

$$\bullet$$ Finally show the above conclusion is true for all $$\lambda.$$ The rest follows. (How?)

$$\rm NB.$$ The approach is based on the assumption that $$\mu\in \mathbb R, ~\lambda \in \mathbb R_{+}.$$ Can OP modify (Is it necessary or not? Check.) the argument for $$\mu\in\mathbb R_{+}?$$

## Reference:

$$\rm [I]$$ Theory of Point Estimation, E. L. Lehmann, George Casella, Springer-Verlag, $$1998,$$ sec. $$1.6,$$ p. $$43.$$

• (+1) neet & sufficiently complete! Commented Jan 2, 2023 at 9:28
• When $n=2$, $(T_1,T_2)=(Y_{(1)},Y_{(2)}-Y_{(1)})$. Commented Jan 2, 2023 at 13:03