# Whether the minimal sufficient statistic is complete for a translated exponential distribution

Let $$X_1, X_2..., X_n$$ follows iid negative exponential distribution with pdf

$$f(x) = \frac{1}{\theta^2} \: e^{-\frac{(x-\theta)}{\theta^2}} \: \: I_{(x>\theta)}$$

I have to show whether the minimal sufficient statistic for this pdf is complete or not? I have found that the minimal sufficient statistic is $$T=\left( X_{(1)}, \sum_{i=1}^{n} (X_i - X_{(1)}) \right)$$. If this minimal sufficient statistic is not complete then there exists a function $$h(T)$$ of the minimal sufficient statistic such that

$$E_\theta [h(T)] =0$$ for all $$\theta>0$$ where $$h(T)$$ is not identically zero.

Is this minimal sufficient complete or not? How can I find the function $$h(T)$$ of the minimal sufficient statistic?

Note that, $$X_{(1)}$$ is the first order statistic i.e., $$min\{X_1,..X_n\}$$.

I have calculated the pdf of $$X_{(1)}$$. Let $$Y= X_{(1)}$$ then the pdf of $$Y$$ is given by,

$$f(y) = \frac{n}{\theta^2} \: e^{-\frac{n(y-\theta)}{\theta^2}} \: \: I_{(y>\theta)}$$

I have also calculated

$$\mathbb{E}[X]= \theta^2 + \theta$$ and $$\mathbb{E}[Y] = \mathbb{E}[X_{(1)}] = \frac{\theta^2}{n} + \theta$$

Now, please help me to find out $$h(T)$$ for which $$E_\theta[h(T)] = 0$$ for all $$\theta>0$$ if the minimal sufficient statistic is not complete or any other way to prove or disprove its completeness.

• The pair$$T=\left( X_{(1)}, \sum_{i=1}^{n} (X_i - X_{(1)}) \right)$$or equivalently$$T=\left( X_{(1)}, \bar X_n\right)$$is indeed minima since any change in this pair modifies the likelihood functionl. Given that the family has a single parameter, it is most likely not complete. Oct 24, 2020 at 7:23
• It is not possible to find the $h(T)$ in this case? Oct 24, 2020 at 7:47
• You have to understand that there is no generic constructive way to derive the function $h$ in completeness exercises. The approach is to try to find combinations of the components of the statistic that are parameter free. For instance,$$\dfrac{X_{(2)}-X_{(1)}}{X_{(3)}-X_{(1)}}$$is parameter free, but this ratio is not a function of the sufficient statistics. Oct 24, 2020 at 7:54
• Thanks for your effort. I think we won't find a $h(T)$ for this problem. We have to use an alternative way. I had also tried to do some calculations, but can not find the desired $h(T)$. Oct 24, 2020 at 14:50
• Right you are. Thank you so much. You have done great work. Oct 26, 2020 at 13:56

Lemma The minimal sufficient statistic $$\left(X_{(1)},\sum_{i=2}^n \{X_{(i)}-X_{(1)}\}\right)$$ is not complete.
Proof. The joint distribution of $$\left(X_{(1)},\sum_{i=2}^n \{X_{(i)}-X_{(1)}\}\right)$$ is the product of an Exponential $$\mathcal E(n/\theta^2)$$ translated by $$\theta$$ and of a $$\mathcal Ga(n-1,1/\theta^2)$$ [the proof follows from Sukhatme's Theorem, 1937, recalled in Devroye's simulation bible (1986, p.211)]. This means that $$X_{(1)}$$ can be represented as $$X_{(1)}=\frac{\theta^2}{n}\varepsilon+\theta\qquad\varepsilon\sim\mathcal E(1)$$ that $$Y$$ is scaled by $$\theta^2$$ since $$Y=\sum_{i=2}^n \{X_{(i)}-X_{(1)}\}=\theta^2 \eta\qquad\eta\sim\mathcal Ga(n-1,1)$$ and that $$\mathbb E_\theta\left[ Y^\frac{1}{2}\right]=\theta \frac{\Gamma(n-1/2)}{\Gamma(n-1)}$$ Therefore, $$\mathbb E_\theta\left[X_{(1)}-\frac{\Gamma(n-1)}{\Gamma(n-1/2)}Y^\frac{1}{2}\right]=\frac{\theta^2}{n}$$ eliminates the location part in $$X_{(1)}$$ and suggests dividing by $$Y$$ to remove the scale part: since $$\mathbb E_\theta\left[ Y^\frac{-1}{2}\right]=\theta^{-1} \frac{\Gamma(n-3/2)}{\Gamma(n-1)}\qquad \mathbb E_\theta\left[ Y^{-1}\right]=\theta^{-2} \frac{\Gamma(n-2)}{\Gamma(n-1)}$$ we have (for an arbitrary $$\gamma)$$ that $$\mathbb E_\theta\left[\frac{X_{(1)}-\gamma Y^\frac{1}{2}}{Y}\right]=\frac{\Gamma(n-2)}{n\Gamma(n-1)}+\frac{\theta^{-1}\Gamma(n-2)}{\Gamma(n-1)}- \frac{\gamma \theta^{-1}\Gamma(n-3/2)}{\Gamma(n-1)}$$ Setting $$\gamma=\frac{\Gamma(n-2)}{\Gamma(n-3/2)}$$ leads to $$\mathbb E_\theta\left[\frac{X_{(1)}-\gamma Y^\frac{1}{2}}{Y}\right]=\frac{\Gamma(n-2)}{n\Gamma(n-1)}$$ which is constant in $$\theta$$. Therefore this concludes the proof.
As pointed out by Sextus Empiricus, this is not the only transform of the sufficient statistic with constant expectation. His proposal $$\mathbb E_\theta\left[ X - \frac{1}{n(n-1)}Y- \frac{\Gamma(n-1)}{\Gamma(n-1/2)}Y^{1/2}\right] = 0$$is an alternative.
• Smart solution. Maybe more direct would be to use these three $$\mathbb E_\theta\left[ Y^\frac{1}{2}\right]=\theta \frac{\Gamma(n-1/2)}{\Gamma(n-1)}$$ $$\mathbb E_\theta\left[ Y\right]=(n-1)\theta^2$$ $$\mathbb E_\theta\left[ X\right] =\theta+\frac{1}{n}\theta^2$$ to argue that $$\mathbb E_\theta\left[ X - \frac{1}{n(n-1)}Y- \frac{\Gamma(n-1)}{\Gamma(n-1/2)}Y^{1/2}\right] = 0$$ Oct 26, 2020 at 14:17
• @Tan: Sukhtame's result that $X_{(1)}$ is an Exponential Exp$(n/θ^2)$ translated by $θ$ is why $X_{(1)}$ writes like this. I provided a reference for the proof. Jan 7, 2021 at 13:18
• @Tan: If $X_i\sim f(x)$ provided in the question, $X_i-\theta\sim\text{Exp}(\theta^2)$ and $X_i-X_{(1)}=(X_i-\theta-(X_{(1)}-\theta)$ means this is also a difference of Exponential variates. Jan 7, 2021 at 13:20