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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.

Xi'an
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n1234
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    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. – Xi'an Oct 24 '20 at 07:23
  • It is not possible to find the $h(T)$ in this case? – n1234 Oct 24 '20 at 07:47
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    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. – Xi'an Oct 24 '20 at 07:54
  • I know that there is no constructive way to derive the function $h$ but in many cases, I have found $h$ just doing some calculations. For this particular problem, I could not find a $h(t)$ whose expectation is zero. That's why I was asking for. But I think for this problem, I have to find another way to verify that it is not complete that you mentioned(the parameter space does not include a 2-dimensional rectangle). – n1234 Oct 24 '20 at 08:33
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    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 $h(T)$. – n1234 Oct 24 '20 at 14:50
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    Right you are. Thank you so much. You have done great work. – n1234 Oct 26 '20 at 13:56

1 Answers1

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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.

Xi'an
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    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 $$\mathbb E_\theta\left[ X - \frac{1}{n(n-1)}Y- \frac{\Gamma(n-1)}{\Gamma(n-1/2)}Y^{1/2}\right] = 0$$ – Sextus Empiricus Oct 26 '20 at 14:17
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    This method should work relatively general. Whenever the two statistics are independent then a trick which may often work is to try finding a transformation or sum of transformed variables such that their expectations have the same dependency on $\theta$ (up to a factor) and then subtract them. – Sextus Empiricus Oct 26 '20 at 14:23
  • @SextusEmpiricus I think, If we let $W=\sum_{i=1}^{n}(X_i -X_{(1)})$ then the value of $\mathbb{E}\left[ W^{\frac{1}{2}}\right]$ will be same as the value of $\mathbb{E}\left[Y^{\frac{1}{2}} \right]$. But is there any easy way to get the pdf of $W$ so that we can calculate the desired expectation? – n1234 Oct 26 '20 at 16:23
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    @n1234 Xi'an already gave these distributions. The minimum of the exponential distribution is exponential distributed and the sum of the other's is gamma distributed and independent of the distribution of the minimum. In the other (deleted) answer Xi'an related this to some theorem's and gave some links for it. I guess that you could also derive it somehow directly.... – Sextus Empiricus Oct 26 '20 at 16:38
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    ...the derivation will be similar to https://en.m.wikipedia.org/wiki/Order_statistic#Order_statistics_sampled_from_an_exponential_distribution (which gives you the distribution of an order statistic) and to https://stats.stackexchange.com/questions/48496/conditional-expectation-of-exponential-random-variable (the derivation for the memorylessness of the exponential distribution can be used in a similar way to say that the distribution of the other variables in the sample minus the value of the lowest variable in the sample will be independent from the value of that minimum). – Sextus Empiricus Oct 26 '20 at 16:40
  • @Xi'an and SextusEmpiricus. Thanks to both of you. I got the point of the pdf just now. – n1234 Oct 26 '20 at 18:04
  • Hello, can you tell me why $X_{(1)}=\frac{\theta^2}{n}\text{Exp}(1)+\theta$? – Tan Jan 07 '21 at 03:52
  • Also, I understand that when $X_{i}\overset{i.i.d}{\sim}\text{Exp}(1)$, we have $\sum_{i=1}^n\{X_i-X_{(1)}\}\sim\text{Gamma}(n-1,1)$. But in OP, $X_{i}$ is not from $\text{Exp}(1)$ (since $X>\theta$), why this is also correct, namely from $\text{Gamma}(n-1,1)$? – Tan Jan 07 '21 at 04:24
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    @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. – Xi'an Jan 07 '21 at 13:18
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    @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. – Xi'an Jan 07 '21 at 13:20