6
$\begingroup$

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.

$\endgroup$
6
  • 2
    $\begingroup$ 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. $\endgroup$
    – Xi'an
    Oct 24, 2020 at 7:23
  • $\begingroup$ It is not possible to find the $h(T)$ in this case? $\endgroup$
    – n1234
    Oct 24, 2020 at 7:47
  • 2
    $\begingroup$ 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. $\endgroup$
    – Xi'an
    Oct 24, 2020 at 7:54
  • 1
    $\begingroup$ 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 $h(T)$. $\endgroup$
    – n1234
    Oct 24, 2020 at 14:50
  • 1
    $\begingroup$ Right you are. Thank you so much. You have done great work. $\endgroup$
    – n1234
    Oct 26, 2020 at 13:56

1 Answer 1

5
$\begingroup$

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.

$\endgroup$
10
  • 3
    $\begingroup$ 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 $$\mathbb E_\theta\left[ X - \frac{1}{n(n-1)}Y- \frac{\Gamma(n-1)}{\Gamma(n-1/2)}Y^{1/2}\right] = 0$$ $\endgroup$ Oct 26, 2020 at 14:17
  • 1
    $\begingroup$ 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. $\endgroup$ Oct 26, 2020 at 14:23
  • 2
    $\begingroup$ @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.... $\endgroup$ Oct 26, 2020 at 16:38
  • 2
    $\begingroup$ ...the derivation will be similar to en.m.wikipedia.org/wiki/… (which gives you the distribution of an order statistic) and to stats.stackexchange.com/questions/48496/… (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). $\endgroup$ Oct 26, 2020 at 16:40
  • 1
    $\begingroup$ @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. $\endgroup$
    – Xi'an
    Jan 7, 2021 at 13:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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