1
$\begingroup$

I have a conceptual doubt in the subject of Sufficient Statistics, based on the next problem:

Let $X_1, X_2, ...,X_n$ be a (random) sample of independent and identically distributed random variables of the next poblation:

$f(x|\theta)=I_{[\theta,\theta +1]} (x) \quad (\theta>0$).

a) Find a sufficient statistic for $\theta$.

Here, I computed the joint density:

$ f(\textbf{x})|\theta)=\prod_{i=1}^{n} I_{[\theta,\theta +1]} (x_i) $

Plus: $\theta\leq X_i\leq\theta+1 \quad \forall i=1,2,...,n$

So: $\theta\leq X_{(1)}\leq X_i\leq X_{(n)}\leq \theta+1 \quad \forall i=1,2,...,n$ .

Then the natural thing is to use the statistic:

$T(\textbf{X})=(X_{(1)},X_{(n)})=(t_1,t_2)$.

So my doubt is: which of these are the correct equivalent expressions to use the Factorization theorem:

  1. $ f(\textbf{x})|\theta)=I_{[X_{(1)},X_{(n)}]} (x_i) \cdot 1$

(Is the expression at the left of the "1" a function of $T$ and $\theta?$

  1. $f(\textbf{x})|\theta)=I_{[\theta,X_{(n)}]} (t_1) \cdot I_{[X_{(1)},\theta +1]} (t_2)\cdot 1$

(Is the thing at the left of the "1" a functon of $\theta$, even thought $\theta$ only appears at the interval of the indicator function?)

And my last question would be:

In general, the dimension of the minimal sufficient statistic is greater or equal than the dimension of $\theta$?

$\endgroup$
2
$\begingroup$

The joint density is completely described by the function you wrote, without needing to restrict the observations, since by definition, $$\mathbb 1_S(x) = \begin{cases} 1, & x \in S \\ 0, & x \not \in S. \end{cases}$$ So the expression $$f(\boldsymbol x \mid \theta) = \prod_{i=1}^n \mathbb 1_{[\theta, \theta+1]}(x_i) = \mathbb 1_{[\theta,\theta+1]}(x_{(1)}) \mathbb 1_{[\theta,\theta+1]}(x_{(n)})$$ is adequate. Then the factorization theorem allows us to choose a function $\boldsymbol T(\boldsymbol x)$ such that $$f(\boldsymbol x \mid \theta) = h(\boldsymbol x) g(\boldsymbol T(\boldsymbol x) \mid \theta).$$ Then $\boldsymbol T$ is called sufficient for $\theta$. In this case, $h(\boldsymbol x) = 1$: there are no factors of the joint density that are independent of the parameter. if we choose $\boldsymbol T(\boldsymbol x) = (x_{(1)}, x_{(n)})$, then the choice $$g(t_1, t_2 \mid \theta) = \mathbb 1_{[\theta,\theta+1]}(t_1) \mathbb 1_{[\theta,\theta+1]}(t_2)$$ yields the desired factorization. It is not necessary to restrict the intervals on which the indicator functions are $1$ any further than the original $[\theta, \theta+1]$, because $x_{(1)}$ is really just a shortcut for $\min(x_1, x_2, \ldots, x_n)$ and similarly $x_{(n)} = \max(x_1, x_2, \ldots, x_n)$; i.e., $$\boldsymbol T(\boldsymbol x) = (\min \boldsymbol x, \max \boldsymbol x),$$ and it is always true that $x_{(1)} \le x_{(n)}$.

Regarding your last question, I suppose it may not be true if $\boldsymbol \theta = (\theta_1, \ldots, \theta_p)$ exhibits some kind of dependence among the parameters; but perhaps such a case could be regarded as contrived.

$\endgroup$
  • $\begingroup$ Thanks a lot heropup, I've understand the dependence of the joint density with theta. I suppose this yields to the fact that if I have something like: $1_{[t_1,t_2]}(something \quad not \quad depending \quad on \quad \theta)$ It would also be a function of the transformation but not of $\theta$, right? $\endgroup$ – User117E29 Feb 24 '17 at 4:48

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.