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I'm trying to solve a problem in problem list of qualifying exam.

Here is my problem : Let $X_1, \ldots, X_n$ be a random sample from a p.d.f $f(x,\theta) = \theta f_1(x) 1_{(-\infty,0)} + (1-\theta) f_2(x) 1_{(0,\infty)}$ where $f_1 \geq 0 , f_2 \geq 0$ and $\int_{-\infty}^0 f_1=0$, $\int_0^\infty f_2=0.$ Prove or disprove that there exists a complete sufficient statistic for $\theta$.

I'm trying to show that there is only one sufficient statistic $T(X_1, \ldots, X_n) =(X_1, \ldots, X_n)$ for $\theta$ and that such $T$ is not complete. But I cannot even show the uniqueness of such sufficient statistic. Anyone can help me?

Michael Hardy
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user88914
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  • You have $$\int_{-\infty}^\infty f(x,\theta) , dx = \theta \int_{-\infty}^\infty f_1(x) 1_{(-\infty,0)}(x) ,dx + (1-\theta) \int_{-\infty}^\infty f_2(x) 1_{(0,\infty)}(x) , dx$$ $$ = \theta \underbrace{ \int_0^\infty f_1(x) 1_{(-\infty,0)}(x) ,dx }{\text{since } \int{-\infty}^0 f_1 = 0} {} + (1-\theta) \underbrace{ \int_{-\infty}^0 f_2(x) 1_{(0,\infty)}(x) ,dx }{\text{since } \int_0^\infty f_2=0} $$ $$= \theta \int_0^\infty 0,dx + (1-\theta) \int{-\infty}^0 0, dx =0.$$ So $x\mapsto f(x,\theta)$ is not a probability density. $\qquad$ – Michael Hardy Nov 23 '16 at 14:05

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I'm going to guess that where you wrote $\int_{-\infty}^0 f_1=0,$ $\int_0^\infty f_2=0,$ you meant $\int_{-\infty}^0 f_2=0,$ $\int_0^\infty f_1=0.$

Thus you have \begin{align} \int_{-\infty}^\infty f_1 = \int_{-\infty}^0 f_1 = 1, \\[10pt] \int_{-\infty}^\infty f_2 = \int_0^\infty f_2 = 1. \end{align} Thus for $i=1,\ldots,n$ you have $\Pr(X_i<0)=\theta$ and $\Pr(X_i>0) = 1-\theta.$

Let $Y$ be the number of negative observations in $\{X_1,\ldots,X_n\}.$ Then $$ Y\sim\operatorname{Binomial}(n,\theta). $$ The conditional density of $X_i$ given that $X_i<0$ is $f_1$, and given that $X_i>0$ is $f_2$.

See if you can show that the conditional distribution of $X_1,\ldots,X_n$ given that $Y=y$ is the same as the distribution of $X_1,\ldots,X_n$ when $\theta= y/n.$ Thus this conditional distribution does not depend on $\theta.$

Next there's the question of completeness. You have $$ \operatorname{E}(g(Y)) = \sum_{y=0}^n g(y) \binom n y \theta^y (1-\theta)^{n-y}. $$ If a polynomial in $\theta$ is equal to $0$ regardless of the value of $\theta$ then all of the coefficients are $0$. See if you can use that to show that $g(y)=0$ for every $y\in\{0,1,2,\ldots,n\}.$

Michael Hardy
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  • Very nice! Thank you and sorry for my mistake. Have a good day! – user88914 Nov 23 '16 at 16:32
  • @user88914 : I'm glad you like it. $\qquad$ – Michael Hardy Nov 24 '16 at 01:03
  • Could you explain me why the conditional density of $X_1...X_n$ given $Y=y$ is the same as the distribution of $X_1...X_n$ when $\theta = y/n$ ?? I cannot show this part of your proof. Thanks for your help :) – user88914 Nov 24 '16 at 01:31
  • @user88914 : The intuitive idea is that $\theta$ is the probability that an observation will be negative, and if in a particular case $y$ of the observations are negative and $n-y$ of them are positive, then the conditional probability given that that event has occurred, that an observation will be negative is then $y/n,$ and the conditional distribution of each observation, given that event, is then the same as the distribution in the case where $\theta = y/n.$ $[\text{ to be continued } \ldots] \qquad$ – Michael Hardy Nov 24 '16 at 03:30