Let X1,…XnX_1,\dots X_n be an i.i.d sample from the distribution with frequency function P(X=x)=(θ2)|x|(1−θ)1−|x|P(X=x) = \left(\frac{\theta}{2}\right)^{|x|}(1-\theta)^{1-|x|} for x=−1,0,1x = -1,0,1.

Use the factorization theorem to find a sufficient statistic for θ\theta.

Does the distribution belong to the one-parameter exponential family.

I need help in these 2 questions. Im not sure how to apply the factorization theorem for question 1.

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1 Answer

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The conditional distribution of (X1,…,Xn)(X_1,\ldots,X_n) given |X1|,…,|Xn||X_1|,\ldots,|X_n| does not depend on θ\theta, since Pr(X1=1∣|X1|=1)=Pr(X1=−1∣|X1|=1)=1/2.\Pr(X_1 = 1\mid |X_1|=1) = \Pr(X_1 = -1 \mid |X_1| = 1) = 1/2. Therefore (|X1|,…,|Xn|)(|X_1|,\ldots,|X_n|) is a sufficient statistic, but it’s not a minimal sufficient statistic. To see that, show that |X1|+⋯+|Xn|∼Binomial(n,θ).|X_1|+\cdots+|X_n|\sim\operatorname{Binomial}(n,\theta).

Note that

n∏k=1(θ2)|xk|(1−θ)1−|xk|=(θ2)|x1|+⋯+|xn|(1−θ)n−(|x1|+⋯+|xn|).

\prod_{k=1}^n \left(\frac \theta 2 \right)^{|x_k|} (1-\theta)^{1-|x_k|} = \left( \frac \theta 2\right)^{|x_1|+\cdots+|x_n|} (1-\theta)^{n – (|x_1|+\cdots+|x_n|)}.