# Complete and Sufficient Statistic for Discrete Distribution

I have a single observation X from the following distribution: $$𝑃(𝑋=−1)=\dfrac{𝑝}{3},𝑃(𝑋=0)=(1−𝑝),𝑃(𝑋=1)=\dfrac{2𝑝}{3}$$ I'm trying to find a complete and sufficient statistic for p based on the single observation X. My current attempt is $$T=X^{2}$$ but I'm unsure of my approach/reasoning: $$f_{\theta}(X) = \left(p \right)^{1_{(X^{2}=1)}}\left(1-p \right)^{1_{(X^{2}=0)}}$$ By factorization $$h(X) = 1$$ and $$g_{\theta}(T) = \left(p \right)^{1_{(X^{2}=1)}}\left(1-p \right)^{1_{(X^{2}=0)}}$$, meaning that $$T=X^{2}$$ is sufficient. Then to show completeness I have: $$E_{p}[a(T)] = P(X^{2}=1)a(1) + P(X^{2}=0)a(0)$$ $$=pa(1) + (1-p)a(0)=0$$ which we can see implies that $$a(T)=0$$ and thus $$T=X^{2}$$ is complete. I just wanted to make sure that this looks okay, I'm a bit unsure about showing a statistic is sufficient and complete.

## 1 Answer

Your explanation for completeness is correct. For sufficiency, I think you need to be a little more careful with the probability mass function for $$X$$. What you wrote down, $$f_\theta(x)$$, is the p.m.f of $$X^2$$, not $$X$$ itself. It should be simple enough to show that,

$$f_\theta(x) = \left(\frac{2}{3}\right)^{\frac{x(x+1)}{2}}\left(\frac{1}{3}\right)^{\frac{x(x-1)}{2}}p^{x^2}(1-p)^{1-x^2}, \;\; x = -1, 0, 1$$

Now you can use your Factorization Theorem argument and change $$h(x)$$ and $$g_\theta(T(x))$$ accordingly.