I have a single observation X from the following distribution: $$𝑃(𝑋=βˆ’1)=\dfrac{𝑝}{3},𝑃(𝑋=0)=(1βˆ’π‘),𝑃(𝑋=1)=\dfrac{2𝑝}{3}$$ 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 1


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.


Your Answer

By clicking β€œPost 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.