Let $X_1,X_2,X_3$ be a iid sample of the Bernoulli ($p$) distribution. Consider the statistic $T = X_1X_2 + X_3$. Show that $T$ is not sufficient for p.
I did $$P(X=x|T=t)=\frac{P(X_1=x_1,X_2=x_2,X_3=t-x_1x_2)}{P(T=t)}$$I'm stuck because I don't know how to calculate ${P(T=t)}.$
Best Answer
Use Law of Total Probability, partitioned on $X_3$
$\begin{align}\mathsf P(T{=}t)~ = & ~ \mathsf P(X_3{=}0)~\mathsf P(X_1X_2{=}t)+\mathsf P(X_3{=}1)~\mathsf P(X_1X_2{=}t{-}1)\\[1ex] = & ~ (1-p)~\mathsf P(X_1X_2{=}t)+p~\mathsf P(X_1X_2{=}t{-}1) \\[2ex]\mathsf P(X_1X_2 {=} z) ~ = & ~ \begin{cases}1-p^2 & : z=0 \\ p^2 & : z=1\\ 0 & :\textsf{otherwise}\end{cases} \\[2ex] \mathsf P(T{=}t) ~ = & ~ \begin{cases} (1-p)(1-p^2) + 0 & : t=0 \\ (1-p)p^2+p(1-p^2) & : t=1 & \\ 0+p^3 & : t=2\\0 & :\textsf{otherwise} \end{cases} \\[1ex] = & ~ \begin{cases} 1-p-p^2+p^3 & : t=0 \\ p+p^2-2p^3 & : t=1 & \\ p^3 & : t=2\\0 & :\textsf{otherwise} \end{cases} \\[3ex] \mathsf P(X_1{=}x, X_2{=}y, X_3{=}t-xy) ~= & ~\begin{cases} (1-p)^3 & : (x,y,t)=(0,0,0) \\ (1-p)^2p & : (x,y,t)\in\{(1,0,0),(0,1,0),(0,0,1)\} \\ (1-p)p^2 & : (x,y,t)\in\{ (1,0,1),(0,1,1),(1,1,1)\} \\ p^3 & : (x,y,t)=(1,1,2) \\ 0 & : \textsf{otherwise} \end{cases} \end{align}$