# Sufficient estimator for Bernoulli distribution using the likelihood function theorem for sufficiency

Let $$(X_1,X_2)$$ be a random sample of two iid random variables, $$X_1\sim Ber(\theta),\theta\in (0,1)$$. Use the following theorem to show that $$\hat{\theta}=X_1+2X_2$$ is sufficient.
Likelihood theorem for sufficiency:
Assume that $$\hat{\theta}(X_1,...,X_n)$$ and the iid random variables $$X_1,...,X_n$$ are discrete with likelihood functions $$L(x_1,...,x_n,\theta)=P_{\theta}(X_1=x_1,...,X_n=x_n)$$ and $$L_{\theta}(t,\theta)=P_{\theta}(\hat{\theta}(X_1,...,X_n)=t)$$. Then the estimator $$\hat{\theta}$$ is sufficient with respect to $$\theta$$ if and only if $$\frac{L(x_1,...,x_n,\theta)}{L_{\theta}(\hat{\theta}(x_1,...,x_n),\theta)}$$ does not depend on $$\theta$$,
for all $$(x_1,...,x_n)\in supp\ L:\hat{\theta}(X_1,...,X_n)\in supp\ L_{\theta}$$

So far I used the law of total probability to compute $$P_{\theta}(\hat{\theta}(X_1,X_2)=t)$$.Note that $$X_1,X_2\in\lbrace0,1\rbrace$$ almost surely. Then:
$$P_{\theta}(\hat{\theta}(X_1,X_2)=t)=P_{\theta}(X_1+2X_2=t)=P(X_1=t|X_2=0)P(X_2=0)+P(X_1=t-2|X_2=1)P(X_2=1)=\frac{P(X_1=t,X_2=0)P(X_2=0)}{P(X_2=0)}+\frac{P(X_1=t-2,X_2=1)P(X_2=1)}{P(X_2=1)}=P(X_1=t)P(X_2=0)+P(X_1=t-2)P(X_2=1)\\ = \theta^t(1-\theta)^{1-t}(1-\theta)+\theta^{t-2}(1-t)^{3-t}\theta=(\theta + (1-\theta))(1-\theta)^{2-t}\theta^{t-1}=(1-\theta)^{2-t}\theta^{t-1}$$
Now if we plug in $$x_1+2x_2$$ for t and compute the ratio $$\frac{P(X_1=x_1,X_2=x_2)}{P_{\theta}(\hat{\theta}(X_1,X_2)=x_1+2x_2)}$$ we get $$\frac{\theta^{x_1+x_2}(1-\theta)^{2-x_1-x_2}}{(1-\theta)^{2-x_1-2x_2}\theta^{x_1+2x_2-1}}=\frac{1}{\theta^{x_2-1}(1-\theta)^{-x_2}}$$, which obviously depends on $$\theta$$. Can someone tell me where I made a mistake ? Because the estimator should be sufficient.

• This is a wee bit of a silly problem as the vector $(X_1,X_2)$ is a one-to-one transform of the statistic $X_1+2X_2$ (which should not be written as $\hat\theta$ since this is not a reasonable estimator of $\theta$, given that $X_1+2X_2\in\{0,1,2,3\}$). Commented Dec 8, 2020 at 15:47