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.

  • 1
    $\begingroup$ 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 $X_1+2X_2\in\{0,1,2,3\}$). $\endgroup$
    – Xi'an
    Commented Dec 8, 2020 at 15:47

1 Answer 1


enter image description here I hAVE tried to find the conditional distribution for two cases only...you should try to find the conditional distribution for other two cases.It will clear your doubt.

  • 2
    $\begingroup$ Please type in your answer rather than giving a screenshot which is hard to read here. $\endgroup$
    – mdewey
    Commented Dec 24, 2020 at 17:31

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