0
$\begingroup$

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.

$\endgroup$
1
  • 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 Dec 8 '20 at 15:47
-1
$\begingroup$

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.

$\endgroup$
1
  • 2
    $\begingroup$ Please type in your answer rather than giving a screenshot which is hard to read here. $\endgroup$ – mdewey Dec 24 '20 at 17:31

Your Answer

By clicking “Post 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.