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Let $X$ be a random variable from the following distribution $$f(x;\theta) = \left\{\begin{array}{ccc} \theta & , & x = -1 \\ (1 - \theta)^2\theta^x & , & x = 0,1,2,\ldots\end{array}\right.$$

where $0\leq \theta \leq 1$. Find a sufficient statistic for the parameter $\theta$.

Answer: I started by creating an indicator function, where $$I_1(x) = \left\{\begin{array}{ccc} 1 & , & x = -1 \\ 0 & , & {\rm otherwise}\end{array}\right.$$

Then, $$f(x;\theta) = \theta^{I_1(x)}(1 - \theta)^2\theta^x = \theta^{I_1(x) + x}(1 - \theta)^2\cdot 1$$

By the Factorization Theorem: $k_1(x;\theta) = \theta^{I_1(x) + x}(1 - \theta)^2$ and $k_2(x) = 1$.

Therefore, $T(x) = I_1(x) + x$ is a sufficient statistic.

Have I proceeded in the correct manner? What role does $0\leq \theta \leq 1$ play here?

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The representation $$f(x;\theta) = \theta^{I_1(x)}(1 - \theta)^2\theta^x$$ is incorrect, even when restricted to $x\in\{-1,0,1,\ldots\}$. It would for instance lead to $$f(-1;\theta) = \theta^{I_1(-1)}(1 - \theta)^2\theta^{-1}=(1-\theta)^2$$ (Notice the confusing impact of the notation $I_1(x)$ with $I_1(-1)=1$ and $I_1(1)=0$...)

Since the part $(1 - \theta)^2\theta^x$ should only appear when $x\ge 0$, it need be replaced with $1$ when $x=-1$, i.e., when $I_1(x)=1$, which is the case for $$\{(1 - \theta)^2\theta^x\}^{1-I_1(x)}$$

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  • $\begingroup$ So should I introduce a second indicator function for $(1 - \theta)^2\theta^x$? $\endgroup$
    – Chesso
    Commented Nov 19, 2020 at 20:57
  • $\begingroup$ Any suggestions on how I can restrict the $x$ in $(1−\theta)^2\theta^x$ to only be used if $x = 0, 1, 2, \ldots.$ $\endgroup$
    – Chesso
    Commented Nov 19, 2020 at 21:18
  • $\begingroup$ So in this case: \begin{eqnarray*} f(x;\theta) & = & \theta^{I_1(x)}\Big[(1 - \theta)^2\theta^x\Big]^{1 - I_1(x)} \\ & = & \theta^{I_1(x)}(1 - \theta)^{2(1 - I_1(x))}\theta^{x(1 - I_1(x))} \\ & = & \theta^{I_1(x) + x(1 - I_1(x))}(1 - \theta)^{2(1 - I_1(x))}\cdot 1 \\ & = & k_1(S(X);\theta)k_2(X). \end{eqnarray*} So by the factorization theorem $T(X) = (I_1(x) + x(1 - I_1(x)), 2(1 - I_1(x)))$ is a sufficient statistic of $\theta$. $\endgroup$
    – Chesso
    Commented Nov 20, 2020 at 12:40
  • $\begingroup$ I would think the homework/problem requires a sufficient statistic for an $n$ sample rather than for a single observation, $x$, since $x$ itself is obviously sufficient. Note also that $I_1(x)+(1-I_1(x))x=|x|$. $\endgroup$
    – Xi'an
    Commented Nov 20, 2020 at 14:13
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    $\begingroup$ In that case, $X$ itself is sufficient. $\endgroup$
    – Xi'an
    Commented Nov 20, 2020 at 15:35

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