Let $X_1, \ldots, X_n$ be a random sample of size $n$ from the following distribution: $$f(x;\theta) = \left\{\begin{array}{ccc} \frac{1 - \theta}{6} & , & x = 1 \\ \frac{1 + \theta}{6} & , & x = 2 \\ \frac{2 - \theta}{6} & , & x = 3 \\ \frac{2 + \theta}{6} & , & x = 4\end{array}\right.$$

where $-1 < \theta < 1$. Find a minimal sufficient statistic for the parameter $\theta$.

Answer: I am attempting to use the Neyman Theorem: $$f(x_1;\theta)\cdots f(x_n;\theta) = k_1\Big[u_1(x_1,\ldots, x_n); \theta\Big]k_1(x_1,\ldots, x_n)$$

So, \begin{eqnarray*} f(x_1;\theta)\cdots f(x_n;\theta) & = & \prod\limits_{i = 1}^n \left(\frac{1 - \theta}{6}\right)^{n_1}\left(\frac{1 + \theta}{6}\right)^{n_2}\left(\frac{2 + \theta}{6}\right)^{n_3}\left(\frac{2 - \theta}{6}\right)^{n_4} \end{eqnarray*}

where $n = n_1 + n_2 + n_3 + n_4$.

However, I do not seem to be able to form $k_1$ and $k_2$ from this, neither am I able to obtain the sufficient statistic $u_1$. Do the $x$-values 1, 2, 3, 4 even play a role here?

  • 3
    $\begingroup$ Re your final question about the $x$ values: if they play no role, then please explain what the $n_i$ are! $\endgroup$
    – whuber
    Commented Nov 14, 2020 at 19:24
  • $\begingroup$ I am gessing that $n_1$ is the number of $X_i$ equal to 1, etc. I don't think you can do better than $n_1$, $n_2$, $n_3$ as a sufficient statistic. $\endgroup$
    – Pohoua
    Commented Nov 14, 2020 at 22:22

1 Answer 1


When looking at the joint density as [note the erroneous inclusion of the product symbol!] \begin{eqnarray*}\require{cancel} f(x_1;\theta)\cdots f(x_n;\theta) & = & \cancel{\prod\limits_{i = 1}^n} \left(\frac{1 - \theta}{6}\right)^{n_1}\left(\frac{1 + \theta}{6}\right)^{n_2}\left(\frac{2 + \theta}{6}\right)^{n_3}\left(\frac{2 - \theta}{6}\right)^{n_4} \end{eqnarray*} the factorisation is already achieved, starting from the joint density as \begin{align*} f(x_1;\theta)&\cdots f(x_n;\theta) = \prod\limits_{i = 1}^n \left(\frac{1 - \theta}{6}\right)^{\mathbb I_1(x_i)}\left(\frac{1 + \theta}{6}\right)^{\mathbb I_2(x_i)}\left(\frac{2 + \theta}{6}\right)^{\mathbb I_3(x_i)}\left(\frac{2 - \theta}{6}\right)^{\mathbb I_4(x_i)}\\ &= \left(\frac{1 - \theta}{6}\right)^{\sum_{i=1}^n\mathbb I_1(x_i)}\left(\frac{1 + \theta}{6}\right)^{\sum_{i=1}^n\mathbb I_2(x_i)}\left(\frac{2 + \theta}{6}\right)^{\sum_{i=1}^n\mathbb I_3(x_i)}\left(\frac{2 - \theta}{6}\right)^{\sum_{i=1}^n\mathbb I_4(x_i)}\\ &= \left(\frac{1 - \theta}{6}\right)^{n_1}\left(\frac{1 + \theta}{6}\right)^{n_2}\left(\frac{2 + \theta}{6}\right)^{n_3}\left(\frac{2 - \theta}{6}\right)^{n_4} \end{align*} which only depends on the four counters $n_1(\mathbf x),\ldots,n_4(\mathbf x)$, the factorisation exhibits the statistic $$S(X_1,\ldots,X_n)=\left(\sum_{i=1}^n\mathbb I_1(x_i),\sum_{i=1}^n\mathbb I_2(x_i),\sum_{i=1}^n\mathbb I_3(x_i),\sum_{i=1}^n\mathbb I_4(x_i)\right)$$ as being sufficient (if not minimal) since the product only depends on these four quantities. To find a further decomposition as $k_1(S(\mathbf X);\theta)k_2(\mathbf X)$ is somewhat moot (for instance $k_2(\mathbf x)=1$).

  • 2
    $\begingroup$ Because $S$ sums to $n,$ it is not minimal... Intuitively, any three components of $S$ will be minimal sufficient -- but that remains to be proven. To see that this is not an entirely trivial issue, consider the variant of the problem where all the occurrences of "$6$" in the definition of $f$ are replaced by "$4$" and all the occurrences of "$2$" are replaced by "$1.$" Now $\pmatrix{1&0&1&0\\0&1&0&1}S(X_1,\ldots,X_n)^\prime=(n_1+n_3, n_2+n_4)^\prime$ is (minimal) sufficient. $\endgroup$
    – whuber
    Commented Nov 20, 2020 at 18:07
  • 1
    $\begingroup$ @whuber: I did not pay attention to the "minimal" part, indeed. What would be required is to show that two different $(n_1,n_2,n_3)$ produce two different polynomials in $\theta$. $\endgroup$
    – Xi'an
    Commented Nov 20, 2020 at 18:38
  • $\begingroup$ @whuber in your variant, would $n_1+n_3$ be minimal sufficient? Or $n_1-n_2+n_3-n_4$? Or even $\frac{n_1-n_2+n_3-n_4}{n_1+n_2+n_3+n_4}$ (which seems like a plausible unbiased estimator for $\theta$) $\endgroup$
    – Henry
    Commented Nov 20, 2020 at 20:07

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