This question is in regards to this post where it asks if a certain statistic is sufficient for the parameter or not. My query is specifically with this problem:

Let $X_1,X_2,X_3$ be i.i.d Bernoulli variables with parameter $\theta$ where $0<\theta<1$. Verify whether $2X_1+3X_2+4X_3$ is a sufficient statistic for $\theta$ or not.

While the first statistic $X_1+2X_2+X_3$ in the linked post is definitely not sufficient for $\theta$ as shown in detail in this thread, the second statistic $2X_1+3X_2+4X_3$ is sufficient for $\theta$ by my calculations.

Suppose $H(X_1,X_2,X_3)=2X_1+3X_2+4X_3$ and $T(X_1,X_2,X_3)=X_1+X_2+X_3$.

For the binary variables $X_1,X_2$ and $X_3$, we have the following $2^3$ possible choices of $(X_1,X_2,X_3)$ and the corresponding values of $H$ and $T$ :

\begin{array}{|c|c|c|} \hline (X_1,X_2,X_3)&H(X_1,X_2,X_3)&T(X_1,X_2,X_3)\\ \hline(0,0,0)&0& 0\\ \hline(0,0,1)&4&1\\ \hline(0,1,0)& 3&1\\ \hline (0,1,1)&7&2\\ \hline (1,0,0)&2&1\\ \hline (1,0,1)&6&2\\ \hline (1,1,0)&5&2\\ \hline (1,1,1)&9&3\\ \hline \end{array}

If I am not wrong, the eight possible values assumed by $H$ can occur in only one way corresponding to the tuples in the first column and the conditional probability $P\{(X_1,X_2,X_3)\mid H\}$ equals $1$ every single time. This alone would mean that $H$ is sufficient for $\theta$, the conditional distribution of $\boldsymbol X$ given $H(\boldsymbol X)$ being independent of $\theta$.

Since the statistic $T$ is minimal sufficient for $\theta$, by definition it must be a function of the sufficient statistic $H$. While I cannot write down an explicit functional form of the map $H\to T$, I think I can say from the above table that there is a rule of correspondence that maps the values of $H$ to the values of $T$. In other words, $T$ is some function of $H$. It doesn't matter whether this function is bijective or not.

I would like to clarify if this logic is correct or not.

Related question:

Now what if there is another random variable, say $X_4$ and I am to verify whether some arbitrary linear combination $h(X_1,X_2,X_3,X_4)$, say, of the sample is sufficient for $\theta$ or not? I wouldn't want to go through all the sixteen cases like this one if $h$ is indeed sufficient for $\theta$. What would be an alternate option then? If my reasoning was correct, I don't think I can simply rule out that $h$ is sufficient before doing anything just because I could not find an explicit function of $h$ which gives me the minimal sufficient statistic.


1 Answer 1


Since$$H:(0,1)^3\longrightarrow \{0,2,...,7,9\}$$is bijective, observing $(X_1,X_2,X_3)$ or $H(X_1,X_2,X_3)$ is equivalent. Hence, $H(X_1,X_2,X_3)$ is a sufficient statistic for the very same reason that $(X_1,X_2,X_3)$ is a sufficient statistic.

The connection between $H$ and $T$ does not matter for this result. That $T(x_1,x_2,x_3)$ is a function of $H(x_1,x_2,x_3)$ is equally obvious for the very same reason that $H$ is sufficient and $T$ is minimal sufficient. Note that a direct transform is $$T(x_1,x_2,x_3)=\{H(x_1,x_2,x_3)+1\} \text{mod} 3$$

To address the "related question", ways to show that $H$ is sufficient one could (a) establish that $(X_1,..)$ given $H$ has a parameter-free distribution, that is, for any possible value $h$ taken by $H$, the inverse image of $H(x_1,\ldots)=h$ has a distribution that is independent from the parameter, (b) show that a minimal sufficient statistic $T$ is a function of $H$, or (c) find any sufficient statistic $S$ such that $S$ is a function of H...

  • $\begingroup$ I mentioned the connection between $H$ and $T$ as it was used to claim that $H$ is not sufficient for $\theta$ in the answer of the linked thread. So I was a little confused. $\endgroup$ Commented Jul 21, 2018 at 14:27
  • $\begingroup$ Could you clarify my other question? If I am given an arbitrary linear combination of more than 3 variables, and if I can make out that it is not a bijective function of the sample itself, then does that mean I have to check all cases to see if it is sufficient by drawing a table like that? Also assuming that I do not see a functional relationship between the given statistic and a minimal sufficient statistic. $\endgroup$ Commented Jul 21, 2018 at 19:34
  • $\begingroup$ Yes. I was asking about the possible ways to show (a) if I cannot show (b) and (c). $\endgroup$ Commented Jul 22, 2018 at 5:14
  • $\begingroup$ For (a) you have to show that for a given value of $H$, $X$ takes values with the same probability. $\endgroup$
    – Xi'an
    Commented Jul 22, 2018 at 8:23

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