# Sufficiency of linear combinations of $(X_i)_{i\ge 1}$ where $X_i\stackrel{\text{i.i.d}}\sim \text{Bernoulli}(\theta)$

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.

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...
• 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. – StubbornAtom Jul 21 '18 at 14:27
• For (a) you have to show that for a given value of $H$, $X$ takes values with the same probability. – Xi'an Jul 22 '18 at 8:23