Problem on sufficient statistics

Question

Let the distribution of $X_1,X_2,...X_n$ depend on two parameters $a, b$ such that there exists a single sufficient statistic, for either parameter when the other is fixed/known.

Show that there is necessarily a pair of jointly sufficient statistics when both parameters are unknown.

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I tried by employing the factorization theorem in each when when one is known and the other is unknown, Say the parameters are $a,b$ then, by Factorisation theorem, $$\begin{array}{} f(x,a)=g_1(t_1,a)h(x) \\ f(x,b)=g_2(t_2,b)h'(x) \end{array}$$ Now, $$f(x,a)f(x,b)=f^2(x;a,b)=g_1(t_1,a)h(x)g_2(t_2,b)h'(x)$$

But, after that I can't think of anything to proceed with

Answer 1

As stressed in most of the comments, the notation is deficient and possibly explains why you cannot solve the problem.

The distribution of the sample $X_1,\cdots,X_n$ is parametrised by $(a,b)$, hence its density should be written as $$f(x_1,\ldots,x_n;a,b)$$ The assumptions made in the question about this density are that there exist two univariate functions $\hat a(x_1,\ldots,x_n;b)$ and $\hat b(x_1,\ldots,x_n;a)$ such that \begin{align} f(x_1,\ldots,x_n;a,b) &= f_a(\hat a(x_1,\ldots,x_n;b);a,b)g_a(x_1,\ldots,x_n;b)\\ &= f_b(\hat b(x_1,\ldots,x_n;a);a,b)g_b(x_1,\ldots,x_n;a) \end{align} by the factorisation theorem applied to both cases when $a$ and $b$ are the parameters (since all functions still depend on the other parameter even if it is considered as fixed).