# Problem on sufficient statistics

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.

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

• I'm voting to close this question because there are ubiquitous spelling errors that show a lack of care by the OP. – Reinstate Monica Aug 16 '19 at 7:31
• Aside from spelling, I actually found the format of the question not so bad (for instance, the question/intentions are clear and, while the question seems self-study, the OP has explained what has been tried by themselves). Therefore I took the effort to improve the question. (also I could not quickly imagine a solution/answer to this question myself, so it got my interest) – Sextus Empiricus Aug 16 '19 at 12:17
• Thank you for the support, Sir! I'll be more careful from next time! – The Three Muskets Aug 16 '19 at 12:19
• What happens when you take the square root of the final line? – Sextus Empiricus Aug 16 '19 at 12:23
• Your notation may be misleading you: it does not involve two parameters, but only one, to which you variously give names "$a$" and "$b$". Correct notation would be of the form $f(x; a,b).$ – whuber Aug 16 '19 at 12:36

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).