$$  \pi(\tau|x)= \frac{ B(S_{\tau}+a,b) \,_2F_1(S_{\tau}+\tau r, S_{\tau}+a, S_{\tau}+a+b; -\frac{1}{r})}
{\sum_{\tau=1}^{n} B(S_{\tau}+a,b) \,_2F_1(S_{\tau}+\tau r, S_{\tau}+a, S_{\tau}+a+b; -\frac{1}{r})}  $$

  \times \frac{B(S_{n-\tau}+a_1,b_1) \,_2F_1(S_{n-\tau}+(n-\tau) r, S_{n-\tau}+a_1, S_{n-\tau}+a_1+b_1;  -\frac{1}{r}) }{B(S_{n-\tau}+a_1,b_1) \,_2F_1(S_{n-\tau}+(n-\tau) r, S_{n-\tau}+a_1, S_{n-\tau}+a_1+b_1;  -\frac{1}{r})}
  • 3
    What is the question? – Jack Nov 14 '16 at 12:45
  • 2
    Welcome to TeX.SX! Even if the title gives hint on what your asking, being a bit more explicit would help us help you! – ebosi Nov 14 '16 at 12:50

I assume, that you wish to merge both equation in one, two line equation. In this the use of package amsmath or mathtools (improved version of amsmath) and their math environments like multline, align etc are handy:

enter image description here


\usepackage{showframe}% for show page layout, in real use had to be omitted

\pi(\tau|x) = 
    \frac{B(S_{\tau}+a,b)_2 F_1(S_{\tau}+\tau r, S_{\tau}+a, S_{\tau}+a+b; -\frac{1}{r})}
          {\sum\limits_{\tau=1}^{n} B(S_{\tau}+a,b)_2 
                             F_1(S_{\tau}+\tau r, S_{\tau}+a, S_{\tau}+a+b; -\frac{1}{r})} 
    \frac{B(S_{n-\tau}+a_1,b_1)_2F_1(S_{n-\tau}+(n-\tau)r, S_{n-\tau}+a_1, S_{n-\tau}+a_1+b_1;  -\frac{1}{r}) }
        {B(S_{n-\tau}+a_1,b_1)_2F_1(S_{n-\tau}+(n-\tau)r, S_{n-\tau}+a_1, S_{n-\tau}+a_1+b_1;  -\frac{1}{r})}
\pi(\tau|x) =
    \frac{B(S_{\tau}+a,b)_2 F_1(S,\tau,r)}
          {\sum\limits_{\tau=1}^{n} B(S_{\tau}+a,b)_2 F_1(S,\tau,r)} \times
F_1(S,\tau,r)   & = F_1(S_{\tau}+\tau r,      S_{\tau}+a,     S_{\tau}+a+b;       -\frac{1}{r})\\
F_1(S,n,\tau,r) & = F_1(S_{n-\tau}+(n-\tau)r, S_{n-\tau}+a_1, S_{n-\tau}+a_1+b_1; -\frac{1}{r})     

Let mi noted, that second part of equation is equal 1 :) (or you have some error in it).

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.