Eq. 45-46 in Gibbs Sampling for the Uninitiated

Question

I am trying to figure out how Eq. 45 simplifies to Eq. 46 in the paper - "Gibbs Sampling for the Uninitiated" by Resnik and Hardisty.www.cs.umd.edu/~hardisty/papers/gsfu.pdf (page 15)

Eq. 45

$$ \frac{\Gamma{(N + \gamma_{\pi1} + \gamma_{\pi0})} \Gamma{(C_x + \gamma_{\pi x} -1)}}{\Gamma{(C_x + \gamma_{\pi x}) \Gamma{(N + \gamma_{\pi1}+ \gamma_{\pi0} -1)}}} $$

Eq. 46

$$\frac{C_{x} + \gamma_{\pi x}}{N+ \gamma_{\pi1} + \gamma_{\pi0} -1} $$

The paper says Eq.45 simplifies to Eq. 46 by using the fact that $$ \Gamma{(a + 1)} = a\Gamma(a) $$

However when I apply the given identity, Eq. 45 simplifies to this

$$\frac{N+ \gamma_{\pi1} + \gamma_{\pi0} -1}{C_{x} + \gamma_{\pi x} -1} $$

The above equation and Eq.46 doesn't seem to be equivalent. Am I missing something?

Answer 1

Well the trick is that you can write $$ \frac{\Gamma(C_{x}+\gamma_{\pi x}) \Gamma(N + \gamma_{\pi 1} + \gamma{\pi 0}-1)}{\Gamma(N+\gamma_{\pi 1}+\gamma_{\pi 0}) \Gamma(C_{x}+\gamma_{\pi x} -1)} $$ which is the correct equation (45) in the paper as

$$ \frac{(C_{x}+\gamma_{\pi x}-1)\Gamma(C_{x}+\gamma_{\pi x}-1)\Gamma(N+\gamma_{\pi 1} + \gamma_{\pi 0} - 1)}{(N+\gamma_{\pi 1}+\gamma_{\pi 0}-1)\Gamma(N+\gamma_{\pi 1}+\gamma_{\pi 0}-1)\Gamma(C_{x}+\gamma_{\pi x} -1)} $$ This simplifies to the derire result

$$ \frac{C_{x}+\gamma_{\pi x}-1}{N+\gamma_{\pi 1}+\gamma_{\pi 0}-1}. \quad (46) $$

The thing is that equation 45 is the inverted fraction of what you seem to write in your question so the the result is also inverted.