I am given an exercise about the beta distribution, with a solution:
EXAMPLE 4.11 A gasoline wholesale distributor has bulk storage tanks that hold fixed supplies and are filled every Monday. Of interest to the wholesaler is the proportion of this supply that is sold during the week. Over many weeks of observation, the distributor found that this proportion could be modeled by a beta distribution with $\alpha = 4$ and $\beta = 2$. Find the probability that the wholesaler will sell at least $90\%$ of her stock in a given week.
Solution If $Y$ denotes the proportion sold during the week, then
$$f(y) =
\begin{cases}
\frac{\Gamma(4 + 2)}{\Gamma(4)\Gamma(2)}y^3 (1 – y), & 0 \le y \le 1, \\
0, & \text{elsewhere,}
\end{cases}$$
and
$$P(Y \lt .9) = \int_.9^\infty f(y) dy = \int_.9^1 20(y^3 – y^4)dy \\
= 20 \left\{ \left. \frac{y^4}4 \right]_.9^1 – \left. \frac{y^5}5 \right]_.9^1 \right\} = 20(.004) = .08.$$
It is not very likely that $90\%$ of the stock will be sold in a given week.
In this exercise they claim that $$\frac{\Gamma(4+2)}{\Gamma(4)\Gamma(2)}=20$$
without any explicit calculations of the integral. Moreover, it seems that the result is simply $(4+2-1)(4-1 + 2-1)=5*4=20$.
What shortcut did they use to compute the gamma distribution without the need for explicit computation?
Best Answer
$$ \Gamma(z+1) = z\Gamma(z) $$
That means
$$ \frac{\Gamma(4 + 2)}{\Gamma(4)} = \frac{5\Gamma(4 + 1)}{\Gamma(4)} = \frac{4\times 5 \Gamma(4)}{\Gamma(4)} = 20 $$