How is the following integral $B(k,n)$ - binomial distributed??

Question

ok, so the assignment is quite simple up to the point where I see the following transition, which I hope somebody can clear up :

$$n\int_{0}^{1}\sqrt{u}(1-u)^{n-1}du= nB(\frac{3}{2},n)$$ there was a substitution before this : $t^2=u$ but I doubt that's important..

Answer 1

That's a Beta function, not a binomial distribution (an integral can hardly be equal to a probability distribution anyway). The definition of the Beta function is $$ B(a,b) = \int_{0}^1 x^{a-1}(1-x)^{b-1} \, dx, $$ and you can show using polar coordinates or similar that it is equal to $$ \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}, $$ where $\Gamma$ is the Gamma-function.

(So in fact the $B$ is meant to be a capital $\beta$, which happens to look the same as a Roman B.)

Answer 2

The definition of BETA function is

$$B(x,y)=\int_0^1u^{x-1}(1-u)^{n-1}du$$

for arbitrary complex numbers $x$ and $y$ with positive real part.

Your integral is

$$n\int_0^1u^{\frac{3}{2}-1}(1-u)^{n-1}du.$$

Comparing the two definitions you get what you need.