Why does $\int_0^1\binom{2n}{n}x^n(1-x)^ndF(x)=\frac{1}{2n+1}$?

Question

Why does $$\int_0^1\binom{2n}{n}x^n(1-x)^ndF(x)=\frac{1}{2n+1},$$

Where $F(x)$ is uniform over the interval $[0,1]$?

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It's not clear from simple factoring:

$$\binom{2n}{n}\int_0^1x^n(1-x)^ndF(x)$$ $$=\frac{(2n)!}{n!^2}\int_0^1x^n(1-x)^ndF(x).$$

Answer 1

Hint.

$$\int_0^1 x^n(1-x)^n\ dx = \frac{\Gamma (n+1)^2}{\Gamma (2 n+2)}$$

And for $n\in\mathbb{N}$ we also have $\Gamma(n+1) = n!$

Put all together and after a little algebra you will notice that

$$\frac{(2n)!}{(n!)^2}\frac{\Gamma (n+1)^2}{\Gamma (2 n+2)} = \frac{1}{1 + 2n}$$

Answer 2

By the binomial theorem,

\begin{align*} \sum_{k=0}^{n} \left( \int_{0}^{1} \binom{n}{k} x^k(1-x)^{n-k} \, dx \right) z^k &= \int_{0}^{1} (1 + (z-1)x)^n \, dx \\ &= \frac{1}{n+1} \cdot \frac{z^{n+1} - 1}{z-1} \\ &= \sum_{k=0}^{n} \frac{z^k}{n+1}. \end{align*}

So we have

$$ \int_{0}^{1} \binom{n}{k} x^k(1-x)^{n-k} \, dx = \frac{1}{n+1} $$

for all $0 \leq k \leq n$.

Answer 3

Claim

$$ \int_{0}^x (x-t)^a t^b\, dt=\frac{\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+2)}x^{a+b+1} \quad(a>-1, b>-1, x>0)\tag{0} $$ Proof

Let $f(x)=x^a$ and $g(x)=x^b$. Then $$ L[f\star g]=L[f]\times L[g]=\frac{\Gamma(a+1)}{s^{a+1}}\frac{\Gamma(b+1)}{s^{b+1}}=L\left[ \frac{\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+2)}x^{a+b+1} \right]\tag{1} $$ where $$ L[f](s)=\int_{0}^\infty f(x)e^{-sx}\,dx $$ and $$ (f\star g)(x)=\int_{0}^x f(x-t)g(t)\,dt. $$ The uniqueness of Laplace transforms for continuous functions yields the result. $\blacksquare$

Now put $x=1$ in equation (0) to get that $$ \int_{0}^1 (1-t)^a t^b\, dt=\frac{\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+2)} \quad(a>-1, b>-1).\tag{2} $$ Now put $a=n$ and $b=n$ in equation 2 to yield that $$ \int_0^1(1-t)^nt^n\, dt=\frac{\Gamma(n+1)^2}{\Gamma(2n+2)}=\frac{(n!)^2}{(2n+1)!} $$ whence $$ \binom{2n}{n}\int_0^1(1-t)^nt^n\,dt=\frac{(2n)!}{(n!)^2}\times\frac{(n!)^2}{(2n+1)!}=\frac{1}{2n+1}. $$

Answer 4

This is easy and does not require any complicated computations. Moreover, it is possible to prove a more general formula, as in @SangchulLee's answer.

Let $U_1,\dots,U_{n+1}$ be independent $U[0,1]$ random variables. Consider the event $$A = \{U_{1}\text{ is $(k+1)$th in value}\},\ k=0,\dots,n.$$ Then, by symmetry, $P(A) = \frac1{n+1}$.

On the other hand, given $U_1 = x$, the conditional probability of $A$ is $$ P(A\mid U_1 = x) = P(\text{exactly $k$ variables of $U_2,\dots,U_{n+1}$ are less than $x$})\\ = {n \choose k} x^k (1-x)^{n-k}.$$

Therefore, by the law of total probability, $$ \frac1{n+1} = P(A) = \int_{\mathbb R} P(A\mid U_1 = x)dF(x) = \int_0^1 {n \choose k} x^k (1-x)^{n-k} dx, $$ as required.