Past exam of the mathematics course of the University of Cambridge / 2021 / ia / Paper 2 / 12F (Probability)

The Chebyshev inequality states that for a random variable $Y$ with finite variance and every $\delta >0$,Indeed, Markov inequality applied to the nonnegative variable $(Y-\mathbb{E}Y{)}^2$ gives

The sum $S_n=X_1+\cdots +X_n$ has the binomial distribution, soFor fixed $n$ this is a finite sum of polynomial functions of $p$, hence is a polynomial.

The sample mean $S_n/n$ has expectation $p$ and varianceThe sum in the question is exactlyChebyshev's inequality therefore gives, uniformly in $p$,

Because $f$ is continuous on the compact interval $[0,1]$, it is uniformly continuous and bounded; write $|f|\le M$. Given $ϵ>0$, choose $\delta >0$ such thatThenThe bound is independent of $p$. Taking the supremum, then $n\to \infty$, gives a limit superior at most $ϵ$. Since $ϵ$ is arbitrary,This is the Bernstein polynomial proof of the Weierstrass approximation theorem.