Let $f\left(x\right)$be a continuous function defined for$0\le x\le 1$. Consider the functions

$B_n\left(x\right)=\sum _{k=0}^{n}f\left(\frac{k}{n}\right)\left(\frac{n}{k}\right)x^k{(1-x)}^{n-k}$(called Bernstein polynomials) and prove that

$\underset{n\to \infty }{\lim }{B}_n\left(x\right)=f\left(x\right)$.

Hint: Let $X1,X2,...$be independent Bernoulli random variables with mean$x$. Show that

$B_n\left(x\right)=E\left[f\left(\frac{x_1+...+x_n}{n}\right)\right]$

and then use Theoretical Exercise$8.4$.

Since it can be shown that the convergence of $B_n\left(x\right)$to $f\left(x\right)$is uniform$x$, the preceding reasoning provides a probabilistic proof of the famous Weierstrass theorem of analysis, which states that any continuous function on a closed interval can be approximated arbitrarily closely by a polynomial.

$f\left(x\right)$be a continuous function defined for$0\le x\le 1.$

Let$f\left(x\right),0\le x\le 1$, be a continuous function. It is then also bounded on this interval. Let's define$B_n\left(x\right)$, it as

$B_n\left(x\right)=\sum _{k=0}^{n}f\left(\frac{k}{n}\right)\left(\begin{array}{c}n\\ k\end{array}\right)x^k(1-x{)}^{n-k}$

We want to show that$\underset{n\to \infty }{\lim }{B}_n\left(x\right)=f\left(x\right)$.

Let$X_1,X_2,\dots$be independent Bernoulli random variables with mean$E\left[X_i\right]=x$. Notice that then$X_1+X_2+\cdots +X_n$is a Binomial random variable with parameters ( $n,x$). Therefore, since

$P\left\{\sum _{i=1}^n{X}_i=k\right\}=\left(\begin{array}{c}n\\ k\end{array}\right)x^k(1-x{)}^{n-k}$

we have that

$$\begin{gathered} E\left[f\left(\frac{1}{n}\sum _{i=1}^n{X}_i\right)\right]=\sum _{k=0}^{n}f\left(\frac{k}{n}\right)P\left\{\sum _{i=1}^n{X}_i=k\right\}= \\ \sum _{k=0}^{n}f\left(\frac{k}{n}\right)\left(\begin{array}{c}n\\ k\end{array}\right)x^k(1-x{)}^{n-k}=B_n(x) \end{gathered}$$.

Using The central limit theorem we have that $\frac{\frac{1}{{\sum }_{i-1}^n}{X}_i-x}{\frac{\pi }{\sqrt{n}}}$tends to the standard normal variable as$n\to \infty$. Therefore, for each$\epsilon >0$,

$$\begin{gathered} P\left\{\left|\frac{1}{n}\sum _{i=1}^n{X}_i-x\right|>\epsilon \right\}=P\left\{\frac{\left|\frac{1}{n}{\sum }_{i=1}^n{X}_i-x\right|}{\frac{\sigma }{\sqrt{n}}}>\frac{\epsilon }{\frac{\sigma }{\sqrt{n}}}\right\}= \\ 1-P\left\{\frac{\left|\frac{1}{n}{\sum }_{i=1}^n{X}_i-x\right|}{\frac{\sigma }{\sqrt{n}}}\le \frac{\epsilon \sqrt{n}}{\sigma }\right\}= \\ 1-P\left\{\left|\frac{\frac{1}{n}{\sum }_{i=1}^n{X}_i-x}{\frac{\sigma }{\sqrt{n}}}\right|\le \frac{\epsilon \sqrt{n}}{\sigma }\right\}\approx 2\Phi \left(-\frac{\epsilon \sqrt{n}}{\sigma }\right) \end{gathered}$$.

On the other hand,

$\begin{array}{r}\underset{n\to \infty }{\lim }\left(-\frac{\epsilon \sqrt{n}}{\sigma }\right)=-\infty \Rightarrow \underset{n\to \infty }{\lim }\Phi \left(-\frac{\epsilon \sqrt{n}}{\sigma }\right.\\ \underset{n\to \infty }{\lim }2\Phi \left(-\frac{\epsilon \sqrt{n}}{\sigma }\right)=0\end{array}$

Hence, for each$\epsilon >0$,

$P\left\{\left|\frac{1}{n}\sum _{i=1}^n{X}_i-x\right|>\epsilon \right\}\to 0,\text{as}n\to \infty$

Now, according to the Theoretical Exerciserole="math" localid="1649857976771" $4$, we have that

$$\begin{gathered} E\left[f\left(\frac{1}{n}\sum _{i=1}^n{X}_i\right)\right]\to f\left(x\right),\text{as}n\to \infty \text{, i.e.} \\ {B}_n\left(x\right)\to f\left(x\right),\text{as}n\to \infty . \end{gathered}$$