91Ó°ÊÓ

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

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

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

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

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

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

we have that

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

Using The central limit theorem we have that $\frac{\frac{1}{{∑}_{i-1}^n}{X}_i-x}{\frac{\mathrm{Ï€}}{\sqrt{n}}}$tends to the standard normal variable as$n→∞$. Therefore, for each$\mathrm{Î\mu }>0$,

$$\begin{gathered} P\left\{\left|\frac{1}{n}\underset{i=1}{\overset{n}{∑}}{X}_i-x\right|>\mathrm{Î\mu }\right\}=P\left\{\frac{\left|\frac{1}{n}{∑}_{i=1}^n{X}_i-x\right|}{\frac{\mathrm{σ}}{\sqrt{n}}}>\frac{\mathrm{Î\mu }}{\frac{\mathrm{σ}}{\sqrt{n}}}\right\}= \\ 1-P\left\{\frac{\left|\frac{1}{n}{∑}_{i=1}^n{X}_i-x\right|}{\frac{\mathrm{σ}}{\sqrt{n}}}≤\frac{\mathrm{Î\mu }\sqrt{n}}{\mathrm{σ}}\right\}= \\ 1-P\left\{\left|\frac{\frac{1}{n}{∑}_{i=1}^n{X}_i-x}{\frac{\mathrm{σ}}{\sqrt{n}}}\right|≤\frac{\mathrm{Î\mu }\sqrt{n}}{\mathrm{σ}}\right\}≈2\mathrm{Φ}\left(-\frac{\mathrm{Î\mu }\sqrt{n}}{\mathrm{σ}}\right) \end{gathered}$$.

On the other hand,

$\begin{array}{r}\underset{n→∞}{\lim }\left(-\frac{\mathrm{Î\mu }\sqrt{n}}{\mathrm{σ}}\right)=-∞⇒\underset{n→∞}{\lim }\mathrm{Φ}\left(-\frac{\mathrm{Î\mu }\sqrt{n}}{\mathrm{σ}}\right.\\ \underset{n→∞}{\lim }2\mathrm{Φ}\left(-\frac{\mathrm{Î\mu }\sqrt{n}}{\mathrm{σ}}\right)=0\end{array}$

Hence, for each$\mathrm{Î\mu }>0$,

$P\left\{\left|\frac{1}{n}\underset{i=1}{\overset{n}{∑}}{X}_i-x\right|>\mathrm{Î\mu }\right\}→0,\text{as}n→∞$

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

$$\begin{gathered} E\left[f\left(\frac{1}{n}\underset{i=1}{\overset{n}{∑}}{X}_i\right)\right]→f\left(x\right),\text{as}n→∞\text{, i.e.} \\ {B}_n\left(x\right)→f\left(x\right),\text{as}n→∞. \end{gathered}$$