# On probabilistic extension for Bernstein polynomials

Suppose $$X_m\sim p_m(x)$$ is a discrete distribution on $$[0,1]$$ where the value takes multipliers of $$\frac{1}{m}$$ (e.g., $$p_m(x=\frac{k}{m})=\frac{1}{m+1})$$. Suppose $$p(x)=\lim\limits_{m\rightarrow\infty}p_m(x)$$, that is, $$p_m$$ converge weakly to $$p$$ on $$[0,1]$$. (e.g., $$p(x)$$ is the uniform distrubtion on $$[0,1]$$).

I wonder the following generalization on bernstein polynomials is true. $$\lim\limits_{n\rightarrow\infty}\sum\limits_{l}p_n{(x=\frac{l}{n})}\cdot\left(\sum\limits_{k=0}^n\binom{n}{k}x^k(1-x)^{n-k}u(\frac{k}{n})\right)=u\left(\int\limits_{0}^1 xp(x)dx\right)$$ for any continuous $$u$$.

• Writing $p(x) = \lim_{m \to \infty} p_m(x)$ makes it look like you want pointwise convergence, which doesn't make sense, but I suppose from context you actually want weak convergence or something similar. If so, then in probabalistic notation the question looks like the following: suppose $0 \le X_n \le 1$ are random variables with $X_n \to X$ weakly (i.e. in distribution). Suppose that conditional on $X_n$, the random variables $Y_n$ have conditional distribution $\mathrm{Bin}(n, X_n)$. For a continuous function $u$, is it the case that $E[u(Y_n/n)] \to u(E[X])$? – Nate Eldredge Feb 21 at 21:25
• Yes it should be weak convergence. Sorry for the original vagueness, i have changed it. – ZUN LI Feb 21 at 21:56

The limit is not $$u(\int_0^1 x p(x)\,dx)$$ but rather $$\int_0^1 u(x) p(x)\,dx$$.
I prefer probabilistic notation, so let $$X_n \sim p_n$$ and $$X \sim p$$. We are then supposing that $$X_n \Rightarrow X$$ in distribution. Let $$Y_n$$ have conditional distribution $$\mathrm{Bin}(n, X_n)$$ given $$X_n$$. The question is then about $$\lim_{n \to \infty} E[u(Y_n / n)]$$. Your conjecture is that it equals $$u(E[X])$$ but it actually equals $$E[u(X)]$$.
Fix $$\epsilon > 0$$. By uniform continuity there exists $$\delta > 0$$ such that if $$|s-t| < \delta$$ then $$|u(s)-u(t) < \epsilon|$$. Now by Chebyshev's inequality, we have \begin{align*} P\left(\left|\frac{Y_n}{n} - X_n\right| \ge \delta\right) &\le \delta^{-2} \operatorname{Var}\left( \frac{Y_n}{n} - X_n\right) \\ &= \delta^{-2} E\left[\operatorname{Var}\left(\frac{Y_n}{n} \mid X_n \right)\right] \\ &= \delta^{-2} n^{-1} E[X_n (1-X_n)] \\ &\le \delta^{-2} n^{-1}.\end{align*} Hence $$P\left(|u(\frac{Y_n}{n}) - u(X_n)| \ge \epsilon\right) \le \delta^{-2} n^{-1}$$, so $$|u(\frac{Y_n}{n}) - u(X_n)| \to 0$$ in probability. By dominated convergence (since $$u$$ is bounded), $$E[u(\frac{Y_n}{n})] - E[u(X_n)] \to 0$$ also. But since $$X_n \Rightarrow X$$, we have $$E[u(X_n)] \to E[u(X)]$$.