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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 $u$.

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  • $\begingroup$ 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 $E[u(Y_n/n)] \to u(E[X])$? $\endgroup$ – Nate Eldredge Feb 21 at 21:25
  • $\begingroup$ Yes it should be weak convergence. Sorry for the original vagueness, i have changed it. $\endgroup$ – ZUN LI Feb 21 at 21:56
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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 $E[u(X_n)] \to E[u(X)]$.

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