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Consider a probability distribution F on the interval $[0,1]$ then it can be approximated by a Bernstein polynomial $$ p_n(x) = \sum_{k=0}^n \dbinom{n}{k} F \left( \frac{k}{n}\right) x^k (1-x)^{n-k} $$

By calculating the derivative of the polynomial $p_{n+1}(x)$ we can find a Bernstein polynomial of degree $n$ that approximates the density relative to $F$ (assuming the density exists)

$$ P_n^f(x)=\sum\limits_{k=0}^n {{n}\choose{k}} x^k (1-x)^{n-k}(n+1)\int\limits_{k/(n+1)}^{(k+1)/(n+1)} f(t) dt. $$ then by exploiting the convegrence of Bernstein polynomial we know that $$ \lim_{n\rightarrow \infty} P_n^f(x)=f(x) $$ One can rewrite $P_n^f(x)$ as a mixture of Beta densities $$ P_n^f(x)=\sum\limits_{k=0}^n w_k \beta(k+1,n-k+1;x) $$ where $\beta(\alpha_k,\beta_k;x)\propto x^{\alpha_k-1} (1-x)^{\beta_k-1}$ is the Beta density.

In $P_n^f(x)$ the parameters $\alpha_k,\beta_k$ are integers greater than one, but the Beta density is also defined for $0<\alpha_k,\beta_k<1$. For instance $\beta(\alpha_k,\beta_k;x)\propto x^{-1/2} (1-x)^{-1/2}$. My question is the mixture $$ P_n^f(x)=\sum\limits_{k=0}^n w_k \beta(\alpha_k,\beta_k;x) $$ with $0<\alpha_k,\beta_k$ and $\alpha_k+\beta_k=M$ for each $k$ has also the property $$ \lim_{n\rightarrow \infty} P_n^f(x)=f(x) $$ For instance we may consider $\alpha_k=Mr$ and $\beta_k=M(1-r)$ with $r=\min(\max(\epsilon,k/n),1-\epsilon)$ for some $\epsilon>0$ (small). The $\epsilon$ is introduced to avoid the cases $\alpha_k=0$ or $\beta_k=0$, where the Beta is not integrable.

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Sorry but the pointwise convergence of $P_n^f$ to $f$ needs some justification more serious than what you write (or some restriction on $f$). – Did Jun 8 '12 at 3:40
You are right, I mean uniform convergence as explained here About probability density f, the condition is that it is the indefinite integral of the distribution F – vatna Jun 11 '12 at 7:35
Uniform convergence is hopeless in general: consider $f$ with some discontinuity points. – Did Jun 19 '12 at 7:52

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