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In the literature the multivariate Bernstein polynomial of a function $f:[0,1]^m\rightarrow\mathbb{R}$ is often defined as the following:

$$B_{f,n}(x_1,\dots,x_m)=\sum_{\mathbf{k}\in \{0,\dots,n\}^m} f\left(\frac{k_1}{n},\dots,\frac{k_m}{n}\right) \frac{n!}{k_1!\dots k_n!}\prod_{j=1}^m x_j^{k_j}$$

However, I have also seen an alternative representation:

$$B_{f,n}(x_1,\dots,x_m)=\sum_{\mathbf{k}\in \{0,\dots,n\}^m} f\left(\frac{k_1}{n},\dots,\frac{k_m}{n}\right)\prod_{j=1}^m \binom{n}{k_j} x_j^{k_j}(1-x_j)^{n-k_j}$$

The first corresponds to the expectation of $f$ over a vector of random variables that have a joint multinomial distribution and second is the expectation of $f$ over a vector of independent binomial random variables.

Does this alternative representation go by a different name? Or is there some equivalency between these two representations? Could anyone point me to some papers that discuss the second representation (the papers I have seen all discuss the first representation)?

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  • $\begingroup$ the first one might be wrong en.wikipedia.org/wiki/Bernstein_polynomial $\endgroup$ Mar 14, 2013 at 15:29
  • $\begingroup$ I just added a reference to the first one $\endgroup$
    – Hugh Medal
    Mar 14, 2013 at 16:16
  • $\begingroup$ they are not equivalent. the second one converges to f, while the first one is a convex function related to f $\endgroup$ Mar 14, 2013 at 17:02
  • $\begingroup$ Do you mean that the first one is a convex function in the vector x? If so, that would be great because I am trying to prove just that. Do you have any references or a proof? $\endgroup$
    – Hugh Medal
    Mar 14, 2013 at 22:05

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