Let $T_n$ denote the $n^\text{th}$ Chebyshev polynomial, defined by the recursion

\begin{align} T_0(x) &= 1,\\ T_1(x) &= x,\\ T_n(x) &= 2x \cdot T_{n-1} (x) - T_{n-2} (x). \end{align}

Now, supposed that I have $m \geqslant n$ samples $X_1, \ldots, X_m$ from a probability measure $p$, and I want to construct an unbiased estimator of the quantity

\begin{align} \tau_n = T_n ( \mathbf{E}_p [X]). \end{align}

I know that this is possible: one can easily construct rudimentary unbiased estimators for $\{ \mu_d = \mathbf{E}_p [X]^d \}_{0 \leqslant d \leqslant n}$ by writing $\mathbf{E}_p [X]^d = \mathbf{E}_{p^{\otimes d}}[X_1 X_2 \cdots X_d]$ , and then use a Rao-Blackwell-type argument to average over permutations of the various $X_i$ to obtain estimators of minimal variance. By linearity, it follows that unbiased estimation of $\tau_n$ is also possible.

Now, I could apply this linearity algorithmically as well, i.e. write out $T_n$ in terms of monomials, estimate the monomials as applied to $\mathbf{E}[X]$, and then aggregate them into a final estimator for $\tau_n$. However, this could get a bit hefty, particularly if I am interested in estimating $\tau_n$ for several values of $n$.

As such, I want to ask whether there is a known way of forming unbiased estimates of $\tau_n$ recursively. I know that this can be done for the sequence $\mu_n = ( \mathbf{E}_p [X])^n$, using e.g. the Newton-Girard identities, which give a highly efficient solution. My hope is that the recursive definition of the Chebyshev polynomials will enable a spiritually similar solution for $\tau_n$, but I am open to the possibility that this is too optimistic.

  • $\begingroup$ What do you mean by "efficient"? $\endgroup$
    – Xi'an
    Aug 14, 2021 at 13:33
  • $\begingroup$ For "computationally efficient", I mean that given $N$ samples from $p$, and the first $(n - 1)$ values of $\tau$, then it should take $\mathcal{O} ( N )$ time to compute $\tau_n$. For "statistically efficient", I mean that the estimator should have the minimal possible variance given the samples (i.e. I do not just want some unbiased estimator, but the optimal one). $\endgroup$
    – πr8
    Aug 14, 2021 at 13:44
  • $\begingroup$ Are you sure there exists a uniformly minimal variance estimator in this setting? $\endgroup$
    – Xi'an
    Aug 14, 2021 at 20:04
  • $\begingroup$ Not entirely. I am going by the intuition that the best estimator for $\mu_d$ is obtained by averaging $X_1 X_2 \cdots X_d$ over all permutations of the samples (I would guess that this can be made rigorous), and then that the best estimator for $\tau_n$ ought to be obtained by writing $\tau_n$ as a linear combination of $\{ \mu_d \}_{d \leqslant n}$ and then combining the "optimal" estimators for $\mu_d$ linearly in the corresponding way. But certainly this reasoning may not be air-tight! $\endgroup$
    – πr8
    Aug 14, 2021 at 20:35
  • $\begingroup$ A Monte Carlo Rao-Blackwellisation would preserve unbiasedness. $\endgroup$
    – Xi'an
    Aug 15, 2021 at 9:18


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.