I've struck an unpleasant problem with the noncentral $\chi^2$ distribution.

I work with random variables, distributed as $\chi^2_{\nu}(\lambda)$, where $\nu$ is the degree of freedom and $\lambda$ is non-centrality parameter (which should be somehow estimated).

In "Continuous Multivariate Distributions Vol.2 by S. Kotz, N. Balakrishnan, N. L. Johnson) it is said that the Cramer-Rao bound for variance of unbiased estimators of $\hat{\lambda}$ is (page 453, eq. 29.39b): $$CRB_{\hat{\lambda}}=4(\theta\lambda^{-1}-1)^{-1}n^{-1},$$ where $n$- is the sample size, $\lambda$-true value of the noncentrality parameter that is under estimation and $\theta$ is the parameter which has the following upper and lower estimates (page 452, eq. 29.36a): $$1+\lambda^{-1}-\frac{1}{2}\nu\lambda^{-2}+\frac{1}{4}\nu^2\lambda^{-3}\leq \theta\lambda^{-1}\sigma^{-2}\leq 1+\frac{5}{4}\lambda^{-1}-\frac{20\nu-13}{39}\lambda^{-2},$$ and $\sigma$ is the noise variance.

I tried to analyse the behavior of $CRB_{\hat{\lambda}}$ when $\lambda \to 0$. But the problem is that it goes to zero to! But in my opinion it is not what we should get. I use an analogy from physics or mathematical statistics . For example this problem arises when one estimates the square of the Euclidean distance between some determined vector and its empirical sample contaminated by additive white Gaussian noise with zero mean and equal variances for all the points of the vector. And the result that I've got ($CRB_{\hat{\lambda}}\to 0$ when $\lambda \to 0$) pushes me to conclude that for a fixed level of noise ($\sigma=\mbox{const}$) it is possible to estimate infinitely close vectors with practically infinite precision (almost with zero variance of estimation).

But that shouldn't be true. It should be the opposite way. So where's the problem?


1 Answer 1


For small $\lambda$, lower bound for $\theta$ is $1/4 \times\nu^2 \lambda^{-3} \le \theta \lambda^{-1} \sigma^{-2}$

simplifying, $\theta$ varies with $\sigma^2 / \lambda^2$, thus $\theta\lambda^{-1}$ varies with $\sigma^2/\lambda$.

The inverse of $\theta \lambda^{-1} - 1$ will thus still increase with a lower value for $\lambda$.

  • $\begingroup$ On the contrary, it appears that since $\theta\lambda^{-1}-1=\sigma^2(\theta\lambda^{-1}\sigma^{-2}-\sigma^{-2})= O(\lambda^{-3})$, the OP is correct that $(\theta\lambda^{-1}-1)^{-1}$ goes to $0$ as $\lambda\to 0$. $\endgroup$
    – whuber
    Jul 3, 2014 at 20:17
  • $\begingroup$ I've edited your answer to use mathjax formatting. Please double-check that I haven't inadvertently introduced errors. $\endgroup$
    – Sycorax
    Sep 7, 2014 at 17:11

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