I'm doing some research into the Cramer-Rao bound for time of arrival localization and have come across a rather strange result: the FIM is singular, but there exists an unbiased estimator. My supervisor insists I'm wrong (I'm sure I must be), but I can't seem to figure out what I'm doing wrong. Can someone please help guide me in the right direction? Sorry for the long development of the problem below.

A transmitter $T$ is at an unknown location $\theta=(x, y)^T$. There are $s$ sensors in the region $\Omega$. The $j^\text{th}$ sensor positioned at $\theta_j=(x_j, y_j)^T$ measures a range to $T$ that is a random variable dependent on the actual distance of $T$ from the sensor ($d_j$), as shown below:

Target and anchor image

The range measurement vector for all anchors is therefore: $$ r = [d_1 + \varepsilon_1, \ldots, d_j + \varepsilon_j]^T = d(\theta)+\varepsilon=\|\theta-\theta_j\|_2 + \varepsilon $$

where the error is assumed to be normally distributed $\epsilon \sim N\left(0, \Sigma_{s_{\mathrm{x}}s} \right)$.

For a multivariate Gaussian distribution, the $(m,n)$ element of the FIM is:

$$ FIM_{m,n}=\frac{\partial d^T}{\partial \theta_m} \Sigma^{-1} \frac{\partial d}{\partial \theta_n}+\frac{1}{2} \operatorname{TR}\left(\Sigma^{-1} \frac{\partial \Sigma}{\partial \theta_m} \Sigma^{-1} \frac{\partial \Sigma}{\partial \theta_n}\right) $$

Assuming independent noise at the sensor measurement and a constant variance for the noise at each sensor, the covariance matrix is diagonal, i.e. $\Sigma=\operatorname{diag}(\sigma^2_1, \sigma^2_2,\ldots, \sigma^2_s)$. Thus, the FIM can be expanded for $\theta=(x, y)^T$ as:

\begin{equation} \left[\begin{array}{cc} FIM_{1,1} & FIM_{1,2} \\ FIM_{2,1} & FIM_{2,2} \end{array}\right] \end{equation}

\begin{align} \label{FIM11} FIM_{1,1} &= \sum_{j=1}^s \frac{1}{2} \left(\sigma_j^2\right)^{-2} \left(\frac{\partial \sigma_j^2}{\partial x} \right)^2 + \left(\sigma_j^2\right)^{-1}\left(\frac{\partial d_j}{\partial x}\right)^2 \\ & = \sum_{j=1}^s \left(\sigma_j^2\right)^{-1} \cos^2 (\phi_j) \end{align}

\begin{align} \label{FIM12} FIM_{1,2}=FIM_{2,1}&=\sum_{j=1}^{s} \frac{1}{2} \left(\sigma_j^2\right)^{-2} \frac{\partial \sigma_j^2}{\partial x} \frac{\partial \sigma_j^2}{\partial y} + \left(\sigma_j^2 \right)^{-1} \frac{\partial d_j}{\partial x} \frac{\partial d_j}{\partial y}\\ &=\sum_{j=1}^s \left(\sigma_j^2 \right)^{-1} \cos(\phi_j) \sin(\phi_j) \end{align}

\begin{align} \label{FIM22} FIM_{2,2} &= \sum_{j=1}^s \frac{1}{2}\left(\sigma_j^2 \right)^{-2} \left(\frac{\partial \sigma_j^2}{\partial y}\right)^2 + \left(\sigma_j^2 \right)^{-1} \left(\frac{\partial d_j}{\partial y}\right)^2 \\ &=\sum_{j=1}^s \left(\sigma_j^2 \right)^{-1} \sin(\phi_j)^2 \end{align}

where $\phi_j$ is the angle between the $j^\text{th}$ sensor and T, since

\begin{align} \frac{\partial d_{j}}{\partial x}&=\frac{\partial}{\partial x} \left[(x-x_j)^{2}+(y-y_j)^{2}\right]^{\frac{1}{2}}=(x-x_j)d_{j}^{-1}\\ &=cos(\phi_j) \end{align}

And similarly $\frac{\partial d_j}{\partial y}=sin(\phi_j)$.

Consider three sensors situated at $(-1,0),$ $(0,0)$ and $(1,0),$ and $T$ at $(2,0).$ The FIM is clearly singular as $\phi_j=0 \, \forall j$ and so the CRB does not exist. The negative log-likelihood function (NLL) can be written as:

\begin{align} -\mathcal{L}&=-\log\left(\prod_{j=1}^{s} P_{r}\left(r_{j} ; d_{j}\right)\right) = -\log\left(\prod_{j=1}^{s} \frac{1}{\sqrt{2 \pi \sigma_{j}^{2}}} e^{-\frac{\left(r_{j}-d_{j}\right)^{2}}{2 \sigma_{j}^{2}}}\right)\\ &= \sum_{j=1}^{s}\left[\frac{1}{2}\log (2 \pi)+\frac{1}{2} \log \left(\sigma^{2}_{j}\right)+\frac{(r-d_{j})^{2}}{2 \sigma^{2}_{j}}\right] \end{align}

I've done an MLE simulation in MATLAB for this.

The circles for two different range measurements vectors are shown below, overlayed with the negative log-likelihood (NLL) function for the measurement.

Range circles and negative log-likehood function

There are two situations that happen with the range circles:

  1. When 2 or more circles intersect, they do so at two locations (the two bluish regions in the left plot) which have reflection symmetry along the $y=0$ line, and so the NLL function has a minima at these two points.

  2. When no circles intersect, the NLL function has a minima on the $y=0$ line (think of this in a least-squares sense: when the circles don't intersect, the least squares estimate is where they are closest, which is the bluish region in the right plot).

Therefore, the $y$-coordinate MLE (i.e. estimator that minimizes NLL function) is unbiased and has finite variance.

This is the output of 1500 MLE estimates I computed in a simulation:

MLE estimates from simulation

I can't seem to figure out where I've gone wrong.

EDIT: I asked this question in math.stackexchange, but I figure this is probably a better place to ask it since it's PhD research that I'm doing. Also as background, I'm an engineer but part of my supervisory team looking at this part of my work is composed of mathematicians.

  • 1
    $\begingroup$ i) could you possibly define the $\phi_j$ for clarity?, ii) could you possibly write down the likelihood explicitly?, iii) a cursory thought is that it might be that the FIM is singular in terms of your given parametrisation, but non-singular under a simple reparametrisation, e.g. for estimating the success probability of a Bernoulli random variable, the FIM for $p$ becomes singular at $0$ and $1$, but there is a simple function $f$ such that the FIM for $f(p)$ is constant. the non-Gaussianity of your estimates suggests the MLE may be close in law to a nonlinear transformation of a Gaussian. $\endgroup$
    – πr8
    Commented Jun 12, 2023 at 10:20
  • $\begingroup$ Thanks for your suggestions in improving the question @πr8. I hope it is clearer now with the definitions of $\phi$ and the likelihood function. Sorry, I'm not sure I understand the reparameterisation you mentioned: do you mean that the Gaussian FIM would not apply in this case? The noise I am adding in the simulations is Gaussian, and I think the two situations that arise from the intersecting and non-intersecting circles show that the estimates would lie along y=0. Is there a condition on the applicability of the FIM or CRB that I might be missing maybe? $\endgroup$
    – JNL
    Commented Jun 14, 2023 at 13:38
  • 1
    $\begingroup$ I was a bit unclear, sorry. When the FIM is non-singular, one expects that the MLE asymptotically behaves like MLE ~ True_Parameter + Approximately_Gaussian_Noise. Your plots seem to suggest that this is not happening in your case (which is natural, due to the singularity of the FIM). However, it looks plausible that instead MLE ~ Nonlinear_Function( True_Parameter + Approximately_Gaussian_Noise), or even that Nonlinear_Function(MLE) ~ Nonlinear_Function(True_Parameter) + Approximately_Gaussian_Noise. The reparametrisation idea relates to this last point; I will try to write something soon. $\endgroup$
    – πr8
    Commented Jun 15, 2023 at 10:35
  • $\begingroup$ Thanks @πr8, I think I understand the concept of the reparamatrisation now. In this case, I might have to consider how to reparametrise the FIM. Do you have any reference for the Bernoulli example you gave? I tried looking it up, but didn't find anything very helpful. Also, I have another question: you mentioned that when the FIM is non-singular, one expects the MLE to approach MLE~(True_param+Gaussian_noise). I was wondering if you know whether FIM non-singularity is a sufficient condition for this? I.e. is it possible for a parameter to be non-identifiable but the FIM to be non-singular? $\endgroup$
    – JNL
    Commented Jun 23, 2023 at 1:27
  • $\begingroup$ What is the question here, specifically? What are the plots, specifically (formally) described? What are the 'circles"? ... $\endgroup$ Commented Jul 3, 2023 at 2:37

1 Answer 1


$\newcommand\th\theta\newcommand\ol\overline$Responsding to your comments:

"About the collinear sensors in my question: I understand that for $y\ne0$, the model is non-identifiable. But, what I find unusual is that for $y\ne0$, the Fisher information matrix is full rank, but at the only $y$-coordinate value where the model is identifiable (i.e. $y=0$), the Fisher information matrix is singular.

And further, at $(x,0)$, the $y$-coordinate MLE is asymptotically unbiased, even though the Fisher information matrix is singular at $(x,0)$.

All this is the consequence of the simple fact that, for collinear sensors, $(x,y)$ is a wrong choice of a parameter, because then the model is not identifiable.

To illustrate this, here is a simpler model that exhibits all the features that confused you. Let $X_1,\dots,X_n$ be independent $N(\th^2,1)$ random variables, where the parameter $\th$ may take any real value. Similarly to your model, the latter, simpler model is "identifiable only at $\th=0$". Also similarly to your model, in the simpler model the $1\times1$ Fisher information matrix is singular only at $\th=0$.

Only $\th^2$ is identifiable and estimable in the simpler model, just as only $(x,y^2)$ is identifiable and estimable in your collinear-sensors model. The MLE of $\th^2$ in the simpler model if $\widehat{\th^2_n}=\max(0,\ol X_n)$, where $\ol X_n:=\frac1n\,\sum_{i=1}^n X_i$. Similarly to your model, in the simpler model the estimators $\pm\sqrt{\widehat{\th^2_n}}=\pm\sqrt{\max(0,\ol X_n)}$ are "consistent and asymptotically unbiased at $\th=0$", in the sense that $\pm\sqrt{\widehat{\th^2_n}}\to0$ in probability and $\pm E\sqrt{\widehat{\th^2_n}}\to0$ as $n\to\infty$ if $\th=0$ is the true value of the parameter.

Summarizing, we can say again that the cause of all the confusion is that, for collinear sensors, $(x,y)$ is a wrong choice of a parameter. Letting, for collinear sensors, $(x,u)$ be the parameter with $u:=y^2$, we get an identifiable model, with an everywhere non-singular Fisher information matrix.

  • $\begingroup$ Ah I see, thanks for that explanation Iosif! It is clear to to me now why the parameters are a poor choice for the collinear sensor case and how that ties in to the singular Fisher information matrix. Really appreciate it :-) $\endgroup$
    – JNL
    Commented Jul 4, 2023 at 1:46

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