0
$\begingroup$

So, I got this problem

I need to estimate via MLE the position of a user given the distance of a certain number of fixed points. I know that (in my exercises) the distance between a point and the user can be modeled as Gaussian distribution $N(d_{i},\sigma )$ where the $_i$ indicates each point.

$d_i$ and $\sigma$ are known

Now, I know I've to maximize the likelihood function which is the production of all mine Gaussian density functions

$\prod_{i=1}^{n} \frac{1}{\sqrt{2\pi }\sigma }e^{- \frac{\sqrt{x-d_i}}{2\sigma^{2}}}$

but, I haven't really understand what's the link between MLE and the actual estimated distance

$\endgroup$
1
$\begingroup$

So you observe a sample of distances $\{d_i\}_{i=1}^n$ perturbed by normal noise with a known standard deviation of $\sigma$. The true location is $x_0$ and you want an estimator $\hat{x}$. Great.

The likelihood function is $$ \mathcal{L}(x;d,\sigma) = \prod_{i=1}^n \dfrac{1}{\sqrt{2\pi}\sigma} e^{-\dfrac{(d_i-x)^2}{2\sigma^2}} $$ if the signals are normally distribute. Like you get a data point $d_i$ from google every hour $i$, but there is noise in the measurement, so $d_i = x + \sigma^2 \varepsilon$ where $\varepsilon \sim N(0,1)$.

First take the log to simplify the calculations: $$ \log \mathcal{L} = \sum_{i=1}^N \left\lbrace -\log(\sqrt{2\pi}\sigma) - \dfrac{(d_i-x)^2}{2\sigma^2} \right\rbrace. $$ We want to pick the $\hat{x}$ that maximizes the log likelihood. The reason is that this estimator is consistent, so as $n$ becomes large, $\hat{x}_n \rightarrow x_0$. The best explanation for this is the study of $M$-estimators, which really makes the identification and convergence arguments clear without relying too much on the structure of any given statistical model.

Anyway, we get candidate maximizers by taking the first derivative and setting it equal to zero: $$ \sum_{i=1}^N \dfrac{(d_i-\hat{x})}{2\sigma^2} = 0 $$ with second order condition $$ \sum_{i=1}^N\dfrac{-1}{2\sigma^2} <0 $$ so the problem is globally concave: there is only one maximizer and it is the solution to the FONC.

Then $$ \hat{x} = \dfrac{1}{N} \sum_{i=1}^N d_i $$ so the MLE is the average of the measurements. Since $\mathbb{E}[d_i] = \mathbb{E}[x+\sigma^2 \varepsilon_i] = x$, it follows that $\mathbb{E}[\hat{x}] = x$, and your estimator is even unbiased. Congratulations!

$\endgroup$
2
  • $\begingroup$ The fact is, I don't observe samples... I know, for example, I have 4 fixed points which distance from the user has a Gaussian distribution $N(12.18,0.01)$ $N(2.07,0.01)$ $N(9.25,0.01)$ $N(8.20,0.01)$ ; also, from the estimated distance I've to find the x,y coordinates but I think it should be an easy trigonometric problem $\endgroup$
    – NicoCaldo
    Apr 12 '20 at 8:35
  • $\begingroup$ What are you trying to estimate? Seriously, think about it. $\endgroup$
    – user762914
    Apr 12 '20 at 14:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.