2
$\begingroup$

This question is an extension to my previous problem (Data association with ekf). My problem here is in the line 16 in the aforementioned link.

16.     $ j(i) = \underset{k}{\operatorname{arg\,max}} \ \ det(2 \pi S^{k})^{-\frac{1}{2}} \exp\{-\frac{1}{2} (z^{i}-\hat{z}^{k})^{T}[S^{k}]^{-1} (z^{i}-\hat{z}^{k})\} $

When I compute this line, I'm getting huge number 1.0e+09 * 3.5230. This is probability density function. Why is the pdf getting bigger than 1 in a huge way?

$\endgroup$
2
$\begingroup$

Getting a value larger than 1 in a pdf is normal. Remember that the pdf does not actually evaluate to a probability itself, but to a density. Only the integral over the function has to evaluate to 1. For a continuous variable the probability of getting exactly one particular value approaches zero. In this case you can only give the probability that the variable is within a certain interval, or you can compare the relative probability.

$\endgroup$
2
  • $\begingroup$ I'm getting zeros for other features. This makes the filter to diverge. I have no idea why this is the case. Matlab is telling me the covariance is not positive definite however in the book the covariance is semidefinite which means the covariance is greater than or equal zero. $\endgroup$ – CroCo Jul 2 '14 at 20:05
  • $\begingroup$ Sounds like another question to me :) $\endgroup$ – Jakob Jul 3 '14 at 7:35
0
$\begingroup$

Just like Jacob said. The integral of pdf from $\int_{-\infty}^{\infty}{f(t) dt}=1$ for some pdf $f(t)$. This does not mean the pdf cannot be more than one at some point (i.e. infinitesimal interval). Thinking about dirac delta function might help.

$\endgroup$

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.