I was going through the derivation of the Kalman filter and it mentions that since noise (v) is uncorrelated to the state (x) and the state estimate (xbar), the following quantity is zero:

E((x - xbar) * v); where E is the expected value

I would like to know the meaning of uncorrelated intuitively as well? Thank you.


2 Answers 2


"Uncorrelated" means that as one of the two variables increases, on average the other one does not increase or decrease.

That does not mean they are independent: for example, suppose the two variables are the $x$- and $y$-coordinates of the point $(x,x^2)$ where $x\in\{-1,0,1\}$, each with equal probability. If you know $x$, that determines $y$, so they are not independent, but they are uncorrelated.

The images below are scatterplots with respective correlations $0.99$, $0.95$, $0.9$, $0.7$, $0.5$, $0.3$, $0$, $-0.3$, $-0.5$, $-0.7$, $-0.9$, $-0.95$, $-0.99$. The one case in which the correlation is $0$ is the case in which the two variables are uncorrelated. If the correlation were exactly $1$ or exactly $-1$, then all of the points would lie on a single straight line.

correlation $=0.99$:

correlation <span class=$=0.99$" />

correlation $=0.95$:

correlation <span class=$=0.95$" />

correlation $=0.9$:

correlation $=0.9

correlation $=0.7$:

correlation <span class=$=0.7$" />

correlation $=0.5$:

correlation $=0.5

correlation $=0.3$:

correlation <span class=$=0.3$" />

correlation $=0$:

correlation <span class=$=0$" />

correlation $=-0.3$:

correlation <span class=$=-0.3$" />

correlation $=-0.5$:

correlation <span class=$=-0.5$" />

correlation $=-0.7$:

correlation <span class=$=-0.7$" />

correlation $=-0.9$:

correlation <span class=$=-0.9$" />

correlation $=-0.95$:

correlation <span class=$=-0.95$" />

correlation $=-0.99$:

correlation <span class=$=-0.99$" />

  • $\begingroup$ does X and Y being completely uncorrelated also imply that pdf of Y doesn't affect pdf of X? $\endgroup$
    – Aditya
    Commented Sep 10, 2021 at 20:35
  • 1
    $\begingroup$ @Aditya No. Consider $X = 0 \text{ or }{+1}\text{ or }{-1}$ each with probability $1/3,$ and let $Y = X^2.$ Then $X,Y$ are uncorrelated, but $X$ completely determines $Y.\qquad$ $\endgroup$ Commented Sep 10, 2021 at 22:25

My interpretation of "uncorrelated" is that knowing information about one variable does not supply us with any information about the other variable.

(My apologies for not have any plots.)

  • 2
    $\begingroup$ Sorry, that's wrong. That is true if they are independent, but not necessarily true if they are merely uncorrelated. I gave the example that if $X=-1$, $0$, or $1$ each with probability $1/3$, then $X$ and $X^2$ are uncorrelated. But obviously knowing the value of $X$ supplies us with information about $X^2$. ${}\qquad{}$ $\endgroup$ Commented Sep 7, 2014 at 19:25
  • 1
    $\begingroup$ Oh. Thank you. My ignorance has been reduced. $\endgroup$ Commented Sep 8, 2014 at 0:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .