Special Topics - The Kalman Filter (24 of 55) Finding the State Covariance Matrix: P=?

[@arthurtarso535]

What you need to do in order to calculate the covariance in this case is: 1) Recall the equation for position: x = x_0 + vt + at^2/2 Here I call velocity v and acceleration a for simplifying notation, since I can't write the x_dot 2)Cov(x,v) = Cov(x_0 + vt + at^2/2 , v) Note that x0 and "a" are constants in this case, so we have that: Cov(x,v) = Cov(x_0,v) + Cov(vt,v) + Cov(vt,at^2/2) = 0 + Cov(vt,v) + 0 = t * Cov(v,v) = t * Var(v) 3) If you consider time variation to be 1 second, we have that the covariance will be: 1*Var(v) = 0.04 Multiplying standard deviations in order to get covariance is a common mistake in statistics.

[@markjurik5548]

Covariance is NOT the product of two standard deviations. The latter will always be positive, but covariance can just as easily be negative as well as positive. For example, the covariance between life expectancy and drug abuse is probably well below zero. Normalize covariance and you get the correlation coefficient, which in this case would likely be - 0.95.

[@jjqqww]

if correlation is 1 then you can

[@dasilvaleandro21]

You can't obtain the covariance just by multiplying the standard deviation of each set of data.

[@KasparJohannes]

Why not?

[@florentinvonfrankenberg7102]

@@KasparJohannes covariance is not equal to standard deviation * standard deviation. I think it is a notation issue in this series, as I noticed it earlier.

[@alessandrozuech61]

@@florentinvonfrankenberg7102 In this video he's doing a process different from the one in video number 22. So is he wrong in this video?

[@matheusbg8]

@@alessandrozuech61 I have the same doubt. Checking the other commentaries I think this math is not correct.

[@gojakangas1]

I agree with others here: the off-diagonal elements are incorrectly computed.

[@vetnetdoo5356]

The covariance is absolutely NOT a product of standard deviations. It is product of standard deviation only if data absolutely covary. The other extreme is that data are independent and covariance is zero.

[@shinobicro]

well they are independent. If you think about it, that are x and y values. x and y vectors are independent one of another. That is basic linear algebra :) . x,y,z axis are defined as axes because they are independent one of another

[@avrtiny]

he corrected it in the next video.

[@EvilSpeculator]

You arrive at the standard deviation by taking the square root of either the variance (for x) or the covariance (for x and y). Thus they are closely related and you can derive one from the other. Take the standard deviation and square it - voila, here's your variance (or covariance).

[@enatouistoria]

shinobicro if they are independent then the covariance is zero

[@105d11]

@@shinobicro If they are independent, then their co-variance will be zero. That's what a zero covariance means - the two variables don't depend on each other.

[@gabrielhuber6453]

this is wrong. You have to include corraltion to get covariance. cov(x,y) = rho_xy * sigma_x * sigma_y

[@awadelrahman]

In a previous video you have shown that considering the variance instead of the standard deviation will broaden the bandwith and then we can consider more data for filtering! is that still the case even whenthe standard deviation is a fraction (less than one) as in this example, then the variance will be less than the standard deviation? Thanks

[@Akash666Akash]

Have the same query!

[@105d11]

In the earlier videos, he was making direct comparisons between sample values and the variance, but that is incorrect/invalid - you cannot do this because they have different units (e.g. m and m^2, respectively).

[@ernest987987]

How can you say that the covariance is just the multiplication of the standard déviations !!!??? You mean it is just an initial value right ? Because it is not true to say that the covariance(1,2) = std1 X std2

[@tongwang5908]

Yes, thanks for asking. I'm also confused as there are some negative numbers in the Video 23.

[@ernest987987]

+Tong Wang Sure, the covariance can clearly be negative, when 2 random variables have a tendency to move in opposite direction. The only way to make the formula cov(x , y) = std(x) times std(y) work, is to have x, y perfectly correlated (correlation of 1). So I am quite sure he is using this as initial values

[@leec8977]

Dear Michel, I see that this list has 55 videos but I can only see 42, where can I see the others? did you explore extended kalman filter? I'm doing a maester's degree and I'm using kalman filter to do modal decomposition from EEG signals to predict seizure episodes. Your videos have helped me a lot!

[@MichelvanBiezen]

We haven't completed the series yet. Perhaps in the future when we have the time.

[@LorenzoMussetti]

Why does the product of the standard deviations become the covariance? Isn't this wrong (apart from when the correlation coeff =1)? What am I missing? I'm really going crazy about this. I noticed no one was answered and I don't know why. Can you please answer professor?

[@gojakangas1]

yeah this is wrong.

[@mehrimakki]

Covariance of two variables is not the product of their standard deviations and it should be modified in this video.

[@sairamaditya9575]

The diagonal elements of a covariance matrix computed for a linearized inverse problem having model parameters m1, m2, m3, m4, m5 are 49, 15, 3, 200, 40, respectively. The standard deviation (uncertainty) in the estimation of model parameters m4 is ________.

[@bryanbocao4906]

How are PROCESS VARIATION STANDARD DEVIATIONs obtained at the beginning?

[@MichelvanBiezen]

That usually depends on the equipment, and sensors that are used, which will result in noise, and measurement errors that need to be taken into account.

[@Akash666Akash]

Hi, you have mentioned the values σx and σxdot are standard deviation caused by the process. Does that mean you have taken it as a random value for the initial t=0? or how did you calculate that?

[@MichelvanBiezen]

They are usually an educated guess.

[@benedicttan502]

Hi, is anyone able to tell me what are the criteria for a P matrix? I understand that it has to be symmetrical and positive semi-definite. I can't quite understand how do i prove that a matrix is positive semi-definite. Thanks!

[@MuhammadAltaf146]

If a matrix has all of it's eigenvalues greater than or equal to zero, the matrix is positive semi definite. In other words, the product x'Px must always be greather than or equal to zero for any non-zero vector x. Hope this answers your question.

[@benedicttan502]

Hello! Yes it does, thank you so much, appreciate it!

[@ahmedmahdi8580]

you make it very simple and understandable

[@lucasguo8090]

hopefully the author updates the video with the errors corrected, I guess it's just a glitch

[@RESC_Eng]

How can I use these equations with coding // could I be able to write a code using MATLAB to process data coming from accelerometer ,after these amazing lessons, to use them the with a mobile robot.

[@HsiaoAllenway]

Thanks a lot

[@상신김-q6w]

와웅

[@MichelvanBiezen]

Glad you liked it.