Special Topics - The Kalman Filter (20 of 55) Example of Covariance Matrix and Standard Deviation

[@ieyazlm9967]

I realy understood the vriance and covariance for the first time. You are realy good teacher proffessor. Please dont stop and continue with ekf and particle filtering.

[@samtree99]

You are simply the best (not "one of") teacher I have ever seen. You went a step further to describe the basic concept first before rush into the more complicated topic. I watch this series because I am building a variometer for paragliding which requires the use of Kalman filter. The videos are so good I didn't even want to finish them all at once. I want to watch it the same way as enjoying my most favor food, slowly any joyfully.

[@bArda26]

it's funny that I was watching your physics videos when I was a freshman. after a decade, here I am again!

[@MichelvanBiezen]

Welcome back!

[@OttoFazzl]

It should be noted that the 68.3% within 1 standard deviation from the mean rule is applicable to the Gaussian distribution, which Kalman filter uses as an assumption.

[@lucasgermano9322]

You're one the best teachers I've ever seen! Congrats!! Hope you find time to do the rest of the series

[@MichelvanBiezen]

Thank you and yes, we have a list of the playlists and topics we need to complete. So many topics, so little time..... :)

[@dougmanatt4317]

You cannot compare the magnitude of the variance to the measurements, the units are different (measurement value squared). Only the square root of the variance can be compared to the range of measurement values

[@joeyshen540]

Exactly, in the case when std is less than 1, the variance is even smaller than the std, there is no sense to compare variance with the measurements. However, this is still THE best series of tutorial for Kalman Filter. Hats off to the author.

[@sperge5673]

@@joeyshen540 Excuse me if I'm wrong, but you might be referring to the extended kalman filter. This refers to numbers with a standard Gaussian distribution.

[@andreastroster285]

Absolutely correct. It is impossible to compare variance and standard deviation numerically, because in general they have different physical dimension (e.g. m^2 vs. m). The professor does not know what he is talking about. :-(

[@vardanmovsisyan18]

Exactly, it is not correct to compare two different units. Also if the measurements are just number, you can not say that variance is bigger than deviation. For example if variance is less than one, the deviation will be bigger.

[@leec8977]

@@andreastroster285 it's an exaregation to say "The professor does not know what he is talking about". He did a fantastic job! a mistake can be made for anyone. If do you think you are better so make your own videos and show us.

[@a3igner]

If the standard deviation is

[@multiplanetarys]

How about the situation, when the STANDARD DEVIATION is less than 1? The VARIANCE will be smaller than STD.

[@mindthomas]

Exactly my question. What would happen then? In general I am used to only using and referring to the 3x Standard deviation to indicate the 99.7% probability.

[@EvilSpeculator]

You are both correct. The variance gets smaller and there is an error in the video. Michel defines the variance as containing the entire range in most cases. But that's incorrect as the variance is not considered to be of the same scale/unit as the input series. You can only compare standard deviation with the input series. If you want to cast a wider net then you need to go 2 or 3 standard deviations out.

[@Nuclear_Man_D]

@@mindthomas Yea, I'm thinking the same. What if the standard deviation is exceptionally large, say 10? then 3*10 = 30 for 99.7%, but the variance would be 100. Either the case of a large or small standard deviation breaks it. The scale/units of the readings will also change the scale of the variance, for instance: 10 cm standard deviation -> 100 cm^2 variance (variance 10x larger) 0.1 m standard deviation -> .01 m^2 variance (variance 10x smaller) 10 cm is the same as 0.1 m of course This makes sense with units actually, it's like calculating the area of a square. Maybe the utility of it is mentioned in the next video.

[@narjesspring4217]

The best that I have ever seen. please continue.

[@absurdengineering]

It should be said that the measurements given in the numerical example are assumed to be “noisy” measurements of the same value. If the value is changing not due to noise/measurement error, we can’t use those data to calculate variances. Variance calculations require multiple measurements of a constant quantity.

[@MichelvanBiezen]

Thank you

[@Hugo.Spain.]

Very straight forward explanation, thanks!

[@MichelvanBiezen]

Glad you liked it.

[@ahmedgaafar5369]

thank you so much Sir, your lectures are quite valuable and your style of teaching is great...thanks again.

[@boriskimelman298]

You can't put the variance and the standard deviation on the same axis, they don't have the same units. If X is meters then standard deviation of X is in meters, but variance is meters squared. You square the number but it changes the meaning. You can't compare X to X+Variance of X. You can't sum them. Variance represents the energy of X as far as I know.

[@sarojdebnath740]

Its a Brilliant Lecture.

[@MichelvanBiezen]

Glad you liked it.

[@charliesarath5770]

Thank you so munch 😊brilliant

[@MichelvanBiezen]

You’re welcome 😊 Glad you liked it.

[@tabhashim3887]

Shouldn't we divide by N-1 because this is a sample mean, not the actual population mean for the variance and covariance?

[@MichelvanBiezen]

You can do that. In the long run, the filter will smooth that out.

[@tabhashim3887]

@@MichelvanBiezen Thank you very much, professor! I love this series by the way!

[@romitjivani4367]

Your videos are so helpful, thank you so much!

[@MichelvanBiezen]

Glad they are helpful. 🙂

[@hamzeasadi671]

I think there is a mistake in 1:50. Generally speaking, you can't say sigma^2 > sigma unless sigma>1.

[@MichelvanBiezen]

Good input.

[@alexandervasilenko1481]

Thanks for lectures! Indeed seems like there is a flaw with SD and variance in this lecture. Commonly 3xSD criteria is used to estimate 99.7% probability instead of variance.

[@hhj989]

But the covariance between X and Y is not the product of variances! The notation sigma_X sigma_Y should be changed into sigma_{X,Y}

[@waynemcrae8097]

sigma_X is the standard deviation. The square of sigma_X is the variance. The equation is correct. The covariance simplifies to the product of the individual standard deviations, as written. The covariance matrix can be written as the product of C . C^T, where C is the standard deviation column vector C = [sigma _X; sigma_Y] and C^T is the transpose. This also makes it easy to generate a multidimensional covariance matrix, and determine which covariance pair goes where.

[@105d11]

@@waynemcrae8097 Covariance can be negative (just look at the definition on the white board), so it cannot be equivalent to a product of standard deviations (which are, by definition, always positive).

[@waynemcrae8097]

@@105d11 This is the sort of thing I was referring to datascienceplus.com/understanding-the-covariance-matrix/ It has been over 2 years, so it is difficult to recall why I was researching Kalman filters at the time, but the covariance matrix matches the equations in the video. As for my notation...it has been 2 years. I'll put it down to the difficulty writing math notation in a text box, or a mistype, or perhaps knowing what I meant to say but not writing clearing or proofreading well enough.

[@RachitKumar]

Thanks a lot for explaining the concept of Kalman Filter in such a simple way. I have a question, what happens when the Standard deviation is less than one? In that case variance will be smaller than standard deviation and the explanation that the variance almost incorporates all the measurements will not be true.

[@MrRezarkh]

Thanks for your great lectures. For this lecture (20 of 55), the data 2,4,5,7,7 that you used, are not normally distributed. That means the 68.3% chance is not true in this example. It is not significantly because the number of the data is low.

[@wimplender5633]

+Reza Ghaemi What point do you want to make? This has nothing to do with something being significant. Also, he just made up 5 observations, coming from some distribution he assumes to be the normal distribution with some unkown mean and variance. It is not unlikely to observe these 5 numbers if the true distribution was normal with mean 5 and variance 3.6. Note however, as is said in the video, it is only 5 observation used for illustration, in practice such a statement would not be so useful

[@andrewkelley8196]

Your statements on the percent of data within 1, 2, or 3 standard deviations only apply to normally distributed data, not any type of data. (Granted, a lot of data is normally distributed.)

[@poutnikq]

I have noticed 2 mistakes in context of comparision of sigma^2 and sigma: 1/ sigma^2 > sigma only if sigma > 1. If sigma < 1 than sigma^2 < sigma. 2/ sigma^2 does not have the same dimension than an x range with which it would be eventually compared.

[@AliRaza19052]

just got my concepts clear about the topic

[@vladsokolik1745]

what happens when the variance is smaller than 1? In that case the standard deviation is larger...

[@Surftech09]

sir, in real life situation whereby for instance a human face is to be tracked, how do I get these values?

[@Surftech09]

+sonic sonic Michel van Biezen

[@waqarakhtar7746]

Please update more videos or clear me what is H matrix in your lectures? H matrix is mentioned in Kalman Gain equation and state matrix equation..but not explained anywhere...

[@MichelvanBiezen]

+waqar akhtar This will be coming up in the near future

[@waqarakhtar7746]

thank you so much... actually i hv to present in the near future..n there is no good info on internet :(

[@markjurik5548]

The symbolic representation of covariance is incorrect. He is representing it with (sigma-x)*(sigma-y), which is the product of 2 standard deviations. The correct representation is (sigma-xy)^2.

[@rich5729]

Your average values assume we know all values of x at the beginning, but surely we don’t? Surely that should be a moving average that considers only the present and past x_i values, or have I misunderstood?

[@laiz7305]

what about standard deviation less than 1 then the variance will become smaller

[@kaihuahuang7741]

Except the variance problem, I think when calculate the std from samples, the degree of freedom should be N-1 not N.

[@antontkachev4747]

To me it seems wrong to compare a variable with something derived from its square, namely the state x and its variance which is the std squared. WRONG!!! I support Xinbi Li. What if std < 1?!

[@byatse3013]

very nise thank you

[@asifnizamani7513]

sir we just put initial position in matlab code

[@ahmedmahdi8580]

thank you pro

[@gencalicicek9528]

Thanks,

[@awadelrahman]

Thanks A lot!

[@kaymakamsken]

%68_? but why_?