Variance-covariance matrix using matrix notation of factor analysis
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- Опубликовано: 29 сен 2024
- This video provides an introduction as to how we can derive the variance-covariance matrix for a set of indicator variables, when we use the matrix notation form of factor analysis models. Check out ben-lambert.co... for course materials, and information regarding updates on each of the courses. Quite excitingly (for me at least), I am about to publish a whole series of new videos on Bayesian statistics on youtube. See here for information: ben-lambert.co... Accompanying this series, there will be a book: www.amazon.co....
Mr. Ben Lambert, you are awesome. Very intuitive material. Thanks a lot.
Just to make it more clear, the element Xii would be the element (i,i) in the SAMPLE matrix minus the mean of that sample for all individuals.
I think the beginning of your video was a little confusing. Otherwise the resulting matrix makes no sense.
Sorry, I just wanted to understand that.
Thanks! this confuse me as well.
True Story, I think this post should be pinned so that people understand that Xvi is the sameple observation minus the mean of all data samples belonging to variable v.
4:02 This is the case for standardised variables i.e mean zero and std dev 1.
@Rafael Padilla this answers your question. Thanks for pointing this out @sushan upadhyay
This is important !! Otherwise the covariance matrix would make no sense
An awful video. Spent almost 30 mins to somehow derive Var and Cov from the given matrix operation. Thanks to god some comments pointed out that the original matrix has to be mean normalized.
Your videos make life easy :D
Thank you
very good explanation (Y)
they have to be mean normalized first
yes!
on 3:25, he clearly states that when dealing with standardized variables. By that he implies that variables has already been normalized
shouldn't we multiply the matrix by (N-1)^-1 instead of (N)^-1 to get unbiased estimate of the variance?
Shouldn't you divide by (N-1) rather than N since said that its a sample rather than a population?
Yes you are right. We need unbaised variance and must devide by (N-1). However for large sample N/(N-1) tends to 1.
The matrix must be for mean =0 for each column. Subract the column mean from respective column elements.
just one question: Y do we multiply by N inverse and not (N-1) inverse?
Thanks.
Awesome video! Can I ask why it's divided by N rather than N-1? I can't differentiate these 2 cases.
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 ________.
you have to mean normalize first? if you don't do mean normalization you will get a different result.
on a m x n matrix A, each row is feature, each column is sample, the covariance matrix of A should be A^T*A or A*A^T?
A*AT
To get the covarariance matric ; Xnv must be centered !?
Thank you, very helpful!
Thamks!!!!
Thank you very much!!
Thank you very much Ben. You are awesome :D
Thanks so much! Your material is GREAT! :) thx
Sir, can you please provide any reading material for reference
i love you ❤ u saved me from insanity
Thanks a lot
Do we need to rescale the X's (for calculating the Cov matrix) if they don't have the same units?
Yes
Thank you very much, Sir!
the values need to be centered first
Awesome video, thank you! Also, I think I can hear your watch in the video
You're my hero
Excellent and very helpful!
Excellent presentation, very helpful
Thank you
Thank you man
Thanks a lot for such great explanation.
May I know which tool-software combination used in this tutorial?