at around 52 minutes you wrote that the variance of x1 given x0 is sigma-0 squared plus sigma-W squared. But if X0 is given, shouldn't it just be sigma0-squared? Because if it's given it's not a random variable anymore
For linear relations Y=Ax+b, A is by definition the Jacobian of Y w.r.t x, i.e. the matrix containing the partial derivative for any y_i given any x_j.
Thank a mil for this. I thought space models can also be non-linear as well as hierarchical to allow for different time scales. However, at 4:54 , it seems that you are saying state space models are always linear?! Am I missing something? Thanks.
Rewatching this now... Could you please explain why we have this equation at 11:38 ? You mentioned keeping the probability content the same, but why? I mean, some deterministic matrix A is multiplied by our random variable vector X, some random vector is added and we get a vector, Y. Why are we insisting to hold the equation at 11:38 for X and Y? Why not let Y be whatever it might? Finally, if you could kindly explain, what f(x)dx means in plain English, I would appreciate it :-). Thanks a million for such an amazing video
First, for your second question, for a continuous random variable X, f(x) is not a probability within (0,1), but a probability density within (0,infinity). What is the probability that any value X=x happen? The answer, one, is obtained by integrating f(x) over all possible value x. What is the probability that X=2.83421? The answer, zero, is justified because for continuous domains that specific instance is one out of infinity many possibilities. Therefore, if we want to define a probability for X, it must be over an interval ∆x, where you can see ∆ like a small increment of x that results in a small slice over which you integrate f(x) from x up to x+∆x. For your first question, as the function y=g(x) transforms the space x into a new space y, f(x) cannot be equal to f(y), because as g(x) either contract or expand the space, the integral of f(y) will not equal 1 anymore. On the other hand, for one small interval (x,x+∆x), we know that the associated probability is f(x)dx. We also know that our function modifies the space such that ∆y≈g(x+∆x)-g(x). Finally we know that the probability content around a specific interval x+∆x must be maintained in the transformed space y+∆y. We can write that as f(x)dx=f(y)dy, where ∆x=dx.
@@james_goulet I truly appreciate you taking the time to provide such a comprehensive explanations. Really appreciate it! My main problem is with the final line of your response: 'probability content around a specific interval x+∆x must be maintained in the transformed space y+∆y. '. Why must it be maintained exactly? Are we enforcing this constraint or this is what happens auto-magically? And if this is a constraint we are enforcing, why are we insisting on it? Thanks a lot.
@@MLDawn This is going to be slightly more technical...Given a one-to-one function y=g(x), it comes from the definition of the CDF Pr(X≤x)=F(x) that must be equal in the original space x and for the transformed space F(g(x))=F(y)=Pr(Y≤y), so that F(x)=F(y). Then if we add a perturbation ∆x to x, the equivalent perturbation for y is g(x+∆x)=y+∆y. Therefore, we have F(x+∆x)=F(y+∆y)=F(g(x+∆x)). Finally, if we take the difference F(x+∆x)-F(x), we know by definition that for ∆x->0, it is equal to f(x)∆x. From the equalities defined before, we can thus retrieve that f(x)∆x=f(y)∆y.
Would you be able to refer me to a link, or the keywords so I can see how the posterior expectation and covariance at 24:40 have been calculated? Many thanks.
@MLDawn profs.polymtl.ca/jagoulet/Site/PMLCE/CH4.html, p.39-40. Note that these equations can also be used to train deep neural networks: ruclips.net/video/jqd3Bj0q2Sc/видео.html
Great energy and involvement. Can't be explained more easier than this. Thank you!
Thank you it is really appreciated :)
The best lecture I have ever seen! Great job Prof. JA Goulet!
Thank you! :)
These videos are gold, thank you so much for this.
Thank you 🙏🤙
Your lessons are remarkable with an excellent flow. Thank you!
👌🤙
at around 52 minutes you wrote that the variance of x1 given x0 is sigma-0 squared plus sigma-W squared. But if X0 is given, shouldn't it just be sigma0-squared? Because if it's given it's not a random variable anymore
Here X_0 on the right and side is a random variable. My notation is a shortcut for Var[X_1|X_0]=Var[\int X1|x_0 f_{X_0}(x_0) dx_0].
@@james_goulet I see---thank you!
Hi again... at 21:20 why are we suddenly insisting that A should be the Jacobian? I thought A was supposed to be ANY matrix! Thanks
For linear relations Y=Ax+b, A is by definition the Jacobian of Y w.r.t x, i.e. the matrix containing the partial derivative for any y_i given any x_j.
Thank a mil for this. I thought space models can also be non-linear as well as hierarchical to allow for different time scales. However, at 4:54 , it seems that you are saying state space models are always linear?! Am I missing something? Thanks.
I could not find the passage at 4:54, but indeed SSM can also be non-linear using the extended, unscented or cubature Kalman filter.
best explanation of state space models by far
Thank you 🙌
@@james_goulet You are welcome dear professor Goulet. I had much difficulty in understanding SP models until I found your lecture.
Rewatching this now... Could you please explain why we have this equation at 11:38 ? You mentioned keeping the probability content the same, but why? I mean, some deterministic matrix A is multiplied by our random variable vector X, some random vector is added and we get a vector, Y. Why are we insisting to hold the equation at 11:38 for X and Y? Why not let Y be whatever it might? Finally, if you could kindly explain, what f(x)dx means in plain English, I would appreciate it :-). Thanks a million for such an amazing video
First, for your second question, for a continuous random variable X, f(x) is not a probability within (0,1), but a probability density within (0,infinity). What is the probability that any value X=x happen? The answer, one, is obtained by integrating f(x) over all possible value x. What is the probability that X=2.83421? The answer, zero, is justified because for continuous domains that specific instance is one out of infinity many possibilities. Therefore, if we want to define a probability for X, it must be over an interval ∆x, where you can see ∆ like a small increment of x that results in a small slice over which you integrate f(x) from x up to x+∆x. For your first question, as the function y=g(x) transforms the space x into a new space y, f(x) cannot be equal to f(y), because as g(x) either contract or expand the space, the integral of f(y) will not equal 1 anymore. On the other hand, for one small interval (x,x+∆x), we know that the associated probability is f(x)dx. We also know that our function modifies the space such that ∆y≈g(x+∆x)-g(x). Finally we know that the probability content around a specific interval x+∆x must be maintained in the transformed space y+∆y. We can write that as f(x)dx=f(y)dy, where ∆x=dx.
@@james_goulet I truly appreciate you taking the time to provide such a comprehensive explanations. Really appreciate it! My main problem is with the final line of your response: 'probability content around a specific interval x+∆x must be maintained in the transformed space y+∆y. '. Why must it be maintained exactly? Are we enforcing this constraint or this is what happens auto-magically? And if this is a constraint we are enforcing, why are we insisting on it? Thanks a lot.
@@MLDawn This is going to be slightly more technical...Given a one-to-one function y=g(x), it comes from the definition of the CDF Pr(X≤x)=F(x) that must be equal in the original space x and for the transformed space F(g(x))=F(y)=Pr(Y≤y), so that F(x)=F(y). Then if we add a perturbation ∆x to x, the equivalent perturbation for y is g(x+∆x)=y+∆y. Therefore, we have F(x+∆x)=F(y+∆y)=F(g(x+∆x)). Finally, if we take the difference F(x+∆x)-F(x), we know by definition that for ∆x->0, it is equal to f(x)∆x. From the equalities defined before, we can thus retrieve that f(x)∆x=f(y)∆y.
@@james_goulet I am not gonna lie! This went right over my head! I will read and contemplate again! Thanks a lot
James-A. is a real G
I have not clue what it means but 🤙
Would you be able to refer me to a link, or the keywords so I can see how the posterior expectation and covariance at 24:40 have been calculated? Many thanks.
@MLDawn profs.polymtl.ca/jagoulet/Site/PMLCE/CH4.html, p.39-40. Note that these equations can also be used to train deep neural networks: ruclips.net/video/jqd3Bj0q2Sc/видео.html
@@BayesWorks This is brilliant!!! Thanks a lot for being such an amazing teacher :-)
@@MLDawn 🤙🙏
I'm surprised his hair can grow down past his sharp jawline!