Thank you so much, it is the best video about the Lag operator, I have read the book of Shamway and Brockwell, BUt your explaination is easy to understand it! It is helpful to me !
The final equation at 5:14 does shorten the long equation but doesn't it take away a piece of crucial information on how many lags are there? Looking at your final equation I cannot tell there were three lags in the original equation. So how is that handled?
I'm confused by what's going on at 3.35. It seems like we're 'factorising' the argument to a function, which wouldn't normally be a valid operation. What makes it permissible here?
There is still a thing that I don't understand: how can the lag operator have the same value throughout the time series? Let's say, for argument's sake, that y(t) = 10, y(t-1) = 7 and y(t-2) = 8. If we calculate L as y(t-1)/y(t), we get 7/10. However, if we calculate it as y(t-2)/y(t-1) it will be 9/7, and it is a different value. I understand that if we combine the steps we get 9/10, but each time the value of L changes, so how can we say that it's L^2? It should be L(t)*L(t-1).
Hi Stefano, I think that you are misunderstanding the notation: when one writes Ly(t) this is NOT a product, but the result of applying L to the t-th element of the sequence (y(k))_k. It does not make sense therefore to say that Ly(t)=y(t-1) is equivalent to L=y(t)/y(t-1); instead, you can take the "inverse" of L to say that y(t)=L^(-1)y(t-1)
That is wrong what you are saying at 3:00. You should additionally take into account also the first three start positions in the timeseries which dissappear while lagging. That means, that to the right side of the equation at 4:55 you should add some polynomial of grade 2 whose coefficients are built up the values Y_0, Y_1, Y_2 and ε_0, ε_1 and ε_2. The index t in the final equation should also be omitted, since the equation applies to the whole series.
This is great, but isn't it confusing/misleading to use the terms "squared" and "cubed" here? These aren't really squared and cubed; they're superscripts.
It was my understanding that in the notation, y(t) gets "multiplied" by the lag operator to become y(t-1). So, in reality, we're shifting the series to go from y(t) to y(t-1) but in our notation we write it as y(t-1) equals L times y(t). So to obtain y(t-2), we lag the series twice. We lag once to get y(t-1) and then apply the lag operator to y(t-1) to get y(t-2). Since we're writing the lag operator as though it were a coefficient of y(t), we multiply y(t) x L to get y(t-1) and then multiply it by L again to get y(t-2). y(t-2) = L x y(t-1) = L x L x y(t) = L-squared x y(t). So these are squares and cubes and not just superscipts. Does that make sense?
So if there is 4 lag terms i.e, Theta 4 and Phi 4, still we will write capital Theta and Phi on the sides? this does not make sense at all.... what is this?
So much clearer than my lecturer, thankyou!
You're very welcome!
u r 1 of the best i have ever seen instructors, many thx from Egypt
Thanks for such lucid explanations. I think I'm gonna have to go through all your videos.
The order of videos in the playlist is not right
Thank you so much, it is the best video about the Lag operator, I have read the book of Shamway and Brockwell, BUt your explaination is easy to understand it! It is helpful to me !
If I only see the final equation , how do I know the actual lagged value?
Those capital letters need a number , don't they ?
hey legend can u give each of your video a order number
You explain so well. Thank you!
your videos are amazing!! really helpful!!! :)
The final equation at 5:14 does shorten the long equation but doesn't it take away a piece of crucial information on how many lags are there? Looking at your final equation I cannot tell there were three lags in the original equation. So how is that handled?
In case anyone had a similar question, the operator generally has a subscript to denote the # of lags
I wish the playlist were ordered 😕
how can you tell it has a lag value of 3? do you need to use ARMA(n,m) values?
I'm confused by what's going on at 3.35. It seems like we're 'factorising' the argument to a function, which wouldn't normally be a valid operation. What makes it permissible here?
Nice video. Just a heads up: Capital "Fee" looks like this: Φ and capital theta looks like this: Θ
thanks for the note!
The final simplified equation has no reference to how many lags there were (3). So is this just a generic equation for ARMA, regardless of the order?
Any thoughts? I'm rewatching all of these time series talks and still wondering about this. Thanks
Life saver. 💓
Good video! Very helpful. Can you make a video talking about the difference operator?
It is somehow strange to factor out an "operation", but hey, it is just notation ;-).
There is still a thing that I don't understand: how can the lag operator have the same value throughout the time series? Let's say, for argument's sake, that y(t) = 10, y(t-1) = 7 and y(t-2) = 8. If we calculate L as y(t-1)/y(t), we get 7/10. However, if we calculate it as y(t-2)/y(t-1) it will be 9/7, and it is a different value. I understand that if we combine the steps we get 9/10, but each time the value of L changes, so how can we say that it's L^2? It should be L(t)*L(t-1).
nice observation
Hi Stefano,
I think that you are misunderstanding the notation: when one writes Ly(t) this is NOT a product, but the result of applying L to the t-th element of the sequence (y(k))_k. It does not make sense therefore to say that Ly(t)=y(t-1) is equivalent to L=y(t)/y(t-1); instead, you can take the "inverse" of L to say that y(t)=L^(-1)y(t-1)
@@jdbarreryt I see now, thank you
Thank you so much for this.
No worries!
Can you please recommend a good book for time series?
Please place this video after ARMA
Thanks
Thanks dear !!
1:09 shouldn't there be a timeseries mean (c) term on the rhs?
Can you kindly share the scanned image of your working?
i have question how do we know lag 1 is either represent yesterday or last month same day, or last year same day
WOW THANK YOU
i saw this kind of polynomial function but whit z^-n as the result of z transform, what is the diference?
you legend
shouldn't the phi's have exponents?
What if the sign is different at t-1 and t-2?. How do we represent that polynomial?
That is wrong what you are saying at 3:00. You should additionally take into account also the first three start positions in the timeseries which dissappear while lagging. That means, that to the right side of the equation at 4:55 you should add some polynomial of grade 2 whose coefficients are built up the values Y_0, Y_1, Y_2 and ε_0, ε_1 and ε_2.
The index t in the final equation should also be omitted, since the equation applies to the whole series.
This is great, but isn't it confusing/misleading to use the terms "squared" and "cubed" here? These aren't really squared and cubed; they're superscripts.
It was my understanding that in the notation, y(t) gets "multiplied" by the lag operator to become y(t-1). So, in reality, we're shifting the series to go from y(t) to y(t-1) but in our notation we write it as y(t-1) equals L times y(t). So to obtain y(t-2), we lag the series twice. We lag once to get y(t-1) and then apply the lag operator to y(t-1) to get y(t-2). Since we're writing the lag operator as though it were a coefficient of y(t), we multiply y(t) x L to get y(t-1) and then multiply it by L again to get y(t-2).
y(t-2) = L x y(t-1) = L x L x y(t) = L-squared x y(t).
So these are squares and cubes and not just superscipts. Does that make sense?
So if there is 4 lag terms i.e, Theta 4 and Phi 4, still we will write capital Theta and Phi on the sides?
this does not make sense at all.... what is this?
and why we are sure that such a L existes ( L*y_t =y_(t-1)?
this final equation is not even saying how much lag it is taking, what is this, does not make sense?
thanks