Conditional Expectation of MA(1)

[brian.field]

@David Harper CFA FRM

Can you explain why the Conditional Mean for an MA(1) is not 0? I see the explanation in the previous chapter regarding:



This makes sense. But, if that was the case, then would not be 0 correct? The below seems to indicate to me that =0 and that = .

I don't really understand this.

 

[jairamjana]

The mean of the disturbance term e(t) conditional on information set omega(t-1) ={e(t-1),e(t-2),e(t-3)......} will be the unconditional mean of e(t).. So E[e(t) | omega(t-1)] will be equal to E[e(t)]... The disturbance term being white noise is uncorrelated with past disturbance and hence e(t) = 0.. .. Disturbance is also called innovation... Anyway for the equation part you should try re reading the topic General linear process in the chapter characterising cycles by diebold.. He starts with assuming a stronger independent white noise for using the models... The second indication is simple too..

Is like saying E[e(t-1)|e(t-1)] hence it equals e(t-1)..
That sentence about computers talks about y(t) as computer sales...and y(t-1) as previous quarter sales.. This will not be constant for when dependent on lagged variables unless y(t) is independent white noise..

Hope it's clear

 

[brian.field]

Thank you @jairamjana

E [ (e_t-1) | e_t-1) ] = the expected value of e_t-1 given e_t-1.

This explains it.

I was thinking that the expected value of e_t-1 = 0, which is true as an unconditional mean (i think) by covariance stationarity. But if we are given the actual e_t-1, then the expected value of it is simply it!

 

[jairamjana]

Just remember that it's same for AR(1).. We always assume e(t) is iid WN(1,sigma^2).. Wold decomposition theory should be reread multiple times..and also general linear process

 

[brian.field]

Along these same lines, is Variance(e_t-1 | e_t-1) = 0 since e_t-1 would be known and constant given itself? Make sense?

 

[brian.field]

I think it's e_t distributed as WN(0, sigma^2) rather than WN(1, sigma^2) correct?

 

[jairamjana]

Whether variance is e_t-1 or e_t it's a constant sigma^2..

Yes I made a typo for the et white noise part it's 0 and sigma ^2

 

[brian.field]

Does this make sense?

Var(Yt | Omega_t-1) = Var((e_t + theta*e_t-1) | Omega_t-1) = Var(e_t | Omega_t-1) + theta^2Var(e_t-1 | Omega_t-1) = sigma^2 + theta^2(0) = sigma^2

 

[jairamjana]

That doesn't work

var(Yt |omega_t-1) = E[(Yt - E(Yt | omega_t-1)^2 | omega_t-1 ]= E[(Yt- theta*(e_t-1))^2| omega_t-1] = E[(e_t)^2 | omega_t-1] = sigma^2

 

[jairamjana]

P.S:i edited the equation just now.. Forgot the power 2 ...

 

[brian.field]

Interesting....I need to think about this.

 

[brian.field]

Does this make sense?

Var(Yt | Omega_t-1) = Var((e_t + theta*e_t-1) | Omega_t-1) = Var(e_t | Omega_t-1) + theta^2Var(e_t-1 | Omega_t-1) = sigma^2 + theta^2(0) = sigma^2

I think this works as well.

 

[jairamjana]

I will give my opinion... That way of splitting the components only works if it is a expectation operator.. Conditional Var has to first be converted into a expectation operator form... I may not be sure of your way.. @David Harper CFA FRM.. If does require clarification..
See page 5 http://econweb.tamu.edu/gma/ECMT475/note-5

If it is uncondtional VAR your method is agreeable.. Thats what I feel..

 

[brian.field]

I am relying on V(aX+bY). I do not disagree with you @jairamjana.

 

[jairamjana]

Var(aX+bY) is an unconditional form right.. So that it will become a^2Var(x) + b^2Var(Y) .. I am just saying I am unsure if it holds for conditional VAR.. Await further feedback from others...

 

[brian.field]

@David Harper CFA FRM @jairamjana and I are trying to decide if the V(aX + bY | Z=z) = a^2Var(X|Z=z) + b^2V(Y|Z=z) approach is sound.

 

[jairamjana]

Just for further input... I followed Wikipedia.. https://en.wikipedia.org/wiki/Conditional_variance ...

 

[ami44]

I think var(X + Y) = var(X) + var(Y) is only true if X and Y are independent. In general it is
var(a*X + b*Y) = a^2 * var(X) + b^2 * var(Y) + 2*a*b*Cov(X,Y)

This formula can be derived by writing down the definition of var() and Cov() and using the linearity of the expected value:
E(a*X + b*Y) = a * E(X) + b * E(Y)

Since the conditional expected value has the same linearity property, the general formula for var(a*X + b*Y) should be also true for conditional var (you have to define something like conditional covariance).

The question in what cases the covariance is 0 is difficult. From the independence of X and Y you can not conclude the independence of X | Z and Y | Z.

So to come back to the original question:
var(X + Y | Z) = var(X | Z) + var(Y | Z)
is true, but only if X and Y are conditional independent, which is not the same as ordinary independence.

Im pretty sure, that Jairamjana's calculation above shows that conditional independence is in fact given here.

 

[jairamjana]

Thank you so much for that clarity @ami44.. So linearity is what is important.. So there you go @brian.field it works both ways...

 

[brian.field]

Fantastic! Both @ami44 and @jairamjana are absolutely fantastic! Thank you both! I can't believe I forgot the covariance term....if I remembered that, I probably would have thought about conditional independence.....