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I am a bit puzzled with this problem. I am unsure how to rigorously write the result. I appreciate your help.

if $e_t\sim \mathcal{N}(0,\,\sigma_{e,t}^{2})$ and

$\sigma_{e,t}^{2}= \sigma_{e,t-1}^{2}+u_t$ where $u_t\sim \mathcal{N}(0,\,\sigma_{u}^{2})$

what is the expected variance of $e_{t+2}$ at time t, $ V_t(e_{t+2})=E_t (e_{t+2}^2)$ ?

Since $E_t (\sigma_{e,t+2}^2)=\sigma^2_{e,t}$, is it then $ V_t(e_{t+2})=\sigma^2_{e,t}$?

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  • $\begingroup$ This might be a good fit for Quantitative Finance Stack Exchange. $\endgroup$ Commented Jul 27, 2022 at 18:09

3 Answers 3

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Hi: You have to take the variance of the $u_t$'s into account since the relationship is recursive and the $\sigma^2_{e,t}$ are not observed.

You want $var(e_{t+2})$.

So, using a recursive argument, we have

(1) $\sigma^2_{e,t+2} = \sigma^2_{e,t-1} + u_{t} + u_{t+1} + u_{t+2}$.

But, $e_{t+2} \sim N(0,\sigma^2_{e,t+2})$.

So, using (1) , we need to write $\sigma^2_{e,t+2}$ as an finite sum because we don't have $\sigma^2_{e,t-1}$ because it's not observed. So, starting from $t=1$, (this assumes that $u_1$ is the first error term ), we have

$\sigma^2_{e,t+2} = \sum_{i=1}^{t+2} u_{i}$

Then, using independence of the $u_{i}$, $var\left(\sum_{i=1}^{t+2} u_{i}\right) = (t+2) \times \sigma^2_{u}$.

EDIT: NOTE THAT I WAS READING THIS AGAIN AND AN OBVIOUS QUESTION THAT SOMEONE MIGHT ASK IS: "Well, $\sigma^2_{u}$ is not observed either so what's the difference between that not being observed and $\sigma^2_{e,t}$ not being observed". My answer would be not much except for the fact that $\sigma^2_{e,t}$ has a time subscript so it's harder to estimate because it changes from period to period. $\sigma^2_{u}$ is a constant through the time periods so easier to estimate.

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  • $\begingroup$ Mark, thank you for your response and help. I have to say I am a bit confused about your answer though. It seems that you gave me an answer for the variance of the variance. I think you mistaken two points: 1) I am interested in the expected variance and not the variance of the variance. 2) I was looking for the expected variance at time t, meaning that I know the information till time t. $\endgroup$
    – burgerino
    Commented Jul 27, 2022 at 14:15
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My own answer uses the law of iterated expectation. I think it could work.

$ V_t(e_{t+2})=E_t (e_{t+2}^2)=E_t (E_{t+2} [e_{t+2}^2])=E_t (\sigma_{e,t+2}^2)=E_t (\sigma_{e,t}^2+u_{t+2}+u_{t+1})=\sigma_{e,t}^2$

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  • $\begingroup$ What you have calculated above is fine. I'll get back to you later as far as why there might be a problem with it though. $\endgroup$
    – mark leeds
    Commented Jul 28, 2022 at 14:11
  • $\begingroup$ burgerino: richard has a good point. it may be useful to send your question to quant.stackexchange. but one other thing. if you look at this paper on arch ( it's the 86 paper by engle, bollerslev ), check out 1.7 on 2967. The arch equation is re-written to eliminate $\sigma^2_t$ because there is no way to observe it. It's unobservable so not helpful to be part of the equation because its presence in the equation makes it difficult to estimate parameters. public.econ.duke.edu/~boller/Published_Papers/ben_hand_94.pdf $\endgroup$
    – mark leeds
    Commented Jul 28, 2022 at 14:17
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Hi Burgerino: I think that your answer can be viewed as correct. At the same, I think it's not so clear what is actually wanted. Still, if you really want the expectation of $e^2_{(t+2)}$ at time $t$, then what you have done is fine. But my problem with your approach is that you end up with $\sigma^2_{(e,t)}$ which is not something that can be estimated from the data ( this is what I mean when I claim that it is not observable. Sometimes the literature uses the term latent variable ) so not that useful.

I'm not sure what the context of this question is ( is the teacher discussing stochastic volatility models or ARCH models ) but my understanding ( which doesn't sound correct ) led me to view $\sigma^2_{(e,t)}$ as a random variable. Then, I calculated the variance of this random variable at time $t+2$. My expression for the variance of $\sigma^2_{(e,t+2)}$ is correct but this was not what you were looking for. So, my apologies for that.

You are correct that one way to view what I'm doing is that I'm calculating the variance of a variance. But I wouldn't concern myself with the fact that $\sigma^2_{(e,t+2)}$ is a variance. I think it's easier to view $\sigma^2_{(e,t)}$ as a random variable that takes on different values over time. So, I'm calculating the variance of an RV that happens to represent the variance of another RV, namely, $e_t$. Note that, if you consider the equation for $\sigma^2_{(e,t)}$ ( below ), a possibly more convenient viewpoint is to view it as an RV that follows a random walk.

(1) $\sigma^2_{(e,t)} = \sigma^2_{(e,t-1)} + u_t$

As I said earlier, I agree that $E_t(e^2_{(t+2)}) = \sigma^2_{(e,t)}$. I think it's best to go with your approach and then see what the person who gave this assignment ( it sounds like an assignment from a class, correct ? ) says. Maybe what you have is correct and the problem that it's not observable does not matter ? So, I hope this helped to explain what I was doing and I agree with your tower property ( conditioning twice ) approach which you use to obtain the answer. I just don't see how one uses the answer since $\sigma^2_{(e,t)}$ cannot be estimated in practice, atleast as far as I can tell.

If you find out any more regardng tnhis question, let me know because I'd be curious what the person is getting at with such a question. My email address is [email protected]. Thanks.

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