Raising the bar Conditional-mean-plus-remainder representation Archives

Oct 17

Canonical form for time series

Canonical form for time series

We start with a doubly-infinite time series $\{ Y_t:t=0, \pm 1, \pm 2,... \} .$ At each point in time $t$ , in addition to $Y_t$ , we are given an information set $I_t.$ It is natural to assume that with time we know more and more: $I_t\subset I_{t+1}$ for all $t$ . We want to apply the idea used in two simpler situations before:

1) Mean-plus-deviation-from-mean representation: $Y=\mu+\varepsilon$ , where $\mu=EY$ , $\varepsilon=Y-EY$ , $E\varepsilon=0.$

2) Conditional-mean-plus-remainder representation: having some information set $I$ , we can write $Y=E_IY+\varepsilon$ , where $E_IY=E(Y|I)$ , $\varepsilon=Y-E_IY$ , $E_I\varepsilon=0.$

Notation: for any random variable $X$ , the conditional mean $E(X|I_t)$ will be denoted $E_{t}X$ .

Following the above idea, we can write $Y_{t+1}=Y_{t+1}-E_tY_{t+1}+E_tY_{t+1}$ . Hence, denoting

$\mu_{t+1}=E_tY_{t+1}$ , $\varepsilon_{t+1}=Y_{t+1}-E_tY_{t+1}$ , we get the canonical form

(1) $Y_{t+1}=\mu_{t+1}+\varepsilon_{t+1}.$

Properties

a) Conditional mean of the remainder: $E_t\varepsilon_{t+1}=E_t(Y_{t+1}-E_tY_{t+1})=0$ , because $E_tE_t=E_t$ . This implies for the unconditional mean $E\varepsilon _{t+1}=0$ by the LIE.

b) Conditional variances of $Y_{t+1}$ and $\varepsilon _{t+1}$ are the same:

$V_t(\varepsilon_{t+1})=E_t(\varepsilon_{t+1}-E_t\varepsilon_{t+1})^2=E_t\varepsilon_{t+1}^2=E_t(Y_{t+1}-E_tY_{t+1})^2=V_t(Y_{t+1}).$

c) The two terms in (1) are conditionally uncorrelated:

$Cov_t(\mu_{t+1},\varepsilon_{t+1})=E_t[(\mu_{t+1}-E_t\mu_{t+1})\varepsilon_{t+1}]=E_t[(\mu_{t+1}-\mu_{t+1})\varepsilon_{t+1}]=0$

( $\mu_{t+1}$ is known at time $t$ ).

They are also unconditionally uncorrelated: by the LIE

$Cov(\mu_{t+1},\varepsilon_{t+1})=E[(\mu_{t+1}-E\mu_{t+1})\varepsilon_{t+1}]=EE_t[(\mu_{t+1}-E\mu_{t+1})\varepsilon_{t+1}]=E[(\mu_{t+1}-E\mu_{t+1})E_t\varepsilon_{t+1}]=0.$

d) Full (long-term) variance of $Y_{t+1}$ , in addition to $V(\varepsilon_{t+1})$ , includes variance of the conditional mean $\mu _{t}$ :

$V(Y_{t+1})=V(\mu_{t+1}+\varepsilon_{t+1})=V(\mu_{t+1})+2Cov(\mu_{t+1},\varepsilon_{t+1})+V(\varepsilon_{t+1})=V(\mu_{t+1})+V(\varepsilon_{t+1}).$

e) The remainders are uncorrelated. When considering $Cov(\varepsilon_t,\varepsilon_s)$ for $s\neq t$ , by symmetry of covariance we can assume that $t\leq s-1$ . Then, remembering that $\varepsilon_t$ is known at time $s-1$ , by the LIE we have

$Cov(\varepsilon_t,\varepsilon_s)=E[E_{s-1}(\varepsilon_t\varepsilon_s)]=E[\varepsilon_tE_{s-1}\varepsilon_s] =0.$

Question: do the remainders represent white noise?

Tags: Canonical form for time series, Conditional-mean-plus-remainder representation, time series, University of London International Programmes, uol affiliate centre

Oct 17

Conditional-mean-plus-remainder representation

category: FN3142 Quantitative Finance, Quantitative Finance no comments

Conditional-mean-plus-remainder representation: we separate the main part from the remainder and find out the remainder properties. My post on properties of conditional expectation is an elementary introduction to conditioning. This is my first post in Quantitative Finance.

A brush-up on conditional expectations

Notation. Let $X$ be a random variable and let $I$ be an information set. Instead of the usual notation $E(X|I)$ for conditional expectation, in large expressions it's better to use the notation with $I$ in the subscript: $E_IX=E(X|I).$
Generalized homogeneity. If $f(I)$ depends only on information $I,$ then $E_I(f(I)X)=f(I)E_I(X)$ (a function of known information is known and behaves like a constant). A special case is $E_I(f(I))=f(I)E_I(1)=f(I).$ With $f(I)=E_I(X)$ we get $E_I(E_I(X))=E_I(X).$ This shows that conditioning is a projector: if you project a point in a 3D space onto a 2D plane and then project the image of the point onto the same plane, the result will be the same image as from single projecting.
Additivity. $E_I(X+Y)=E_IX+E_IY.$
Law of iterated expectations (LIE). If we know about two information sets that $I_1\subset I_2,$ then $E_{I_1}E_{I_2}X=E_{I_1}X.$ I like the geometric explanation in terms of projectors. Projecting a point onto a plane and then projecting the result onto a straight line is the same as projecting the point directly onto the straight line.

Conditional-mean-plus-remainder representation

This is a direct generalization of the mean-plus-deviation-from-mean decomposition. There we wrote $X=EX+(X-EX)$ and denoted $\mu=EX,~\varepsilon=X-EX$ to obtain $X=\mu+\varepsilon$ with the property $E\varepsilon=0.$