Expectation of a Product of Martingales

Given a non-negative martingale process $(X_t)$ with its natural filtration, what can be said about
$$\mathbb{E}(X_t X_{t+1} X_{t+2} X_{t+3} | \mathcal{F}_{t-1})$$ specifically can we say $$\mathbb{E}(X_t X_{t+1} X_{t+2} X_{t+3} | \mathcal{F}_{t-1}) > (X_{t-1})^4$$ i.e. does it behave as a submartingale, without making further assumptions about covariances

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I have made the typesetting better, is this the expression you have intended? – Jonas Teuwen May 27 '11 at 15:52
Indeed it was the expression I intended thanks a bunch – Can May 30 '11 at 7:24

The answer is YES if one replaces $\gt$ by $\geqslant$ and if one assumes that $(X_t)$ has finite fourth moments. This is based on three simple properties of conditional expectation, namely:

(C) Convexity. (T) Tower property. (M) Taking out what is measurable.

(C) means that if $\varphi$ is convex and if $Z$ and $\varphi(Z)$ are integrable, then $$E(\varphi(Z)|G)\geqslant \varphi(E(Z|G)).$$ (T) means that if $H\subseteq G$ and if $Z$ is integrable, then $$E(Z|H)=E(E(Z|G)|H).$$ (M) means that if $U$ is $G$ measurable and if $UZ$ and $Z$ are integrable, then $$E(UZ|G)=UE(Z|G).$$

Now we must estimate $Y=E(Y_1|F_0)$ with $Y_1=X_1X_2X_3X_4$.

First step

(i) By (T) for $Z=Y_1$, $G=F_3$ and $H=F_0$, $Y=E(E(Y_1|F_3)|F_0)$.

(ii) By (M), since $U=X_1X_2X_3$ is $F_3$ measurable, $E(Y_1|F_3)=X_1X_2X_3E(X_4|F_3)$.

(iii) Since $E(X_4|F_3)=X_3$, $Y=E(Y_2|F_0)$ with $Y_2=X_1X_2X_3^2$.

Second step

(i) By (T) for $Z=Y_2$, $G=F_2$ and $H=F_0$, $Y=E(E(Y_2|F_2)|F_0)$.

(ii) By (M), since $U=X_1X_2$ is $F_2$ measurable, $E(Y_2|F_2)=X_1X_2E(X_3^2|F_2)$.

(iii) By (C) for $\varphi:z\mapsto z^2$, $E(X_3^2|F_2)\geqslant X_2^2$.

(iv) Because every $X_i$ is nonnegative, $E(Y_2|F_2)\geqslant Y_3$ hence $Y\geqslant E(Y_3|F_0)$, with $Y_3=X_1X_2^3$.

Third step

Apply (T) once again, this time to $Z=Y_3$, $G=F_1$ and $H=F_0$, then (C) once again, this time with $\varphi:z\mapsto z^3$ (which is convex on $z\geqslant0$ and only there), then (M) once again with $U=X_1$, and once again the nonnegativity of every $X_i$, and you are done.

Likewise, for every $n\geqslant1$, if $(X_t)$ is integrable enough, $$E(X_{t+1}X_{t+2}\cdots X_{t+n}|F_t)\geqslant E((X_{t+1})^n|F_t)\geqslant (X_t)^n.$$ The idea of synthetizing properties (C), (T) and (M) is adapted from Probability with Martingales by David Williams (see Section 9.7 Properties of conditional expectation: a list).

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If you are making explicit use of the tower property, you should also state that you are using the fact that if $X$ is $\mathcal{F}$-measurable, then $E(XY|\mathcal{F})=XE(Y|\mathcal{F})$ – Aaron May 27 '11 at 18:55

I dont know much about martingales, but I'd say yes. Calling $a,b,c,d$ your $X_t ... X_{t+3}$ conditioned to the past $F_{t-1}$, and applying the formula $E(a b) = E(a E(b|a))$ we get that the LHS is

$E(a \; E(b \; E(c \; E( d | a b c ))))$ (the variables in each term are assumed to be contitioned on the past values).

The most inner term is $E( d | a b c ) = c$

The next resulting term is $E( c^2 | a b )$ . As we know that $E(c | ab) = b$, this must be greater (or equal) than $b^2$

By the same reasoning, the next term must be greater than $a^3$, etc.

Update: Didier's answer came while I was finishing this; he says the same, basically - and better.

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