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Let us consider a Markov process in discrete time $X = (X_n)_{n\geq 0}$ which values belong to $\mathbb{R}$ and $X_0 = x$. For each $x\in\mathbb{R}$ there is a probability measure $\mathsf P_x$. When one would like to make $X$ be a martingale, the change of measure can help - namely the process $Z_n \in \mathcal F_n$ such that $$ \frac {dQ}{dP}|_{\mathcal F_n} = Z_n $$ and $X$ is a $Q$-martingale. I am a bit confused since for a Markov process there is family of measures $(\mathsf P_x)_{x\in\mathbb{R}}$ rather than a unique one. How one defines a change of measure for a Markov process?

I guess that here $Z_n = f(X_n)$ - but I am looking for the formal definition of change of measure.

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1 Answer 1

One asks that properties (1), (2) and (3) below hold:

(1) The process $(X_n)$ is a $Q$-martingale, that is, $E_Q(X_{n+1}|F_n)=X_n$.

(2) There exists a $P$-martingale $(Z_n)$ such that $E_Q(U)=E(UZ_n)$ for every $F_n$-measurable random variable $U$.

(3) The process $(Z_n)$ is nonnegative.

Condition (1) means that $E_Q(X_{n+1}U)=E_Q(X_nU)$ for every $F_n$-measurable random variable $U$. Condition (2) implies that $$ E_Q(X_{n+1}U)=E(X_{n+1}UZ_{n+1}),\qquad E_Q(X_nU)=E(X_nUZ_n). $$ The two must coincide for every $F_n$-measurable random variable $U$ hence one asks that $$ E(X_{n+1}Z_{n+1}|F_n)=X_nZ_n. $$ Finally, the goal is to find a nonnegative process $(Z_n)$ such that both $(Z_n)$ and $(X_nZ_n)$ are $(F_n)$-martingales. Such a process is called a martingale density for $(X_n)$, see this paper by Martin Schweizer for an introduction in the continuous time setting.


Let us now try to exhibit a martingale density for $(X_n)$. For every nonnegative $n$, write $E_n$ for the conditional expectation conditionally on $F_n$ and introduce the processes $(Y_n)$ and $(R_n)$ defined by $$ Y_{n+1}=X_{n+1}-E_nX_{n+1},\qquad R_{n+1}=\frac{T_nY_{n+1}}{E_n(X_{n+1}Y_{n+1})}. $$ for an $F_n$-measurable random variable $T_n$, to be specified later on.

Then $E_nY_{n+1}=0$ and both $T_n$ and the denominator of $R_{n+1}$ are $F_n$-measurable hence $E_nR_{n+1}=0$. Furthermore, $E_n(R_{n+1}X_{n+1})=T_n$.

Finally, a solution to conditions (1)-(2) is given by $Z_0=1$ and, for every positive $n$, $$ Z_n=\prod_{k=1}^n(1-R_k)\quad\text{with}\ T_n=E_nX_{n+1}-X_n. $$ An equivalent formula for $R_n$ is $$ R_{n}=\frac{(X_{n}-E_{n-1}X_{n})(E_{n-1}X_{n}-X_{n-1})}{E_{n-1}(X_{n}^2)-(E_{n-1}X_{n})^2}. $$ The denominator of $R_n$ is always nonnegative, and one can probably assume it is almost surely positive and that $R_n$ is well defined. But it is not clear (to me) that $R_n\le1$ almost surely, as it should if $Z_n$ is to be a Radon-Nykodym density... In other words I do not know if condition (3) holds for the process $(Z_n)$ defined above (but there might be a simple argument showing either that (3) always holds or, that it may happen that (3) does not hold and that in these cases there is no solution to your problem).

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Thank you for the answer. Usually when I am looking for the density process $Z_t$ in a continuous time setting, I solve appropriate SDE to find a Doléans-Dade exponential which certainly admits (2) and (3) and the only question is to admit (1). Is there a similar approach in the discrete time? –  Ilya Jul 6 '11 at 14:45

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