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Suppose I have a random walk $X_{n+1} = X_n+A_n$ where $A_n$ is an iid sequence, $\mathsf EA_n = A>0$. How to construct a martingale measure for this case?

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  • $\begingroup$ I like your question very much as it forces us to think about things which we use all the time in continuous time in discrete time where we should understand them "easily". I would also like to ask you to provide some more assumptions on the setting. $\endgroup$
    – Richi Wa
    Commented Oct 28, 2013 at 13:34

2 Answers 2

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Similar to the answer aleady given. We can use a measure $Q$ such that $E_Q[A_n] = 0$. Let's reformulate the sequence as $X_0 =x$ and $X_{n+1} = X_n + A_{n+1}$.

First, beause expectation is linear: $$ E_Q[X_{n+1}|F_n] = E_Q[X_n|F_n] + E_Q[A_{n+1}|F_n]. $$ Now assume that $\{F_n\}_{n=0}^\infty$ is the filtration that represents the information of $(X_n)_{n=0}^\infty$ (the sigma-algebra generated) with all the null-sets and the technical assumptions then $$ E_Q[X_n|F_n] = X_n $$ and $E_Q[A_{n+1}|F_n] = E_Q[A_{n+1}] = 0$ by independence and because $ E_Q[A_{n+1}] = 0$. If we choose $Q$ such that it is equivalent to $P$ this should be the solution.

I hope I don't miss simething important here. If $A_n \sim N(1,1)$ under $P$ then $Q$ could be $N(0,a)$ with $a>0$ which is equivalent and has the correct expectation. You can even calculate the change of measure. If I am correct then it turns out that Q is not (!) unique. I am curious about following discussions.

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  • $\begingroup$ Agree that $\mathsf Q$ must satisfy $\mathsf E_\mathsf Q A_n = 0$ for all $n$, however I wondered which shape will the Radon-Nikodym density have in this case. Now, more than 2 years after OP was posted I think I can come up with the answer - in such a case I'll post it here. Regarding your comment on the uniqueness: indeed, one shall not expect it. Even if $A$ is 3-valued, you obtain a tree model where the market is not complete, so there are several martingale measures. $\endgroup$
    – SBF
    Commented Oct 29, 2013 at 10:09
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Edit: Albeit of BFin or entry MFE type, sounds like homework.Answer: In many ways, for example take the countable product of (.-E[A])*(lawofA). More generally if g(x,y) is a function such that E[g(A,E[A])]=0 then g(.,E[A])*lawofA will do. Of course it doesn't have to be equivalent, like if A is deterministic.

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  • $\begingroup$ what should I do if $A$ say $\mathcal N(1,1)$? $\endgroup$
    – SBF
    Commented Jun 23, 2011 at 17:41
  • $\begingroup$ For g, you could use an Hermite polynomial of random (independent) degree, applied to A-1. Use it to pushforward the law of A countably many times. Construct the limit of this compatible system.Finally apply your measurable mapping: Xn+1=Xn+An. X is a martingale. Btw is this homework?The question is useless because too vague: specifying probability space and filtration would help. $\endgroup$
    – imateapot
    Commented Jun 23, 2011 at 18:54

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