Take the 2-minute tour ×
Quantitative Finance Stack Exchange is a question and answer site for finance professionals and academics. It's 100% free, no registration required.

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?

share|improve this question
    
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. –  Richard Oct 28 '13 at 13:34
add comment

2 Answers 2

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.

share|improve this answer
    
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. –  Ilya Oct 29 '13 at 10:09
add comment

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.

share|improve this answer
    
what should I do if $A$ say $\mathcal N(1,1)$? –  Ilya Jun 23 '11 at 17:41
    
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. –  imateapot Jun 23 '11 at 18:54
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.