Constructing an Equivalent Martingale Measure for a Discrete Random Walk

Summary

The document considers a discrete random walk with independent, identically distributed increments that have positive mean under the original probability measure. It explains that a candidate measure can make the process a martingale by giving each next increment conditional mean zero, provided the increments remain independent of the information available at the prior time. Under that measure, the conditional expectation of the next state equals the current state.

It gives a Gaussian illustration in which the increment distribution is changed to one with zero mean, and notes that multiple equivalent measures may be possible. A second answer describes further ways to construct increment distributions with zero expectation, including cases where equivalence is not required. The discussion is informal and leaves technical conditions largely implicit: equivalence, independence, integrability, and whether the proposed measure on the full sequence is valid all require care. It does not establish a unique measure or connect the construction to a specific traded asset or pricing model.

Key ideas

• A random walk is a martingale under a measure when each next increment has conditional expectation zero.
• Independence of the increments from prior information supports the conditional expectation calculation.
• Changing the increment distribution can produce a measure with zero mean increments.
• More than one candidate measure may satisfy the zero mean condition.
• Equivalence and other measure construction conditions need to be checked carefully.

Tags

# Change of measure discrete time


# Change of measure discrete time












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?

## Answer by Richi Wa (score 2)

https://quant.stackexchange.com/a/9285

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.

## Answer by imateapot (score 1)

https://quant.stackexchange.com/a/1352

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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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.