# Ch 9 Brownian motion and martingales page 12 of the chapter

Discussion in 'CM2' started by Adithyan, May 14, 2019.

It follows by the independent increments property of Brownian motion that the log-returns, and
hence the returns themselves, are independent over disjoint time periods.

Can you tell me how returns themselves are independent over disjoint time periods? I understand why log returns are independent.

Please also explain me as to why increments in continuous time log normal model is Markovian?

2. ### Anna BishopActEd TutorStaff Member

There is a statistical property that says: If X and Y are independent random variables, then real-valued functions, g(X) and h(Y) will also be independent random variables.

Therefore, if X = log A and Y = log B are independent random variables then A = exp(X) and B = exp(Y) will also be independent random variables.

X and Y represent the log returns and A and B represent the returns.

(Proof of the general result here if you are interested: https://proofwiki.org/wiki/Functions_of_Independent_Random_Variables_are_Independent)

The Markov property follows from the property of independent increments. At a basic level, if returns in time period t2 to t3 are independent of returns in the time period t1 to t2 where t1 < t2 < t3, then the distribution of returns between t2 and t3 will depend only on t2 and not on the distribution of returns between t1 and t2.

Does this help?
Anna

Don't you call Log(St/Ss) as log return? Would you call Log St and Log Ss as log return each? Aren't St and Ss price of the stocks at time t and time s?

As a result, I don't get the following line in page 6 of Ch 11 as well

It is assumed that returns over non-overlapping intervals are independent of each other.
This is because the normal variables generating the random variation in the log of the
security price are assumed to be independent.

4. ### Anna BishopActEd TutorStaff Member

Yes log(St/Ss) is the log return.
Yes St and Ss are the share prices at times t and s.

Assume s < t < u

We have:
• X = log(St/Ss) is Normal (μ(t-s), σ^2(t-s))
• Y = log(Su/St) is Normal (μ(u-t), σ^2(u-t))
You mentioned above that you are happy with the idea that the log returns are independent. In other words, X and Y are independent.

There is a statistical result that states that if X and Y are independent then g(X) and h(Y) will be independent where g(X) and h(Y) are real valued functions of X and Y.

Therefore, exp(X) = St/Ss and exp(Y) = Su/St will be independent random variables. In other words, the returns themselves are also independent.

Let me know if I'm missing the point Adithyan?
Anna