Let $X_n, n = 0, 1, 2, . . .$ denote an unbiased Normal Random Walk. $X_0 = 10$, and $X_{n+1} = X_n + Y_{n+1}$, with $\{Y_n\}$ are i.i.d. $N(0, 1)$.

Then how can I show that:

A) $M_n = X_n^2-n$ is a martingale

Find the value of:

B) $\mathbb{E}M_{20}$

Does the answer lie in reformulation $X_n = X_{n+1} - Y_{n+1}$ ?

And then:

$$X_n = X_{n+1} - Y_{n+1} \implies X_n^2 = (X_{n+1})^2 - 2X_{n+1}Y_{n+1} + (Y_{n+1})^2$$

And then we proceed with

$$\mathbb{E}[(X_{n+1})^2 - 2X_{n+1}Y_{n+1} + (Y_{n+1})^2 -n| \mathcal{F}_s], s \leq n+1$$

$$\mathbb{E}[(X_{n+1})^2 | \mathcal{F}_s] - 2\mathbb{E}[X_{n+1}Y_{n+1}|\mathcal{F}_s] + \mathbb{E}[(Y_{n+1})^2|\mathcal{F}_s] - \mathbb{E}[n | \mathcal{F}_s]$$

Is this right so far? if not, where I have I made a mistake, and how should I proceed?

  • $\begingroup$ I believe this should carry the self-study tag (and follow the guidelines there). $\endgroup$ – Glen_b -Reinstate Monica May 18 '15 at 0:59

$\mathbb{E} [X_{n+1}^2-(n+1) \space | \space \mathcal{F}_n \space] =\mathbb{E} [X_{n}^2+2X_n Y_{n+1}+Y_{n+1}^2-(n+1) \space | \space \mathcal{F}_n \space] $

This gives $X_n^2 + 2 X_n\mathbb{E} [Y_{n+1} \space | \space \mathcal{F}_n \space]+\mathbb{E} [Y_{n+1}^2 \space | \space \mathcal{F}_n \space]-n-1$.

Because ${\{Y_n\}}$ is iid standard normal, $\mathbb{E} [Y_{n+1}^2 \space | \space \mathcal{F}_n \space]$ and $\mathbb{E} [Y_{n+1} \space | \space \mathcal{F}_n \space]$ are 1 and 0 respectively. Finally, we have $\mathbb{E} [M_{n+1} \space | \space \mathcal{F}_n \space] =X^2_n-n=M_n$ Therefore, we have $\{M_n\}$ being a martigale.

Second question,

$\mathbb{E} M_{20}=M_0=100 $

  • $\begingroup$ Thanks lion, just one question. If I want to show $M_n$ is a martingale, doesn't that mean showing $$\mathbb{E}(M_n | \mathcal{F}_n) = M_n$$ ? Yet here you have shown that $$\mathbb{E}(M_{n+1} | \mathcal{F}_n) = M_n$$ How does this work $\endgroup$ – elbarto May 17 '15 at 13:06
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    $\begingroup$ @elbarto $M_n$ is measurable with respect to $\mathcal{F}_n$ and, thus $$\mathbb{E}[M_n|\mathcal{F}_n] = M_n$$ automatically (and this would hold for any $\mathcal{F}_n$-measurable process, regardless of whether it is a martingale). The idea with a martingale is that future values given the observed information are equal in expectation to what you see now, i.e. what yawning lion wrote. $\endgroup$ – P.Windridge May 17 '15 at 13:27
  • $\begingroup$ Yep, shortly realised this after the fact! One additional question, would it also be fair to say $$\mathbb{E}[M_n | \mathcal{F}_{n-1}] = M_{n-1}$$ satisfies martingale property? or $$\mathbb{E}[M_n | \mathcal{F}_{n-4}] = M_{n-4}$$ $\endgroup$ – elbarto May 17 '15 at 13:34
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    $\begingroup$ $\mathbb{E}[M_n|\mathcal{F}_{n-1}]=M_{n-1}$ is the definition of martingales. And, $\mathbb{E}[M_n|\mathcal{F}_{n-4}]=M_{n-4}$ can be derived from this definition and the tower expectation rule. $\endgroup$ – Yawning Lion May 20 '15 at 8:07

There is no mistake, your demo is right, but you must to go more deeply in the demo. If you want to demonstrate that your expression is a martingale you must express $X_{n}$ as a sum of the $Y_{n}$. After that, you would show the martingale property of your expression.

  • $\begingroup$ But $X_n$ is in $\mathcal{F}_n$, I believe $X_n$ can be treated as if it is constant. Am I right? $\endgroup$ – Yawning Lion May 18 '15 at 17:15
  • $\begingroup$ Sorry for the delay, Yawning Lion May. In this particular case, you get chance that your proof works, but rigorously, you can't say directy that $X_n$ is $\mathcal{F}_n$ measurable. Explictly, nothing says that $X_n$ is $\mathcal{F}_n$ measurable, perhaps $X_n$ is $\mathcal{F}_{n+1}$ measurable. I suggest you check that $X_n$ is $\mathcal{F}_n$ measurable. For instance, if my prof sees your demo, you'll have false to the question. Greetings $\endgroup$ – Harry May 23 '15 at 9:35

The another answer isn't quite rigorous because you can't write that $\mathbb{E}[X_nY_{n+1}]=X_n\mathbb{E}[Y_{n+1}]$ directly.Firstly, you must to write that $X_n=X_0+\sum_{k=0}^{n-1}Y_{k}$. As a result, you can use the independance between the $Y_k$ w.r.t to the filtration $\mathcal{F}_n$.

  • $\begingroup$ This is not something that the other answer writes, is it? $\endgroup$ – Did Jun 29 '15 at 12:51

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