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I am wondering if there is such a sequence of random variables $(X_n)_{n=0}^\infty$ such that $\mathbb{E}(X_{n+1}\mid X_n)=X_n$ for all $n\geq0$ but which is not a martingale with respect to the filtration $\mathcal{F_n}=\sigma(X_0, \dots, X_n)$.

I would really appreciate if you could show me such an example.

saz
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benny
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2 Answers2

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Let $(Y_j)_{j \in \mathbb{N}}$ be a sequence of identically distributed independent random variables such that $\mathbb{E}Y_j=0$. For some fixed $N \in \mathbb{N}$ we define

$$\begin{align*} X_n &:= \sum_{j=1}^n Y_j \qquad \text{for all} \, \, n \leq N \\ X_{n} &:= \sum_{j=1}^N Y_j + Y_1 - Y_2 = X_N+ Y_1-Y_2 \qquad \text{for all} \, \, n >N. \end{align*}$$

For $n \leq N$ and $n>N+1$, the condition

$$\mathbb{E}(X_{n} \mid X_{n-1}) = X_{n-1}$$

is obviously satisfied. For $n=N+1$, we have

$$\mathbb{E}(X_{N+1} \mid X_N) = X_N + \mathbb{E}(Y_1 \mid X_N) - \mathbb{E}(Y_2 \mid X_N). $$

Since $(Y_j)_{j \in \mathbb{N}}$ is identically distributed and independent, we have

$$\mathbb{E}(Y_1 \mid X_N) = \mathbb{E}(Y_2 \mid X_N)$$

and therefore

$$\mathbb{E}(X_{N+1} \mid X_N) = X_N.$$

On the other hand,

$$\begin{align*} \mathbb{E}(X_{N+1} \mid \mathcal{F}_N) &=X_N + 2\underbrace{\mathbb{E}(Y_1 \mid \mathcal{F}_N)}_{\mathbb{E}(X_1 \mid \mathcal{F}_N)=X_1} - \underbrace{\mathbb{E}(Y_1+Y_2 \mid \mathcal{F}_N)}_{\mathbb{E}(X_2 \mid \mathcal{F}_N) = X_2} \\ &= X_{N+1} \neq X_N. \end{align*}$$

Intuition: It is widely known that the process

$$S_n := \sum_{j=1}^n Y_j$$

can be used to model a fair game; the outcome of the $j$-th round is given by $Y_j$. Now we change the rules of our (fair) game: After $N$ rounds the game is stopped; in the final round the player gains the outcome of the first round, but loses the outcome of the second round. There are two cases:

  • The player is very drunk and has already forgotten the outcomes of the first two rounds. In this case, from the point of view of our drunken player, the (changed) game is still fair - it looks like another two rounds of our (original) game.
  • If the gambler is still sober, then he remembers the outcome of the first two rounds and can calculate the outcome of the final round explicitly (there is no randomness, given the information up to time $N$!), i.e. $$\mathbb{E}(X_{N+1} \mid \mathcal{F}_N) = X_{N+1}.$$
saz
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Remember that in the definition of martingale you should have $E|X_n| < \infty$.

So, it is enough to take a sequence which hasn't first moment. Put $X_n = X$ where $X$ follows a Cauchy distribution. It satisfy your condition, trivially, but it is not a martingale because they don't have first moment.

Rodrigo Ribeiro
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    But then $\mathbb{E}(X_{n+1}\mid X_n)$ does not exist hence the condition that $\mathbb{E}(X_{n+1}\mid X_n)=X_n$ fails. – Did Mar 12 '15 at 16:30
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    @Did, Why do not exists? $E[X | X] = X$ once $X$ is trivially measurable with respect the $\sigma$ algebra generated by $X$. Am I missing something? – Rodrigo Ribeiro Mar 12 '15 at 17:25
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    Yes you are: E(X|Y) is defined only when X is integrable. – Did Mar 12 '15 at 17:28
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    In Shiryaev's book, page 213 the definition doesn't demand integrability. He defines for a positive random variable, doesn't need to be integrable, and then extends to any random variable when $min{E[X^+| \mathcal{F}],E[X^-| \mathcal{F}]}< \infty$ almost sure. – Rodrigo Ribeiro Mar 12 '15 at 17:46
  • Sorry, but I really didn't understand what you said. Question: A random variable which follows a Cauchy distribution fits the Siryaev's definition? Seems to me that the answer is "yes". And it is natural to define $E[X|X] = X$... – Rodrigo Ribeiro Mar 12 '15 at 17:55
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    Shiryaev's definition does yield E(X|X)=X for every random variable X, yes. As I said, this definitional choice is unfrequent. (I may add that this is most probably not the kind of counterexample the OP has in mind.) – Did Mar 12 '15 at 18:12
  • Yes, I guess you are right. Thanks! – Rodrigo Ribeiro Mar 12 '15 at 18:31