Let the stochastic process $\{X_n\}$ be constructed inductively as follows:
- $X_0=0$, and
- for $n\ge 1$, and conditionally on $\mathcal{F}_{n-1}=\sigma(X_0,\ldots,X_{n-1})$, we set
$\;\,\,\, X_n=nX_{n-1}$ or $0$ with probabilities $1/n$ and $1-1/n$
My problem is to verify that $\{X_n\}$ is a martingale i.e. $E(X_n\mid \mathcal{F}_{n-1})=X_{n-1}$
I guess I can compute $E(X_n\mid\mathcal{F}_{n-1})$ like this: $$ E(X_n\mid\mathcal{F}_{n-1})=nX_{n-1}\cdot \frac{1}{n}+0\cdot(1-\frac{1}{n})=X_{n-1} $$
However, I can not find any theoretical reason letting me do this.
Could you please help? Thank you!
Let $Y_i$ be a rv with $P(Y=1)=1/n$ and $P(Y_i=0)=1-1/n$. By assumption suppose that $Y_i$ is an iid sequence of rv and moroever, let $Y_i$ be independent of $X_0,X_1,\ldots,X_{i}$. In other words, $\sigma(Y_i)$ and $F_{i}$ are independent for all $i$. Now write $X_{n+1}=nY_nX_n$. Conditioning on $\mathcal{F}_n$,
$E[X_n\mid\mathcal{F}_{n-1}]=E[nY_{n-1}X_{n-1}\mid\mathcal{F}_{n-1}]=nX_{n-1}E[Y_{n-1}\mid\mathcal{F}_{n-1}]=nX_{n-1}E[Y_{n-1}]=nX_{n-1}(1/n)=X_{n-1}$
where we used the fact that $X_{n-1}$ is $\mathcal{F}_{n-1}$ measurable and independence of $Y_{n-1}$. from $F_{n-1}$.