Let $X_1,X_2,…$ be independent random variables:
$$
\
X_n =
\begin{cases}
0 \mbox{ with probability }= 1 – \frac{1}{n} \\
1 \mbox{ with probability } = \frac{1}{2n} \\
-1 \mbox{ with probability } = \frac{1}{2n}
\end{cases}\
$$
Let $Y_1 = X_1$ and for n $\geq 2$
$$
\
Y_n =
\begin{cases}
X_n &\mbox{ if } Y_{n-1} = 0 \\
nY_{n-1}|X_n| & \mbox{ if } Y_{n-1} \neq 0 \\
\end{cases}\
$$
Show that $Y_n$ is a martingale w.r.t. to $\mathcal{F}_n$ and show why the martingale converce theorem is unapplicable.
Martingale:
$$
E[Y_n|F_{n-1}] = E[X_n * 1_{[Y_{n-1} = 0]} + nY_{n-1}|X_n|* 1_{[Y_{n-1} \neq 0]}|F_{n-1}]
$$
Splitting the two conditional expectation and seeing that $E[X_n]$ = 0 and $E[|X_n|] = \frac{1}{n}$ the martingale property is satisfied.
About the martingale convergence theorem I don't know how to prove it. I would think that the sup E$[|X_n|]$ is $\infty$ but Idk how to show it.
Best Answer
Indeed, the fact that we cannot use the martingale convergence theorem is because $\sup_{n\geqslant 1}\mathbb E\left[\left\lvert Y_n\right\rvert\right]$ is infinite. Indeed, denoting $A_n=\{Y_{n-1}=0\}$ and $B_n=\{Y_{n-1}\neq 0\}$ gives $$ \left\lvert Y_n\right\vert=\left\lvert X_{n}\right\vert\mathbf{1}_{A_{n-1}}+n\left\lvert Y_{n-1}\right\vert\left\lvert X_n\right\vert\mathbf{1}_{B_{n-1}} $$ and since $X_n$ is independent of $\mathcal F_{n-1}=\sigma(X_1,\dots,X_{n-1})$ and $Y_{n-1}$ is $\mathcal F_{n-1}$-measurable, we derive that $$ \mathbb E\left[\left\lvert Y_n\right\vert\right] =\mathbb E\left[\left\lvert X_n\right\vert\right]\mathbb P\left(A_{n-1}\right) +n\mathbb E\left[\left\lvert Y_{n-1}\right\vert\right]\mathbb E\left[\left\lvert X_n\right\vert\right]\ $$ and since $\mathbb E\left[\left\lvert X_n\right\vert\right]=1/n$, writting $p_n=\mathbb P(A_n)$, we get $$ \mathbb E\left[\left\lvert Y_n\right\vert\right]=\frac{p_{n-1}}n+\mathbb E\left[\left\lvert Y_{n-1}\right\vert\right] . $$ Moreover, $$ p_n=\mathbb P(Y_n=0,Y_{n-1}=0)+\mathbb P(Y_n=0,Y_{n-1}\neq 0) =\mathbb P(X_n=0,Y_{n-1}=0)+\mathbb P(X_n=0,Y_{n-1}\neq 0)=1-1/n $$ hence letting $A_n=\mathbb E\left[\left\lvert Y_n\right\vert\right]$, we got $$ A_n=\frac 1n-\frac 1{n(n-1)}+A_{n-1} $$ and writing $$ A_N-A_1=\sum_{n=2}^N\left(A_n-A_{n-1}\right)=\sum_{n=2}^N\left(\frac 1n-\frac 1{n(n-1)}\right) $$ combined with the divergence of $\sum_{n\geqslant 2}1/n$ and the convergence of $\sum_{n\geqslant 2}1/(n(n-1))$ shows that $(A_N)_{N\geqslant 1}$ is not bounded.