How do I construct an example of a martingale which is bounded in $L^1$ but not in $L^2$

Question

I know that for a martingale to be bounded in $L^1$ it needs to be $ \sup _{n} E\left[\left|M_{n}\right|\right]<\infty$ but I am not sure how to construct a martingale which is not bounded in $L^2$.

Answer 1

Simply take $M_n=\mathbb E\left[X\mid\mathcal F_n\right]$, where $\left(\mathcal F_n\right)_{n\geqslant 1}$ is a filtration, $X$ is non-negative, integrable, $\sigma\left(\bigcup_{n\geqslant 1}\mathcal F_n\right)$-measurable but not square integrable.

We can take a more interesting counter-example by taking $M_n=\prod_{i=1}^n X_i$, where $\left(X_i\right)_{i\geqslant 1}$ is independent, $\mathbb P(X_i=i)=1/i$ and $\mathbb P(X_i=0)=1-1/i$. Then $\left(M_n\right)_{n\geqslant 1}$ is a martingale with respect to the filtration $\left(\mathcal F_n\right)_{n\geqslant 1}$ where $\mathcal F_n$ is the $\sigma$-algebra generated by $X_1,\dots,X_n$ but for $p>1$, $$\mathbb E\left[\lvert M_n\rvert^p\right]=\prod_{i=1}^ni^{p-1} \geqslant n^{p-1}.$$