Question from a previous exam that I can not figure out:
Let $Y_i$ be mutually independent random variables with $P(Y_i=1)=p=1-P(Y_i=-1)$. Assume $p>1/2$. Define $M_n$ as:
$$M_n=\sum_{i=1}^n (Y_iY_{i-1}-(2p-1)Y_{i-1}), \text{with } Y_0=0.$$
It is easy to show this is a martingale, but I do not know where to start when considering convergence in $L^2$. Of course $M_n$ converges in $L^2$ if $\sup_{n\rightarrow\infty}E(M_n^2)<\infty$. Since $|M_n|<(n-1)\cdot 2p$ this is an $L^2$ martingale. Any hints?
It does not converge in $L^2$. A martingale has orthogonal increments, so $$ E(M_n^2) = \sum_{i=1}^n E \left [ \left ( Y_i Y_{i-1} - (2p-1) Y_{i-1}\right )^2 \right ]. $$ Expanding the square and using independence gives \begin{align*} E \left [ \left ( Y_i Y_{i-1} - (2p-1) Y_{i-1}\right )^2 \right ] & = E(Y_i^2 Y_{i-1}^2) + (2p-1)^2 E(Y_{i-1}^2) - 2(2p-1) E(Y_iY_{i-1}^2) \\ & = 1 + (2p-1)^2 - 2(2p-1)^2 \\ & = 1 - (2p-1)^2, \end{align*} and thus $$ E(M_n^2) = n (1 - (2p-1)^2), $$ which is not bounded.
Now, it is another story if $p$ depends on $n$. Typically, if $p = p_i = 1 - a_i$ with $\sum a_i < + \infty$, then you get $$ E(M_n^2) = \sum_{i=1}^n (1 - (1-a_i)^2), $$ with $(1 - (1-a_i)^2) \sim 2 a_i$, and then the series is convergent, i.e. $\sup E(M_n^2) < + \infty$.