From Williams' Probability w/ Martingales:
Re (iii), why do we need square integrability? I mean, why is integrability not good enough? Based on an answer in my previous question, I think integrability is sufficient for 'taking out what is known'.
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Sign up to join this communityFrom Williams' Probability w/ Martingales:
Re (iii), why do we need square integrability? I mean, why is integrability not good enough? Based on an answer in my previous question, I think integrability is sufficient for 'taking out what is known'.
Based on Lost1's comment:
In the first place, to have conditional expectation, we need integrability.
A product of integrable random variables is not necessarily integrable:
Let $X, Y \in \mathscr L^{1}(\Omega, \mathscr F, \mathbb P)$.
Consider $X$ and $X - Y$ w/ $X$ having an infinite second moment but finite first moment.
Then
$$E[X(X-Y)] = E[X^2] - E[XY] = \infty$$
assuming $-\infty \le E[XY] < \infty$ and indeterminate otherwise. In either case, $X(X-Y)$ is not integrable.
An example is $X$ having a student-t distribution with two degrees of freedom and $Y$ can be anything integrable, I guess.
If $X, Y$ are integrable and independent, then $XY$ is integrable. However,
$C_n$ and $X_n - X_{n-1}$ are not necessarily independent. So if they are not square integrable or bounded,
$$E[Y_n - Y_{n-1} | \mathscr{F_{n-1}}]$$
may not exist.
Based on Nate Eldredge's comment:
For nonnegativity (see previous revision of question)
$X_n = −n$ and $C_n = −1$