# How is integrability used in fundamental principle 'you can't beat the system'?

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'.  • You need to assume $C$ is nonnegative to prove the supermartingale statement, otherwise the last inequality in the proof might fail. For the martingale statement (using the words in brackets), you can drop it. Nov 6, 2015 at 14:48
• @NateEldredge Didn't notice the supermartingale. Thanks. Why do we really need that though? I guess intuitively, it doesn't make sense to have a negative stake (borrowing money?), but why is C.X not a supermartingale?
– BCLC
Nov 6, 2015 at 14:50
• Intuitively, a supermartingale is a gambling game that is biased against the player. You can win at such a game if you are allowed to bet negative amounts, i.e. play as the house. Alternatively, it's a stock that tends to decrease in value - you could make money if you were allowed to sell short. Nov 6, 2015 at 14:50
• As a really simple deterministic counterexample, let $X_n = -n$ and $C_n = -1$. Nov 6, 2015 at 14:51
• For conditonal expectation to exist, do you not need the r.v. to be $L^1$? In this case, you need $C_n(X_n-X_{n-1})$ to be integrable. The easiest condition is to assume that both $C_n$ and $X_n-X_{n-1}$ are both $L^2$ and use Cauchy-Schwarz. Maybe you can try to construct counter examples where both are integrable but the products fail to be? Note Saz's answer assumed X.Y is integrable. If you can show the product is integrable than $L^1$ is okay but in many cases, it is easier to show $L^2$. Nov 24, 2015 at 22:58

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 $C_n = −1$