# Identically distributed, uncorrelated, Bernoulli rvs are independent.

I have a question regarding the proof of the following Lemma, given in the book Quantitative Risk Management by McNeil et. al.

Let $(\Omega, \mathcal F, P)$ be a probability space and $(\mathcal F_t)_{t \in \mathbb Z}$ be a filtration on $(\Omega, \mathcal F, P)$.

A stochastic process $(X_t)_{t \in \mathbb Z}$ defined on $(\Omega, \mathcal F)$ is called a martingale difference sequence with respect to $(\mathcal F_t)$ if for any $t$, $X_t \in L^1(\Omega, \mathcal F, P)$, $X_t$ is $\mathcal F_t$-measurable for all $t$, and $\mathbb E[X_t \mid \mathcal F_{t-1}] = 0$ for each $t$.

A stochastic process $(X_t)$ is called covariance stationary, if the first two moments exist, $\mu(t) := \mathbb E[X_t] = \mu$ for all $t$, and $\mathrm{Cov}[X_{t+k}, X_{s+k}] = \mathrm{Cov}[X_t, X_s]$ for all $t,s, k \in \mathbb Z$. If $(X_t)$ is covariance stationary, we define the autocorrelation function as $\varrho(h) = \varrho[X_h, X_0]$ for all $h \in \mathbb Z$, where $\varrho[X,Y] := \frac{\mathrm{Cov}[X, Y]}{(\mathrm{Var}[X] \mathrm{Var}[Y])^{1/2}}$ is the correlation coefficient, for $X, Y \in L^2(\Omega, \mathcal F, P)$. In this case, $(X_t)$ is called a white noise process if $\varrho(h) = 1$, if $h = 0$, and $\varrho(h) = 0$, if $h \ne 0$.

Let $(Y_t)_{t \in \mathbb Z}$ be a sequence of Bernoulli random variables adapted to a filtration $(\mathcal F_t)_{t \in \mathbb Z}$ and satisfying $\mathbb E[Y_t \mid \mathcal F_{t-1}] = p > 0$ for all $t$. Then $(Y_t)$ is a process of i.i.d. Bernoulli variables.

The proof goes as follows: The process $(Y_t - p)$ is a martingale difference sequence. Furthermore, $\mathrm{Var}[Y_t - p] = p(1 - p)$ for all $t$, thus $(Y_t - p)$ and $(Y_t)$ are white noise processes of uncorrelated random variables. (Since one can show that a martingale difference sequence with constant variance is a white noise process as defined above.) Then he writes: "It is easily shown that identically distributed uncorrelated Bernoulli variables are i.i.d."

I do not really see that. I understand that the uncorrelatedness implies that the variables $(Y_t)$ are pairwise independent, but why are they mutually independent? Why is this true?

• Could you remind the definitions of "martingale difference sequence" and "white noise process of uncorrelated random variables"? Commented Dec 5, 2017 at 13:58
• I have edited the definitions. Commented Dec 5, 2017 at 14:11

For a Bernoulli random variable $Y$: $$f(Y) = f(1)Y + f(0)(1 - Y)$$ This formula and an application of the tower property give for $t \neq s$: $$\mathbb{E}[f(Y_t) \cdot g(Y_s)] = \mathbb{E}[f(Y_t)]\cdot \mathbb{E}[g(Y_s)].$$ So Bernoulli random variable are a special case of r.v.'s in which conditional independence implies independence.
Let us consider the case of three r.v.'s. The case of $n$ random variables follows exactly in the same way (indeed it's also the same argument I implicitly used in the first case). For $t \ge s\ge u$ we have that: $$\mathbb{E}[f(Y_t) \cdot g(Y_s) \cdot h(Y_u)] = \mathbb{E}[ \mathbb{E}[f(Y_t) \cdot g(Y_s) \cdot h(Y_u) | \mathcal{F}_s]] = \mathbb{E}[ \mathbb{E}[f(Y_t) | \mathcal{F}_s]\cdot g(Y_s) \cdot h(Y_u) ]$$ Now using that: $$\mathbb{E}[f(Y_t) | \mathcal{F}_s] = \mathbb{E}[f(Y_t)]$$ together with the already established identity for two random variables, the original expectation is actually equal to the product $$\mathbb{E}[f(Y_t) \cdot g(Y_s) \cdot h(Y_u)] = \mathbb{E}[f(Y_t)] \cdot \mathbb{E}[g(Y_s)] \cdot \mathbb{E}[h(Y_u)].$$
• Could you elaborate please? Obviously for two random variables I do not need the trick since the statement is clear. But how does it work with, lets say, $Y_t, Y_s, Y_r$? I have $E[g(Y_s) h(Y_r) \mid Y_t] = E[E[g(Y_s) h(Y_r) \mid Y_s] \mid Y_t] = E[g(Y_s)] E[h(Y_r)]$ by independence, where the first equality is the tower property. But for the tower property, one needs $\sigma(Y_t) \subset \sigma(Y_s)$ according to wikipedia. Commented Dec 6, 2017 at 17:12
• You condition w.r.t. $\mathcal{F}_s,$ not $\sigma(Y_s)$. I added some details to the answer. Commented Dec 7, 2017 at 11:35