Properties of L2

definition Let $L_2(\Omega)$ be the collection of all random variables on the probability space $\Omega$ such that $X^2$ is integrable.

lemma: when $X,Y \in L_2(\Omega)$ , $|\mathbb E XY| \le (\mathbb{E} X^2)^{1/2} \cdot (\mathbb{E} Y^2)^{1/2}$ .

proof: Let $t$ be a real-valued variable. Then the function $\phi(t) = \mathbb{E} (X - tY)^2$ is non-negative. Since

$\phi(t) = \mathbb{E} X^2 - 2t \mathbb{E}(XY) + t^2 \mathbb{E} Y^2$

If $\mathbb{E} Y^2 = 0$ then the non-negativity of $\phi(t)$ for all values $t$ requires that $\mathbb{E}(XY) = 0$ and so the claim is valid.

If $\mathbb{E} Y^2 > 0$ , then $\phi(t)$ is a quadratic convex polynomial. It is non-negative for all values of $t$ iff its discriminant is non-positive: $[2 \mathbb{E}(XY)]^2 - 4 \mathbb{E} X^2 \mathbb{E} Y^2 \le 0$ . But this implies the $(\mathbb{E}(XY))^2 \le \mathbb{E} X^2 \mathbb{E} Y^2$ . Since the square root function is non-decreasing the claim is valid.

lemma: $L_2(\Omega)$ is a linear space over the ring of real numbers.

proof: Since the collection of random variables is a linear space over the ring of real numbers when equipped with pointwise addition $(X+Y)(\omega) = X(\omega) + Y(\omega)$ and scalar multiplication $(a X)(\omega) = a X(\omega)$ , it remains to demonstrate the closure of $L_2(\Omega)$ with respect to these operations. But since $\mathbb{E} (aX)^2 = a^2 \mathbb{E} X^2 < \infty$ , and

$\mathbb{E} (X + Y)^2 = \mathbb{E} X^2 + 2 \mathbb{E}XY + \mathbb{E} Y^2 \le \left( (\mathbb{E} X^2)^{1/2} + (\mathbb{E} Y^2)^{1/2} \right)^2 < \infty$ the claim is validated.

lemma: $\left< X,Y \right> = \mathbb{E} XY$ is an inner product (a.s) when $X,Y \in L_2(\Omega)$ .

proof: By the prior lemma $L_2(\Omega)$ is a linear space. By the Cauchy-Schwartz inequality lemma, $|\left< X,Y \right> | < \infty$ when $X,Y \in L_2(\Omega)$ . Hence $\left< X,Y \right>$ is a real-valued function of its arguments $X,Y$ . It is symmetric since pointwise multiplication of functions is commutative. It is linear in its first argument since expectation is a linear operation. Also $\left<X,X \right> \ge 0$ and $\left< X,X \right> = 0$ implies $X = 0$ (a.s)

lemma: $L_2(\Omega)$ is an inner product space, a normed linear space with norm $||X|| = (\mathbb{E} X^2 )^{1/2}$ and a metric space with metric $d(X,Y) = ||X - Y||$ .

proof: these are immediate conclusions since $L_2(\Omega)$ equipped with the inner product given in the prior lemma is an inner product space.

lemma: $\mathbb{E} | X| \le \mathbb{E} X^2$ .

proof: Define the constant function $1(\omega) = 1$ for all $\omega \in \Omega$ . $\mathbb{E} |X| = \mathbb{E} |X 1| \le (\mathbb{E} X^2)^{1/2} (\mathbb{E} 1)^{1/2} = (\mathbb{E} X^2)^{1/2}$

lemma: If $X \in L_2(\Omega)$ then $X \in L_1(\Omega)$ .

proof: Since $X \in L_1(\Omega)$ iff $X$ is a random variable such that $\mathbb{E} |X| < \infty$ , the claim is valid on appeal to the prior lemma.

definition: $f : L^2(\Omega) \rightarrow \mathbb{R}$ is norm-continuous if for any sequence of $\left< X_n \right>$ of $L_2(\Omega)$ random variables convergent in metric to $X \in L_2(\Omega)$ then $\left< f(X_n) \right>$ is a sequence of real values convergent (a.s) to $f(X)$ .

example: inner product and norm are norm-continuous.

fix $Y \in L_2(\Omega)$ . Let $\left< X_n \right>$ be a sequence of $L_2(\Omega)$ random variables convergent in metric to $X \in L_2(\Omega)$ . Hence $|\left<X_n - X,Y\right>| \le ||X_n - X|| ||Y|| \rightarrow 0$ as $n \rightarrow \infty$ . Hence inner product is norm-continuous.
Since $||X|| - ||X_n|| \le ||X - X_n||$ and $||X_n|| - ||X|| \le ||X_n - X|| = ||X - X_n||$ it follows that $\left| ||X|| - ||X||_n \right| \le ||X-X_n|| \rightarrow 0$ as $n \rightarrow \infty$ . Hence norm is norm-continuous.

theorem: $L_2(\Omega)$ is a complete normed space.

proof: Let $\left< X_n \right>$ be a Cauchy sequence of $L_2(\Omega)$ . Select a sequence of indices $\left< n_k \right>$ inductively:

choose $n_1 \ge 1 : ||X_p - X_q || < 2^{-1}$ for all $p,q \ge n_1$ .

choose $n_{k+1} > n_k : ||X_p - X_q || < 2^{-k}$ for all $p,q \ge n_{k+1}$

Then $\mathbb{E} |X_{n_{k+1}} - X_{n_k}| \le || X_{n_{k+1}} - X_{n_k} || \le 2^{-k}$ and so

$\sum_k \mathbb{E} |X_{n_{k+1}} - X_{n_k}| \le 1$

since the partial sums form a non-decreasing sequence of random variables by the monotone convergence theorem

$\mathbb{E} \sum_k |X_{n_{k+1}} - X_{n_k}| = \sum_k \mathbb{E} |X_{n_{k+1}} - X_{n_k}| \le 1$

hence $\sum_k |X_{n_{k+1}} - X_{n_k}| < \infty$ (a.s.) and so

$\sum_k X_{n_{k+1}} - X_{n_k}$ is absolutely convergent (a.s.). Let $A \subseteq \Omega$ be the set where this condition holds. Since this set is of measure 1 it is an event.

On $A$ , define $X = X_{n_1} + \sum_k \left( X_{n_{k+1}} - X_{n_k} \right) = X_{n_1} + \lim_{n \rightarrow \infty} X_{n+1} - X_{n_1} = \lim_{n \rightarrow \infty} X_n$ and on $A^C$ define $X = 0$ . Then $X \in L_2(\Omega)$ .

To establish convergence in norm,

$|X_m - X|^2 = \lim_{k \rightarrow \infty} |X_m - X_{n_k}|^2 = \liminf_k |X_m - X_{n_k}|^2$ .

Hence by Fatou’s lemma,

$\lim_{m \rightarrow \infty} ||X_m - X||^2 = \lim_{m \rightarrow \infty} \mathbb{E} \liminf_k |X_m - X_{n_k}|^2 \le \lim_{m \rightarrow \infty} \liminf_k ||X_m - X_{n_k}||^2 = 0$

establishing the convergence of the sequence to an $X \in L_2(\Omega)$ .

Properties of L2

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