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I'm reading a paper in which the expected value of a random variable, $\mathbb{E}[X]$, is characterized as an orthogonal projection. This is on page 10. I've seen the geometric interpretation of conditional expectation but never $\mathbb E[X]$.

$L^2$ contains the one-dimensional subspace $\Delta$ of deterministic variables, i.e. variables which are constant on the state space $\Omega.$ Given a random variable $X$, its expectation is:$$ E(X)=\int_\mathbb{R}xp(x)dx=\langle X,1\rangle.$$

My question is how do we know this one-dimensional subspace consisting of deterministic variables exists?

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  • $\begingroup$ It is the space of variables that are almost surely constant. It's hard to see what more you might be asking for. The only mathematics needed to prove this rigorously is to show the integral of their squares is finite, but that's a trivial consequence of the axioms of probability (total probability is $1$) and linearity of integration. Could you therefore clarify what you mean by "exists" or "know"? $\endgroup$ – whuber Jan 11 at 17:48
  • $\begingroup$ This is the question I found on math stack exchange that inspired the confusion: math.stackexchange.com/questions/18395/… I'm new to Lp spaces so maybe I'm tripping myself up with definitions $\endgroup$ – statian Jan 12 at 2:52
  • $\begingroup$ Basically, I was confused because some of the answers give examples of functions that are in $L^2$ but not $L^1$ $\endgroup$ – statian Jan 12 at 3:16
  • $\begingroup$ @Brutus: Then can you answer your own question so that it does not linger on unanswered? $\endgroup$ – kjetil b halvorsen Mar 29 at 21:09
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Sometimes things look trivial but they're not, so it can pay to be formal.

Recall, then, that

  1. A random variable is a measurable real-valued function defined on a probability space $(\Omega, \mathfrak{F}, \mathbb{P})$ and

  2. $L^2$ is the completion of the space of square-integrable random variables.

The question then is whether for every $y\in\mathbb{R}$ it is the case that the constant function

$$\lambda_{y}: \Omega\to \mathbb{R};\quad \lambda(y)(\omega) = y\text{ for all }\omega\in\Omega$$

satisfies these requirements and whether the set of such functions is a closed vector subspace.

(BTW, axioms of set theory guarantee that such functions exist: as sets, they are $$\lambda_y = \Omega \times \{y\} \subset \Omega \times \mathbb{R}$$ and all axiomatic set theories assure the existence of finite Cartesian products.)

First, $\lambda_y$ is measurable because with respect to any given $y$ there are only two kinds of Borel sets in $\mathbb{R}:$ those that contain $y$ and those that do not. The inverse image of $\lambda_y$ applied to a Borel set is therefore either $\Omega$ or empty. But it is axiomatic that those two sets belong to any sigma algebra on $\Omega.$ Thus, by definition, every $\lambda_y$ is measurable.

Second, we may compute the $L^2$ norm of $\lambda_y$ as

$$||\lambda_y||_2^2 = \int_\Omega |\lambda_y(\omega)|^2 \mathrm{d}\mathbb{P}(\omega).$$

Because integration is linear and $\lambda_y$ is constant, it factors out of the integral. The result is

$$|y|^2 \int_\Omega \mathrm{d}\mathbb{P}(\omega) = |y|^2(1) = |y|^2$$

because (also axiomatically) the probability measure $\mathbb{P}$ integrates to unity. Taking square roots we obtain

$$||\lambda_y||_2 = \sqrt{|y|^2} = |y| \lt \infty$$

for all $y.$

Note that we needed $\mathbb P$ to be a finite measure. It perhaps is the key observation. For instance, the nonzero constant functions are not in $L^2(\mathbb{R})$ with respect to Lebesgue measure, because their integrals all diverge.

Finally, it is immediate from the definition that for all real numbers $x,y,a,b$,

$$a\lambda_x + b\lambda_y = \lambda_{ax+by}$$

because when either side is applied to any $\omega\in\Omega,$ they produce the same value $ax+by.$ This shows the set of $\lambda_y$ is closed under finite linear combinations. Furthermore, this set clearly is finite dimensional: any singleton $\{\lambda_y\}$ for $y\ne 0$ is a basis, because for all real numbers $x,$ the Commutative Law of Multiplication asserts $yx=xy,$ which upon dividing both sides by $y$ can be restated as

$$\lambda_x = (y^{-1}x)\lambda_y.$$

Consequently the constant functions are a closed linear (finite-dimensional) subspace of $L^2(\Omega, \mathfrak{F}, \mathbb{P}).$

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