Random elements and random processes

Question

Suppose that $X$ is a random element with values in $L^2([0,1],\mathbb C)$ such that $\operatorname E\|X\|<\infty$. The expected value of $X$ is equal to $\mu(t)=\operatorname EX(t)$ for almost all $t\in[0,1]$. So we have a random element, but we actually want to have a random process, i.e. a collection of random variables, to evaluate the mean. How can we define the random process $\{X(t):t\in[0,1]\}$?

Strictly speaking, the space $L^2([0,1],\mathbb C)$ consists of equivalence classes of almost everywhere equal square integrable complex functions. So it does not make sense to consider the value of $f\in L^2([0,1],\mathbb C)$ at some point $t\in[0,1]$ since this value is not defined.

However, if we want to define the random variable $X(t)$ for some $t\in[0,1]$ and evaluate its expected value $\operatorname EX(t)$, we need to assign a complex number $X(t,\omega)$ for each $\omega\in\Omega$ (and make sure that this map is measurable).

So it seems that it is not even possible to define the random process $\{X(t):t\in[0,1]\}$ because we do not have information about the values of this function at point $t\in[0,1]$. However, I have seen a few times when the expected value of a random element with values in $L^2([0,1],\mathbb C)$ is evaluated pointwise, i.e. $\operatorname EX(t)$ for $t\in[0,1]$, so maybe it is possible to define the random process $\{X(t):t\in[0,1]\}$ in a sensible way.

Any help is much appreciated!

Answer 1

I would like to make some comments, mostly keeping in mind the functional data analysis framework (even though I'm not a specialist in FDA I found the question interesting and looked into some references).

• There is a syntactic solution to the problem: put $T=[0,1]$, $\mathcal{T}=L^2(T,\tau;\mathbb{C})=\mathcal{H}$, where $\tau$ is a probability measure on $T$. Then one can think of an ($\mathbb{P}$-a.e. defined) measurable selection $X^\bullet:\Omega\to \mathcal{H}$ of $L^2$ $\tau$-a.e. defined paths in $\mathbb{C}$ as an everywhere defined random process (made up of $\mathbb{P}$-a.e. defined random variables):

$$X_\bullet:\mathcal{T}\to L^0(\Omega,\mathbb{P};\mathbb{C}), \phi\mapsto[\omega\mapsto \langle\phi\,,\, X^\omega\rangle_\mathcal{H}].$$

(Note that here I used two different notations for the same space of curves; it's fun to think of a $\tau$-a.e. defined path in $\mathbb{C}$ as a time parameter of a process. Here $\mathbb{P}$ is the suppressed probability measure on the suppressed measurable space $\Omega$.)

(See e.g. Kupresanin, Shin, King and Eubank's paper "An RKHS framework for functional data analysis", p.3628, or (if I'm right that there is a typo) Hsing and Eubank's book Theoretical Foundations of Functional Data Analysis, with an Introduction to Linear Operators, p.185.)

• My first reaction to the question was along the lines of what lonza leggiera suggests above in the comments (if the suggestion in question is to assume that $(\Omega,\mathbb{P})$ is a standard probability space). Three possible lines of attach are as follows: 1) find a measurable (or better) section of the quotient map from the space $\mathcal{L}^2$ of everywhere defined measurable $L^2$ functions to $L^2$, 2) find a measurable lift of $X^\bullet$ to $\mathcal{L}^2$, 3) play the "intersections of full measure subsets over which functions are everywhere defined" game (alternatively one can try using Lusin type arguments, if approximations are allowed). Here I'm thinking of $\mathcal{L}^2$ to be endowed with the Borel $\sigma$-algebra endowed by the $L^2$ pseudonorm. (A related question is to see if the closed subspace of functions that vanish $\tau$-a.e. is complemented in $\mathcal{L}^2$) (By the time of writing this I could not produce rigorous arguments in either direction; I will add an argument (or a counterexample) here if I think of something.) (Also possibly related: https://terrytao.wordpress.com/2020/10/04/foundational-aspects-of-uncountable-measure-theory-gelfand-duality-riesz-representation-canonical-models-and-canonical-disintegration/ , especially to the third argument.)

• From the perspective of probability theory in general admittedly the previous point seems misguided; as far as I understand no assumptions ought to be made on $(\Omega,\mathbb{P})$. Indeed, especially in the FDA theory being able to evaluate the random element $X^\bullet$ at any time $t\in[0,1]$ seems to be important. In fact, more structure seems to be desired: that the random element $X^\bullet$ takes values in some vector subspace $\mathscr{H}$ of the space $\mathcal{F}(T;\mathbb{C})$ of everywhere defined curves on the complex plane endowed with a Hilbert space structure such that for any $t\in\mathbb{T}$: the evaluation map $\delta_t:\phi\mapsto \phi(t)$ that takes a curve to where it's at at time $t$ be continuous on $\mathscr{H}$. This leads to the objects called reproducing kernel Hilbert spaces (RKHS). Thus it seems from the perspective of FDA not being able to evaluate functions everywhere is pathological, and the current solution to this matter is to use an RKHS instead of $L^2(T,\tau;\mathbb{C})$.

(For this point see the aforementioned book or the notes of Paulsen "An Introduction to the Theory of Reproducing Kernel Hilbert Spaces" at https://www.math.uh.edu/~vern/rkhs.pdf , especially the beginning of p.3.)

Answer 2

Here's an elaboration of the idea I suggested in my comments above, in the form of a proof of the following result:

• Let $\ (\Omega,\scr{F},\mathbb{P})\ $ be a probability space, $\ X:\Omega\rightarrow L^2([0,1],\mathbb{C})\ $ a function which is measurable (with respect to the Borel sets of $\ L^2([0,1],\mathbb{C})\ $ and the $\ \sigma$-algebra $\ \scr{F}\ $), and for which $$ \infty>E\big(\|X\|_2\big)=\int_\Omega\|X(\omega)\|_2\,d\mathbb{P}(\omega)\ . $$ Then there is a measurable function $\ \varphi:[0,1]\times\Omega\rightarrow\mathbb{C}\ $, such that $\ \varphi(\,.,\omega)=X(\omega)\ $, for $\ \mathbb{P}\text{-a.e. }\omega\in\Omega\ $. Thus, if we define $\ \psi:[0,1]\rightarrow\mathbb{C}^\Omega\ $ by $$ \psi(t)(\omega)=\phi(t,\omega)\ , $$ then $\ \psi\ $ is a stochastic process such that $\ \psi(.)(\omega)=X( \omega)\ $ for $\ \mathbb{P}\text{-a.e. }\omega\in\Omega\ $.

Here, I will take symbols of the form $\ f(\,.),\, g(\,.,.)\ $ etc. to represent the equivalence classes of functions which are almost everywhere equal to $\ f, g\ $ etc. .

Here's the proof:

First note that $\ L^2([0,1],\mathbb{C})\subseteq L^1([0,1],\mathbb{C})\ $ and $\ \|f\|_1\le\|f\|_2\ $ for all $\ f\in L^2([0,1],\mathbb{C})\ $ (see this post by Ayman Hourieh, for example). Also, since $\ [0,1]\ $ is separable, then so is $\ L^1([0,1],\mathbb{C})\ $. Let $\ \{x_j\,|\,j\in\mathbb{N}\,\}\ $ be a countable dense subset of $\ L^1([0,1],\mathbb{C})\ $, $$ B(f,d\,)\stackrel{\text{def}}{=}\left\{ \left. g\in L^1([0,1],\mathbb{C})\,\right|\|g-f\|_ 1<d\right\} $$ (the open ball of $\ L^1([0,1],\mathbb{C})\ $ with centre $\ f\ $ and radius $\ d\ $), and for each $\ r\in\mathbb{N}\ $ let \begin{align} P_1(r)&=B\big(x_1,r^{-1}\big)\\ P_{j+1}(r)&=B\big(x_{j+1},r^{-1}\big)\setminus\bigcup_{i=1}^jP_i(r)\ \text{ for }\ j=1,2,\dots\ . \end{align} If $\ x\in L^1([0,1],\mathbb{C})\ $ and $\ r\in\mathbb{N}\ $, then since $\ \{x_j\,|\,j\in\mathbb{N}\,\}\ $ is dense in $\ L^1([0,1],\mathbb{C})\ $, there exists $\ j\in \mathbb{N}\ $ such that $\ \|x-x_j\|_1\le r^{-1}\ $—that is $\ x\in B\big(x_j,r^{-1}\big)\ $. Therefore, for every $\ r\ $, $\ L^1([0,1],\mathbb{C})\subseteq\bigcup_\limits{j=1}^\infty B\big(x_j,r^{-1}\big)\ $, $\ \big\{P_j(r)\,|\,j\in\mathbb{N}\big\}\ $ is a partition of $\ L^1([0,1],\mathbb{C})\ $, and $$ \big\{X^{-1}\big(P_j(r)\big)\,\big|\,j\in\mathbb{N}\big\}\subseteq\mathscr{F} $$ is a partition of $\ \Omega\ $. For each $\ i\in\mathbb{N}\ $, let $\ \hat{x}_i:[0,1]\rightarrow\mathbb{C}\ $ be any measurable representative of the equivalence class $\ x_i\ $ (that is, $\ \hat{x}_i(\,.)=x_i\ $), and for each $\ r\in\mathbb{N}\ $ and $\ \omega\in\Omega\ $, let $\ i_r(\omega)\ $ be the (unique) value of $\ i\ $ such that $\ \omega\in X^{-1}\big(P_i(r)\big)\ $. Now let $$ z_r(t,\omega)=\hat{x}_{i_r(\omega)}(t) $$ for $\ r\in\mathbb{N},t\in[0,1]\ $ and $\ \omega\in\Omega\ $. For all $\ \omega\ $ and $\ r\ $, we have $\ X(\omega)\in P_{i_r(\omega)}(r)\subseteq B\big( z_r(\,.,\omega),r^{-1}\big)\ $, so $$ \|\,X(\omega)-z_r(\,.,\omega)\,\|_1\le\frac{1}{r}\ . $$ Also, if $\ S\ $ is any measurable subset of $\ \mathbb{C}\ $, then \begin{align} (t,\omega)\in z_r^{-1}(S)&\Rightarrow \hat{x}_{i_r(\omega)}(t)=z_r(t,\omega)\in S\\ &\hspace{1.5em}\text{and }\ X(\omega)\in P_{i_r(\omega)}(r)\\ &\Rightarrow(t,\omega)\in\hat{x}^{-1}_{i_r(\omega)}(S)\times X^{-1}\big(P_{i_r(\omega)}(r)\big)\\ &\Rightarrow(t,\omega)\in\bigcup_{i=1}^\infty\left(\hat{x}_i^{-1}(S)\times X^{-1}\big(P_i(r)\big)\right)\ . \end{align} Therefore, $$ z_r^{-1}(S)\subseteq\bigcup_{i=1}^\infty\left(\hat{x}_i^{-1}(S)\times X^{-1}\big(P_i(r)\big)\right)\ . $$ Conversely, if $\ (t,\omega)\in\bigcup_\limits{i=1}^\infty\left(\hat{x}_i^{-1}(S)\times X^{-1}\big(P_i(r)\big)\right)\ $, then $\ (t,\omega)\in\hat{x}_i^{-1}(S)\times X^{-1}\big(P_i(r)\big)\ $ for some natural number $\ i\ $. But $\ \omega\in X^{-1}\big(P_i(r)\big)\Rightarrow$$\,i=i_r(\omega)\ $, because $\ i_r(\omega)\ $ is the unique value of $\ i\ $ for which $\ \omega\in X^{-1}\big(P_i(r)\big)\ $. Therefore, $$ (t,\omega)\in\hat{x}^{-1}_{i_r(\omega)}(S)\times X^{-1}\big(P_{i_r(\omega)}(r)\big)\ , $$ and, in particular, $\ t\in\hat{x}^{-1}_{i_r(\omega)}(S)\ $, from which it follows that $\ z_r(t,\omega)=\hat{x}_{i_r(\omega)}(t)\in S\ $, or, equivalently, $\ (t,\omega)\in z_r^{-1}(S)\ $. Hence we have the reverse inclusion $$ \bigcup_{i=1}^\infty\left(\hat{x}_i^{-1}(S)\times X^{-1}\big(P_i(r)\big)\right)\subseteq z_r^{-1}(S)\ , $$ and therefore $$ z_r^{-1}(S)=\bigcup_{i=1}^\infty\left(\hat{x}_i^{-1}(S)\times X^{-1}\big(P_i(r)\big)\right)\ . $$ Thus $\ z_r:[0,1]\times \Omega\rightarrow\mathbb{C}\ $ is measurable.

Also, for all $\ \omega\in\Omega\ $, $$ \|z_r(.,\omega)\|_1\le\|X(\omega)\|_1+\frac{1}{r}\ , $$ and therefore \begin{align} \int_\Omega\int_0^1|z_r(t,\omega)|\ dt\ d\mathbb{P}(\omega)&=\int_\Omega\|z_r( \,.,\omega)\|_1\,d\mathbb{P}(\omega)\\ &\le\frac{1}{r}+\int_\Omega\|X(\omega)\|_1\,d\mathbb{P}(\omega)\\ &<\infty\ . \end{align} Thus $\ z_r(\,.,.)\in L^1([0,1]\times\Omega)\ $.

Now for every $\ r,n\in\mathbb{N}\ $ we have \begin{align} \|\,\,z_r(\,.,\omega)-z_n(\,.,\omega)\,\|_1&\le\|\,X(\omega)-z_r(\,.,\omega)\,\|_1+\|\,X(\omega)-z_n( .,\omega)\,\|_1\\ &\le\frac{1}{r}+\frac{1}{n}\\ &\le\frac{2}{\min(r,n)}\ . \end{align} Therefore, \begin{align} \int_\Omega\int_0^1|z_r(t,\omega)-z_n(t,\omega)|dt\,d\mathbb{P}(\omega) &=\int_\Omega\|\,\,z_r(\,.,\omega)-z_n(\, .,\omega)\,\|_1\,d\mathbb{P}(\omega)\\ &\le\frac{1}{r}+\frac{1}{n}\ , \end{align} and $\ \big\{z_r(\,., ,.)\big\}_{r=1}^\infty\ $ is thus a Cauchy sequence of $\ L^1([0,1]\times\Omega)\ $. It must therefore converge to some element $\ z\ $ of $\ L^1([0,1]\times\Omega)\ $. Let $\ \varphi:[0,1]\times\Omega\rightarrow\mathbb{C}\ $ be any measurable representative from the equivalence class $\ z\ $. For each $\ r\in\mathbb{N}\ $ let $$ M_r=\left\{\,\omega\in\Omega\,\left|\,\|X(\omega)- \varphi(\,.,\omega)\|_1\ge\frac{4}{r}\,\right.\right\}\ . $$ Then $$ \left\{\,\omega\in\Omega\,\left|\,X(\omega)\ne \varphi(\,.,\omega)\right.\right\}=\bigcup_{r=1}^\infty M_r\ . $$ But, for all $\ s\ge r\ $ and all $\ \omega\in M_r\ $, \begin{align} \|\varphi(\ .,\omega)-z_s(\omega,.)\,\|_1&\ge\|\,\varphi(\,.,\omega)-z_r(\,.,\omega)\,\|_1-\|z_r(\,.,\omega)-z_s(\,.,\omega)\|_1\\ &\ge\|\,\varphi(\,.,\omega)-z_r(\,.,\omega)\,\|_1-\frac{2}{r}\\ &\ge\|X(\omega)-\varphi(\,.,\omega)\|_1-\|X(\omega)-z_r(\,.,\omega)\|_1-\frac{2}{r}\\ &\ge\|X(\omega)-\varphi(\,.,\omega)\|_1-\frac{3}{r}\\ &\ge\frac{1}{r}\ , \end{align} so $\ M_r\subseteq\bigcap_\limits{s=r}^\infty\big\{\omega\in\Omega\,\big|\|\,\varphi(\ .,\omega)-z_s(\omega,.)\,\|_1\ge\frac{1}{r}\big\}\ $. Let $$ Q_s(r)=\big\{\omega\in\Omega\,\big|\|\,\varphi(\ .,\omega)-z_s(\omega,.)\,\|_1\ge\frac{1}{r}\big\}\ $$ and $$ T_r=\bigcap_\limits{s=r}^\infty Q_s(r) $$ Since $\ z_n\rightarrow\varphi(\,.,.)\ $, then, for any $\ r\in\mathbb{N}\ $ and $\ \epsilon>0\ $, there exists an $\ n\ge r\ $ such that $\ \|\varphi(\,.,.)-z_n(\,.,.)\|_1\le\frac{\epsilon}{r}\ $, and we have \begin{align} \frac{\epsilon}{r}&\ge\|\varphi(\,.,.)-z_n(\,.,.)\|_1\\ &=\int_\Omega\|\varphi(\,.,\omega)-z_n(\omega,.)\|_1\,d\mathbb{P}(\omega)\\ &\ge\int_{Q_n(r)}\|\varphi(\,.,\omega)-z_n(\omega,.)\|_1\,d\mathbb{P}(\omega)\\ &\ge\frac{\mathbb{P}\big(Q_n(r)\big)}{r}\\ &\ge\frac{\mathbb{P}\big(T_r\big)}{r}\ , \end{align} from which it follows that $\ \mathbb{P}\big(T_r\big)\le\epsilon\ $. Since this holds for all $\ r\ $ and all $\ \epsilon>0\ $, then $\ \mathbb{P}\big(T_r\big)=0\ $ for all $\ r\ $, and hence $\ \mathbb{P}\left(\bigcup_\limits{r=1}^\infty T_r\right)=0\ $. Now since $$ \left\{\,\omega\in\Omega\,\left|\,X(\omega)\ne \varphi(\,.,\omega)\right.\right\}=\bigcup_{r=1}^\infty M_r\subseteq\bigcup_{r=1}^\infty T_r\ , $$ we have $\ \varphi(\,.,\omega)=X(\omega)\ $, for $\ \mathbb{P}\text{-a.e. }\omega\in\Omega\ $. This concludes the proof of the result.