Expected value of a random operator

Question

Suppose that $X$ is a random element with values in a separable Hilbert space $\mathbb H$ such that $\operatorname E\|X\|^2<\infty$. The expected value of $X$ is defined as a unique $\mu\in\mathbb H$ such that $$\operatorname E\langle x,X\rangle=\langle x,\mu\rangle$$ for each $x\in\mathbb H$ (the existence and uniqueness of such an element follows from the Riesz representation theorem).

I am trying to understand the definition of the covariance operator of $X$. Usually, the covariance operator of $X$ is defined as $$\operatorname E[(X-\mu)\otimes(X-\mu)],$$ where $\mu$ is the expected value of $X$ and $x\otimes x:\mathbb H\to\mathbb H$ with $x\in\mathbb H$ denotes an operator defined by $(x\otimes x)(y)=\langle y,x\rangle x$ for each $y\in\mathbb H$. So $(X-\mu)\otimes(X-\mu)$ is a random operator and $\operatorname E[(X-\mu)\otimes(X-\mu)]$ is the expected value of a random operator. How is the expected value of this random operator defined?

Answer 1

Similarly to the expected value of a random vector. If $X$ is a random vector operator then its expected value is the unique operator $T$ such that $\mathbb{E} \langle v, X w\rangle = \langle v, T w\rangle$ for all $v,w \in \mathbb{H}$.

Answer 2

Since you defined the expected value as a linear functional, via $\langle x,\mu\rangle := \operatorname E\langle x,X\rangle$, you can similarly define the covariance as a sesquilinear form, $$ B(x, y) = \operatorname{E}\left(\langle x, X-\mu\rangle \langle X-\mu, y\rangle \right) $$ This is evidently bounded: $|B(x,y)|\le C\|x\|\|y\|$ where $C = \operatorname{E}\|X-\mu\|^2$. So there is a bounded operator $T$ such that $B(x, y) = \langle Tx, y\rangle$, and this is the covariance in operator form.