The inner product that appears in most books on probability is the covariance $\langle X,Y \rangle = E[XY]$ (considering that $X$ and $Y$ are zero mean real random variables).
Are there other inner products that are used on vector spaces of random variables? How are they useful?
(Note: I had earlier asked this on math.stackexchange but had received no answers nor comments and thus felt this question may be more appropriate here.)
One simple example is as follows. Consider the space $L^2(\{-1;1\}^n)$ of functions on the discrete hypercube $\{-1;1\}^n$ (with uniform measure), and the Dirichlet quadratic form $$\langle f,g\rangle_\nabla:=\sum_{i=1}^n\mathbb{E}\left(\nabla_if\nabla_ig\right),$$ where $$\nabla_if(x_1,\dots,x_n)=f(x_1,\dots,-x_i,\dots,x_n)-f(x_1,\dots,x_i,\dots,x_n).$$ Using expansion in the Fourier-Welsh basis $\chi_\omega((x_1,\dots,x_n))=\prod_{i\in\omega}x_i,$ where $\omega$ runs over subsets of $\{1,\dots,n\}$, it is easy to prove Poincaré's inequality: $$ \mathbb{E}f^2-(\mathbb{E}f)^2\leq \langle f,f\rangle_\nabla. $$ This allows to estimate the variance of an arbitrary random variable $f$ if one has bounds on influences of individual variables $\mathbb{E}\left(\nabla_if\right)^2$. A typical application: consider a random metric on $\mathbb{Z}^2$ by declaring every lattice edge to have length $a>0$ or $b>a>0$, with probability $1/2$ independently of each other (a first passage percolation model). Consider the distance between (0,0) and (0,k). Flipping the variable corresponding to one edge can increase the distance only if this edge is in the shortest path, and in any configuration there are at most $bk/a$ such edges. Therefore, the variance of the distance grows at most linearly in $k$. (By much less elementary methods, Benjamini, Kalai and Schramm improved this by estimate by a $\log k$ factor. The correct estimate is believed to be $k^{2/3}$)
A quite general discussion of random vectors and different inner products is given here. You could insert a positive definite linear operator $\langle X,AY\rangle$ and choose $A$ as the inverse of the covariance matrix to transform the covariance into the identity.
