# Vector spaces of random variables having zero expectation

Edit: Robin's comments appear to have made the matter a lot clearer to me. I now suppose that the vector space of random variables with zero expectation are studied in the context of second order stationary processes.

The other question remains: are vector spaces of random variables with non-zero expectation also studied?

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I can't make head nor tail of this. Random variables (sometimes) have expectations, but vector spaces, to my knowledge, usually don't. – Robin Chapman Nov 8 '10 at 7:57
The set of random variables forms a vector space. For any two random variables, the inner product is defined as the joint expectation $E\{XY\}$. Furthermore, it is frequently assumed that $E\{X\} = E\{Y\} = 0$. What I would like to know is if this latter assumption is necessary in all cases? – Olumide Nov 8 '10 at 8:37
@Olumide: the set of random variables with nonzero expectation does not form a vector space. However, if X is a random variable, then X - E(X) has zero expectation, so for most purposes it's enough to study X - E(X), which is often technically easier, to study X. – Qiaochu Yuan Nov 8 '10 at 9:15
The answer to your question is "yes": $L^p(\mu)$ the set of random variables $X$ on the probability space with probability measure $\mu$ with the property that $E(|X|^p)$ is finite, is very well-studied, in probability theory and in functional analysis. – Robin Chapman Nov 8 '10 at 9:33
@Olumide: Surely, X−E(X) has zero mean, if by “mean” you mean expectation, and if the letter E stands for expectation. If not, you'd better start explaining your terminology and notation. Also, Qiaohcu is absolutely right that the variables with nonzero expectation does not form a vector space, as any vector space contains zero, and also the expecation of E(Y)X−E(X)Y is zero. – Harald Hanche-Olsen Nov 8 '10 at 10:08

One can, of course, think of $L^2(\Omega)$ as of a Hilbert space with a scalar product $E[\xi\eta]$. But for random variables much more important is the covariance $E[\xi\eta]-E[\xi]E[\eta]$. Though it looks at first sight as a scalar product, unfortunately it's not, as $\mathrm{cov}(\xi,\xi)=0$ doed not imply $\xi=0$. However, on the space of centered r.v.'s it is a scalar product. And this Hilbertian structure fully determines the laws in some cases, like a Gaussian case, as Shvai Covo already mentioned. And also this Hilbertian structure plays a very important role for (weakly) stationary processes (also noted by Shvai Covo).
Thanks! So is it correct to say that the set of random variables with non-zero mean forms a vector space but not a Hilbert space? By the way doesn't $Cov(X,X) = 0$ if $X \ne 0$ imply that the covariance is a seminorm? -- which makes me wonder why seminorms are used/acceptable in the construction of penalties in the development of the theory of splines but not in this case? – Olumide Nov 8 '10 at 10:49
Suppose we are in the context of second order stationary processes. Then, quoting from Wikipedia (entry on Stationary process), "such a process will be wide sense stationary if the mean and correlation functions are finite". In turn (again, see Wikipedia), the mean function $m_x (t) = {\rm E}\{ x(t)\}$ of a wide-sense stationary process must be constant. This can account for the assumption of zero expectation you indicated. The situation is similar with regard to Gaussian processes, where it is "frequently assumed" that the process has zero mean; then, as is well known, the law of the process is determined by its covariance.