Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
1  
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
2  
@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
1  
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
1  
@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

2 Answers 2

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).

Vector spaces of non-centered random variables are not so popular. One of applications which I think about is financial mathematics, though there you more often work with some cones rather than full vector spaces. Still, a lot of machinery is based (especially in discrete time) on some vector space techniques.

share|improve this answer
    
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
    
Also, would you please clarify if it is correct to speak of the vector space of all random variables (regardless of the process that generated the observation) or the vector space of random variables generated by a process. –  Olumide Nov 8 '10 at 10:54

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.

share|improve this answer
    
I know this. Following Qiaochu's comment, what I would like to know is why random variables with non-zero or even non-constant mean do not form a vector space. In anticipation of an answer to this question I can see why it is important to detrend an observation and make it second-order stationary. This is because the theory of linear predictors on vector spaces is only defined for zero-mean processes. –  Olumide Nov 8 '10 at 10:02

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.