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I just started to learn the inner product and vector space definition for random values , and I could see the following inner product definition for the random values X and Y in the vector space.

< X , Y > = E[XY]

Orthogonality of random vectors and the inner product

The question is that: If the X and Y are single continuous random variables, what is the dimension of the vector space V where random variables X and Y stay. Is Dim(V) = 1 or Dim(V) = infinite ？ I believe it is a very basic question but I can not catch it. I also have searched it on the google but nothing implies on it.

I searched from @Kavi Rama Murthy's comment "X and Y as real valued functions" and found some info on the "function space". From my understanding , X and Y are in the "function space" right ? And because the outputs of continuous random variables X and Y contain infinite choice , the dim(V) is also infinite ? Not sure whether I think incorrectly

I understand it now. The length of the vector which the function random variable outputs is infinite , but there are only 2 basis: X and Y (if the 2 infinite length vectors are linearly independent).

More precisely , the dimension of the subspace which only contains the basis X and Y are 2. The vector space is the function space which holds the random variable X and Y , and the maximum dimension of the continuous function space is infinite.

It is also strange that nobody on the web used to mention that : Random Variable is in the "Function space".

zhfkt
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If you are considering $X$ and $Y$ as real valued functions on the sample space then, as a vector space over $\mathbb R$, the dimension is $0$, $1$ or $2$. It is $1$ iff one of then is scalar multiple of the other and it is $0$ iff $X=Y=0$.

Kavi Rama Murthy
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    Thank you for replying. I searched from your comment "X and Y as real valued functions" and found some info on the "function space". From my understanding , X and Y are in the "function space" right ? And because the outputs of continuous random variables X and Y contain infinite choice , the dim(V) is also infinite ? Not sure whether I think incorrectly – zhfkt Jul 27 '19 at 05:34
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    $X$ and $Y$ are just two elements of the function space. So then can span a space of dimension at most $2$. – Kavi Rama Murthy Jul 27 '19 at 05:45
  • I understand it now. The length of the vector which the function random variable outputs is infinite , but there are only 2 basis: X and Y (if the 2 infinite length vectors are linearly independent). So the answer for this question is your comment. It is also strange that nobody on the web used to mention that : Random Variable is in the "Function space". Thank you for helping ! – zhfkt Jul 27 '19 at 09:36
  • More precisely , the dimension of the subspace which only contains the basis X and Y are 2. The vector space is the function space which holds the random variable X and Y , and the dimension of the continuous function space is infinite. – zhfkt Jul 27 '19 at 09:48