4
$\begingroup$

When we say a random variable is i.i.d., it's often used to describe the dependency between the observations of that random variable, which I call the row dimension, indexed by time if it's a time series.

What does i.i.d. mean when applied to multiple random variables at once? Does it mean that each separate r.v.'s observations are again i.i.d. observation-to-observation? or does it imply that all observations from r.v. 1 are i.i.d. from all the observations from r.v. 2, etc, etc? In other words, does the univariate definition of observation-to-observation change to one of distribution-to-distribution (individual distributions being independent from one another)?

Improve this question
$\endgroup$

2 Answers 2

Reset to default
7
$\begingroup$

It usually means that the p-variate distribution of the vector rv $X_1$ is the same as that of $X_2$, $X_3$, ..., $X_n$ and that they're mutually independent.

It doesn't usually indicate anything about the independence nor identicality of distribution of the components of $ X_i$.

Usually if the discussion is about the components of $X_i$ there will be explicit/unambiguous reference to that fact.

Improve this answer
$\endgroup$
2
  • $\begingroup$ so you're saying the univariate definition of i.i.d. that goes observation-by-observation does change to one of distribution-to-distribution $\endgroup$
    – develarist
    Commented Aug 5, 2020 at 23:50
  • 1
    $\begingroup$ Sorry I don't follow what you're saying. Note that in the multivariate case, the "observation" is a vector of values, with some joint distribution. $\endgroup$
    – Glen_b
    Commented Aug 6, 2020 at 5:05
1
$\begingroup$

"When we say a random variable is i.i.d". No! 'We' don't say that. iid never refers to a single random variable, but always to a family/sequence (finite or infinite) of random variables $X_{1},X_{2},\dots$. And here actually one may rather speak of random elements than variables because this concept, i.e. identical distributions and independence, doesn't need at all the assumption of the $X_{i}$s being real-valued. They may also be vectors, functions, sets, (almost) whatever you like.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Using the term "variable" doesn't assume it is real-valued. There are also discrete or censored random variables, to name a few. $\endgroup$
    – chl
    Commented Nov 2, 2020 at 19:28
  • $\begingroup$ Discrete rv are in particular real valued. I don't know what censored r.v. are. But that wasn't the point anyway. Whatever we want to call it, the concept of iid is not restricted to real valued random entities. $\endgroup$
    – Tobsn
    Commented Nov 2, 2020 at 19:42
  • $\begingroup$ Re "No!" I think this answer (which is not wrong, btw) might just be papering over the issue that troubled the OP: given any sequence $X_1,X_2,\ldots$ of real-valued iid variables we may construct the single random variable $(X_1,X_2,\ldots)$ whose values are in a real vector space and in this sense, I suspect, many authors might freely (and even unconsciously) refer to it as an "iid" variable in the sense that its marginals are independent and identically distributed. $\endgroup$
    – whuber ♦
    Commented Nov 2, 2020 at 20:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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