7
$\begingroup$

Often people put down $X\perp\ Y$ as independence, but this merely means that the expectation of X and Y is zero and does not have any implication on correspondence between their joint PDF, CDF and marginal PDF, CDF, etc...

Is there a widely agreed upon notation for saying that two RVs are independent?

$\endgroup$
  • 5
    $\begingroup$ \perp\!\!\!\perp yields $\perp\!\!\!\perp$ which is often used to denote independence, e.g. $X\perp\!\!\!\perp Y$. $\endgroup$ – Stefan Hansen Oct 6 '14 at 9:54
4
$\begingroup$

$\require{txfonts}$As you say, the use of $\perp$ (\perp) for independence is not good, since it often means orthogonal, which in probabilitry theory translates to correlation zero. Independence is a (much) stronger concept, so needs a stronger symbol, and sometimes I have seen $\perp\!\!\!\perp$ (\perp\!\!\!\perp) used. That seems like a good idea!

OK, seems like math markup here does not like \Perp, but it is defined in $\LaTeX$ packages pxfonts/txfonts. It is like \perp, but with double vertical lines. Above I replace a hack.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Apart from multivariate normal distributions of the kind $(X,Y)$, where one can write $Cov(X,Y)=0$, one writes "$X$ and $Y$ are independent". Why bother with symbols if normal language is already clear and short?

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ It is only for jointly normal random variables that $\operatorname{Cov}(X,Y)=0$ can be used as a substitute for saying $X$ and $Y$ are independent. Is there much saving in stating "$X$ and $Y$ are jointly normal random variables with $\operatorname{Cov}(X,Y)=0$ and means $\mu_X, \mu_Y$ and variances $\sigma_X^2, \sigma_Y^2$ respectively" versus "$X\sim N(\mu_X,\sigma_X^2)$ and $Y\sim N(\mu_Y,\sigma_Y^2)$ are independent random variables"? $\endgroup$ – Dilip Sarwate Oct 6 '14 at 13:59
1
$\begingroup$

$X|Y = X$ does not reflect the symmetry of the (non-)relation but shouldn't it signify independence?

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Welcome to the site, @user78740. Was this intended as an answer to the OP's question, a comment requesting clarification from the OP or one of the answerers, or a new question of your own? Please only use the "Your Answer" field to provide answers to the original question. You will be able to comment anywhere when your reputation is >50. If you have a new question, click the gray ASK QUESTION at the top of the page & ask it there, then we can help you properly. Since you're new here, you may want to take our tour, which has information for new users. $\endgroup$ – gung - Reinstate Monica Jun 2 '15 at 16:29
  • $\begingroup$ I read this as a definite answer--but it's dangerous to pose answers as questions, for people misinterpret the intention and vote to close them. Regardless, there is an interesting distinction operating here: "$X|Y=X$" may indeed imply independence, but it appears to do so by virtue of a mathematical argument: it does not in itself signify independence (unless you actually define independence in terms of conditional distributions, which is a bit unusual). $\endgroup$ – whuber Jun 2 '15 at 17:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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