1
$\begingroup$

If we have two Random Variables X and Y, what is the interpretation for the three cases:

a] p(X=x,Y=y) = $p_1$(X=x) * $p_2$(Y=y)

b] p(X=x,Y=y) < $p_1$(X=x) * $p_2$(Y=y)

c] p(X=x,Y=y) > $p_1$(X=x) * $p_2$(Y=y)

where p -> joint pdf of X and Y,

$p_1$ -> marginal pdf of X

$p_2$ -> marginal pdf of Y

Improve this question
$\endgroup$
5
  • $\begingroup$ Various types of correlation between X and Y $\endgroup$ – Yonatan N Oct 30 '10 at 15:10
  • $\begingroup$ If its not in LaTeX, then it becomes tough for users to understand as to what exactly the OP Wants $\endgroup$ – anonymous Oct 30 '10 at 16:06
  • $\begingroup$ @Chandru1: I apologize. I have formatted the question better. @Yonatan N: Can you please elaborate? Does a] signify X and Y are independent? What is the difference between b] and c] wrt correlations? $\endgroup$ – SkypeMeSM Oct 30 '10 at 16:38
  • $\begingroup$ I would call "b" "negatively correlated". In other words, observing y decreases the probability of observing x. The reverse for c. $\endgroup$ – Yaroslav Bulatov Dec 31 '10 at 10:01
  • $\begingroup$ As pointed out by Yonatan N, these are just different types of correlations. In general, they are defined via the c.d.f. instead of the p.d.f. For example, your case (a) should be $\mathbb P(X\leq x,Y\leq y) = \mathbb P(X\leq x)\mathbb P(Y\leq y)$, for all $x,y\in\mathbb R$, and so on. (a) does not imply independence in general. (c) ($\mathbb P(X\leq x,Y\leq y) \geq \mathbb P(X\leq x)\mathbb P(Y\leq y)$) is often refereed to that $X$ and $Y$ are positively associated. $\endgroup$ – Morning Jan 30 '11 at 15:53
2
$\begingroup$

I assume that this statement applies for specific values of $x,y$ because I'm not sure that (b) or (c) is possible for all $x,y$. Also, the way you've written it here these must be discrete random variables (e.g. by use of the '=') so these are actually pmfs, not pdfs. That being said, first note that $P(X = x, Y = y) = P(X = x | Y = y) P(Y = y)$ (and similarly with the roles of $X$ and $Y$ reversed). So:

(a) implies that $P(X = x | Y = y) = P(X = x)$, which is the definition of independence. That is, knowing that $Y = y$ tells us nothing about the event $X = x$.

(b) implies that $P(X = x | Y = y) < P(X = x)$, which tells us that knowledge that $Y=y$ decreases the probability that $X=x$, meaning that the events $X=x$ and $Y=y$ are negatively ''associated'' with each other (I do not use the term correlation here because that is a statement about two random variables, but the problem stated here is one about probabilities of specific events).

Similarly, (c) gives an indication that $X=x$ and $Y=y$ are two events that are positively associated in some sense.

Improve this answer
$\endgroup$
0
$\begingroup$

Two random variables $X$ and $Y$ are independent if $P(X=x, Y=y) = P(X=x)P(Y=y)$, i.e. their joint pdf factorizes. E.g. $X$ = getting heads on the first toss of a coin, $Y$ = getting heads on the second toss of the coin.

In general, for random variables $X$ and $Y$, $P(X = x, Y = y) \leq P(X=x)$ and $P(X = x, Y=y) \leq P(Y=y)$. This implies that (c) can't hold true. There's no specific interpretation to (b).

Improve this answer
$\endgroup$
5
  • $\begingroup$ Here I am asking about product of marginal probability distribution functions of X and Y wrt joint pdf of X and Y. I apologize for the lack of formatting in the question. $\endgroup$ – SkypeMeSM Oct 30 '10 at 16:35
  • 1
    $\begingroup$ That's not quite right on (b) and (c). Your argument only shows that $P(X = x, Y = y)^2 \leq P(X = x)P(Y = y)$. $\endgroup$ – Mike Spivey Oct 30 '10 at 21:07
  • 2
    $\begingroup$ neither (b) nor (c) can hold as written for all x and y, since both sides of the inequalities sum to 1. perhaps the PO will clarify or restate the question. $\endgroup$ – ronaf Oct 31 '10 at 4:09
  • $\begingroup$ Yes (b) and (c) will not hold for ALL x and y. But for certain values of x,y, (b) will hold and certain others (c) will hold. I was wondering what is the interpretation for both the cases. And again.. I am considering marginal pdfs of x and y in the RHS of inequality. Thanks. $\endgroup$ – SkypeMeSM Oct 31 '10 at 4:53
  • $\begingroup$ I'm wondering for what X and Y (and values), (c) holds? $\endgroup$ – Naga Oct 31 '10 at 17:33
0
$\begingroup$

Since I can't make comments (haven't got enough points yet), I writing them in the form of answer. The word "pdf" stands for "probability density function", not for "probability distribution function". In your particular case, of course, you mean some discrete distributions, which are referred to as "pmf" (probability mass function). So, please do use the correct terminology.

Improve this answer
$\endgroup$

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.