4
$\begingroup$

What is the probability of a constant being between two random variables (i.e. P(X < a < Y)) in terms of the joint probability distribution function of X and Y. X and Y are not independent, otherwise it would be really simple.

$\endgroup$
5
  • $\begingroup$ It is also really simple: Hint: $$\{x<a<y\}=\{x<a\}\cap\{a<y\}$$ $\endgroup$ – Xi'an Nov 20 '17 at 19:19
  • $\begingroup$ Oh, yeah, its $\int_{a}^{0} \int_{0}^{a} f(x,y)dxdy$ right? $\endgroup$ – Sam Baker Nov 20 '17 at 19:25
  • 1
    $\begingroup$ The bounds depend on the support of $(X,Y)$ but $\int_a^0$ is certainly incorrect. $\endgroup$ – Xi'an Nov 20 '17 at 19:31
  • $\begingroup$ Made a mistake again. I meant $\int_{a}^{inf} \int_{-inf}^{a} f(x,y)dxdy$ $\endgroup$ – Sam Baker Nov 20 '17 at 19:32
  • 2
    $\begingroup$ Drawing a picture of this event in the $(X,Y)$ plane will make the answer evident. $\endgroup$ – whuber Nov 20 '17 at 19:34
4
+50
$\begingroup$

The figure (below) answers the question.

It reveals a complication: when the distribution is not continuous at $X=a,$ we have to take care not to include the chance of the event $X=a\lt Y$ in our calculation. This is done by "sneaking up" to the answer as a limit. Therein lies the interest in this question.


By definition, the joint probability function is

$$F_{XY}(x,y) = \Pr(X \le x\text{ and } Y \le y).$$

It determines the marginal probability functions $F_X$ and $F_Y$ as

$$\lim_{y\to\infty} F_{XY}(x,y) = \Pr(X \le x) = F_X(x)$$

and

$$\lim_{x\to\infty} F_{XY}(x,y) = \Pr(Y \le y) = F_Y(y).$$

We can almost get to the solution in the form

$$\Pr(X \le a \lt Y) = \Pr(X\le a) - \Pr(X\le a, Y \le a) = F_{X}(a) - F_{XY}(a,a)$$

because the event $X\le a$ is the disjoint union of the events $X\le a \lt Y$ and the intersection of $X\le a$ with $Y \le a.$ See the figure.

Figure showing the events

In this figure, we seek the probability of the upper left quadrant--which does not include its boundary rays. This quadrant is the left open half-plane (that is, without its vertical boundary) with the lower left quadrant (including its upper boundary) removed.

This solution works when $F_X$ is continuous at $a,$ because the chance that $X=a$ is nil. However, the only way to account for that chance in full generality is to notice that the event $X\lt a$ is the union of the events $X \lt a-\epsilon$ for all $\epsilon \gt 0,$ whence

$$\Pr(X \lt a \lt Y) = \lim_{\epsilon\to 0^+} \left(F_{X}(a-\epsilon) - F_{XY}(a-\epsilon,a)\right).$$

Notice this implicitly involves double limits because $F_X$ is determined from $F_{XY}$ via a limiting process.


BTW, all the limits in this question always exist, because they are all limits of non-decreasing bounded functions: every one of them is a limit of probabilities computed over a sequence of larger and larger sets (each a superset of all the preceding ones) and, of course, probabilities are bounded by $1.$

$\endgroup$

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.