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.

  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ Drawing a picture of this event in the $(X,Y)$ plane will make the answer evident. $\endgroup$ – whuber Nov 20 '17 at 19:34

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

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


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

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 $1.$


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