I am trying to find the value $P( X > 6, Y < 7)$. The main difficulty is that I cannot apply the cdf formula
$$\begin{align}F_{xy}( x_1 < X < x_2 \cap y_1 < Y < y_2) &= F_{xy}(x_2,y_2) - F_{xy}(x_1,y_2) - F_{xy}(x_2,y_1) \\&+ F_{xy}(x_1,y_1) \end{align}$$ as I would have $F_{xy}(-\infty,+\infty)$. Another possible way would be to sum the components in the joint distribution table where the properties are met.

  • $\begingroup$ Could you explain where the infinities come in? They suggest that somehow there are infinite values of $X$ and $Y$ listed in your table, which seems unlikely. $\endgroup$ – whuber Feb 2 at 18:51
  • $\begingroup$ I assumed that if $x_1 < X < x_2$ and $ x_1 = 6$ then $x_2 = +\infty$; $\endgroup$ – Sergiu Talmacel Feb 2 at 19:30

You can write what you ask (instead of $<$, you'll need $\leq$ whenever CDF is concerned, especially with discrete RVs) in terms of CDFs: $$P(X>6,Y\leq7)=P(Y\leq 7)-P(X\leq 6, Y\leq 7)=F_Y(7)-F_{XY}(6,7)$$

And, your formula is correct only when $X\leq x_2$ and $Y\leq y_2$. And, you can calculate the marginal CDF of $Y$, using the joint CDF and the supremum $X$ value that can occur for $x$.

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  • $\begingroup$ Could you explain more clearly how did you this result ? $\endgroup$ – Sergiu Talmacel Feb 2 at 19:33
  • $\begingroup$ First to second is from total probability i.e. $$P(X\leq 6\cap Y\leq 7)+P(X>6\cap Y\leq 7)=P(Y\leq 7)$$ second to third is just definiton of CDF. $\endgroup$ – gunes Feb 2 at 19:35
  • $\begingroup$ I got it. Thank you so much! $\endgroup$ – Sergiu Talmacel Feb 2 at 19:39

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