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Edit 2: Tried to make the question clearer by adding a graph

Edit: Please note: This is not a duplicate of another question I asked earlier (Finding joint density of dependent variables). The difference being that in the other question, $x$ and $y$ are correlated and hence their distributions are not independent -- i.e. $f_{X,Y}\neq f_Xf_Y$. Here, I am assuming that the two variables are independently distributed, but that the probability that I want to compute (i.e. the upper bounds of the interval) is a function of the other variable $a=u(Y), \ \ b=v(X)$.

I'm new to the site so please accept my apologies for asking anything to obvious.

I've got two independent continuous random variables $X$ and $Y$ with pdf's $f_X$ and $f_Y$, such that the joint density function is given by $f_{X,Y}$ and the joint cdf by $F_{X,Y}$.

I am looking for the joint probability $Prob(X \leq a,Y \geq b)$, where each variables' cut-off point is a function of the other variable, i.e. $a=u(Y)$ and $b=v(X)$. Essentially I need to integrate where the interval is dependent on the random variables:

$$\int^a\int_b f_{X,Y}(x,y)\,dy\,dx$$

Graphically, I am looking for the blue area:

enter image description here

which (I believe) should come down to $F_{X,Y} (a,y)-F_{X,Y} (a,b)$, but $a$ and $b$ depend on $X$ and $Y$, respectively.

Is there a way to find an analytical solution to this problem? Many thanks in advance even for a small hint. Much appreciated!

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  • $\begingroup$ Hi Graham, I asked the other question too, but it's in fact a distinct problem. Here the variables are assumed to be distributed independently, only the upper interval bounds are a function of the other variable; the other question concerns dependent distributions of the two variables. $\endgroup$
    – jush2
    May 21, 2014 at 1:56

1 Answer 1

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By definition, $$ P(X\leqslant a(Y),Y\leqslant b(X))=\iint\mathbf 1_{x\leqslant a(y)}\,\mathbf 1_{y\leqslant b(x)}\,f_X(x)\,f_Y(y)\,\mathrm dx\mathrm dy. $$ With no further information on the functions $a$ and $b$, it seems difficult to be more specific.

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  • $\begingroup$ Thanks Did, that's already helpful. So suppose I have just a simple linear function, with a(y) = m + k*y, and likewise for b(x). Can I find an analytical solution for the joint probability simply by integrating in parts? $\endgroup$
    – jush2
    May 22, 2014 at 0:17
  • $\begingroup$ Except for degenerate cases such as constant functions $a$ and $b$, I see no reason to expect close form formulas. To get a grasp of what is going on, you might try to solve completely a concrete case with some explicit functions $a$ and $b$, starting from the identity in my post. $\endgroup$
    – Did
    May 22, 2014 at 5:52

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