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If we are given the following joint CDF: enter image description here

How may I go about finding $P(Y < X^2)$? I am quite unsure. Am I supposed to somehow make use of this formula? \begin{align}%\label{} \nonumber P(x_1<X &\leq x_2, \hspace{5pt} y_1<Y \leq y_2)= \\ \nonumber &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}

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One way is to start by finding the density: \begin{align} f_{X,Y}(x,y) & = \frac{\partial^2}{\partial x\,\partial y} F_{X,Y}(x,y) = \begin{cases} 2y & \text{if } 0<x<1 \text{ and } 0<y<1, \\ 0 & \text{if } x>1 \text{ or } y>1 \\ & \text{or } x<0 \text{ or } y<0. \end{cases} \end{align}

$$ \iint\limits_{0\,<\,y\,<\,x^2\,<1} f(x,y)\, d(x,y) = \int_0^1 \left( \int_0^{x^2} 2y \, dy \right) \, dx. $$

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  • $\begingroup$ Thanks for the reply. Just to confirm, are you essentially finding $P(X \leq 1, Y \leq x^2)$? Also, out of curiosity, is there a way to get the same result by computing $P(Y < x^2) = F_Y(x^2)$? $\endgroup$
    – user812082
    Aug 2 at 21:36
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    $\begingroup$ @user62487108: $P(X \leq 1, Y \lt X^2) = P(Y \lt X^2)$ since $P(X>1)=0$. Meanwhile $P(Y < x^2) = F_Y(x^2)$ is correct but the question is rather different: $P(Y < X^2)$ which is a number rather than a function of $x^2$ $\endgroup$
    – Henry
    Aug 2 at 21:56
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    $\begingroup$ @user62487108 : The inner integral finds $\Pr(Y\le x^2),$ which is a function of (lower-case) $x.$ But then in the outer integral we integrate with respect to $x,$ so there is no $x$ left in the bottom line. $\qquad$ $\endgroup$ Aug 2 at 22:00
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    $\begingroup$ That is to say it is: ${~~\int_0^1 f_X(x)\,\mathsf P(Y{\,<\,}X^2\mid X{\,=\,}x)\,\mathrm d x\\=\int_0^1\int_0^{x^2} f_X(x)\,f_{Y\mid X}(y\mid x)\,\mathrm dy\,\mathrm dx\\=\int_0^1\int_0^{x^2} f_{X,Y}(x,y)\,\mathrm d y\,\mathrm d x}$ $\endgroup$ Aug 2 at 22:50
  • $\begingroup$ @GrahamKemp : That is not what I had in mind. Rather, I was just integrating the joint density over the appropriate set. $\endgroup$ Aug 3 at 18:42