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I am given following joint pdf:

$$ f(x,y)=\frac{2}{3}(x+2y) $$ $$0\leq x \leq1, 0\leq y \leq1 $$

Now I need to find the $P[X\leq0.5,Y\leq0.8]$.

My question is that will I double integrate the expression from 0 till respective limits like this:

$$ \int_{0}^{0.8}\int_{0}^{0.5} \frac{2}{3}(x+2y) \,dx\,dy $$

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    $\begingroup$ Yes, that is okay. $\endgroup$
    – drhab
    Dec 16 '18 at 16:36
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    $\begingroup$ There is no confusion. For any $a,b\in(0,1)$, \begin{align} P(X\le a, Y\le b)&=E\left[\mathbf1_{X\le a,Y\le b}\right] \\&=\iint \mathbf1_{x\le a,y\le b }\,f(x,y)\,\mathrm{d}x\,\mathrm{d}y \\&=\frac{2}{3}\iint\mathbf1_{x\le a,y\le b }(x+2y)\mathbf1_{0<x<1,0<y<1}\,\mathrm{d}x\,\mathrm{d}y \\&=\frac{2}{3}\int_0^b \int_0^a (x+2y)\,\mathrm{d}x\,\mathrm{d}y \end{align} $\endgroup$ Dec 16 '18 at 16:42

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