X and Y are continuous RVs, such that $f(x,y) = 2, 0\leq x\leq 1, 0\leq y\leq 1, 0\leq x+y\leq 1$
I'm trying to find $P(x<1/2,y>1/2)$. So i'm integrating from $\dfrac{1}{2}$ to $1$ for $y$ and $0$ to $\dfrac{1}{2}$ for $x$ on $2\mathrm dx \mathrm dy$. But I'm still not getting the answer. What am I doing wrong?
This shows why it is important to keep track of when the pdf is zero. Using indicator functions, the pdf of $(X,Y)$ can be written as $$ f(x,y)=2\mathbf{1}_{0\leq x\leq 1}\mathbf{1}_{0\leq y\leq 1-x} $$ and hence $$ \begin{align} P(X<\tfrac12,Y>\tfrac12)&=\iint_{\mathbb{R}^2}f(x,y)\mathbf{1}_{x<1/2}\mathbf{1}_{y>1/2}\,\mathrm dy\,\mathrm dx\\ &=2\iint_{\mathbb{R}^2}\mathbf{1}_{0\leq x<1/2}\mathbf{1}_{1/2<y\leq 1-x}\,\mathrm dy\,\mathrm dx\\ &=2\int_0^{1/2}\int_{1/2}^{1-x} \mathrm dy\,\mathrm dx\\ \end{align} $$ showing that the upper limit of the innermost integral should be $1-x$ and not $1$.