Let $X$ and $Y$ have the joint pdf $f(x, y) = 6(1 − x − y)$ for $x + y < 1, \ 0 < x, \ 0 < y$ and zero elsewhere. Compute $P(2X + 3Y < 1)$ and $E[XY + 2X^2]$.
So I have to find $P(2X + 3Y <1)$:
$$ \int_?^?\int_?^{(1-2y)/3} 6(1-x-y) \ dx \ dy $$
I know for sure I have made a mistake on domain of $x$ and $y$. I've heard that it's supposed to be 0 to 1/2 and 0 to $(1-2x)/3$ but I have no idea how they got such domain.
And for $$E[XY+2X^2] = \int_?^?\int_?^? 6(1-x-y) \ dx \ dy $$
How do I find the domain again and is this the right integral?
The probability density function "lives" on the part of the first quadrant that is below the line $x+y=1$. This is a triangle $T$. Draw the triangle. For the expectation question, we want to integrate $(xy +2x^2)(6)(1-x-y)$ over $T$.
Express as an iterated integral. The variable $y$ goes from $0$ to $1-x$, and then $x$ goes from $0$ to $1$.
For the probability that $2X+3Y\lt 1$, draw the line $2x+3y=1$. We want to find the probability that $(X,Y)$ lands in the part of $T$ that is below the line $2x+3y=1$. This is a triangle with corners $(0,0)$, $(1/2,0)$, and $(0,1/3)$.
Express as an iterated integral. Note that $y$ goes from $0$ to $(1-2x)/3$ and then $x$ goes from $0$ to $1/2$.