probabilityprobability distributions
If the joint probability density of X and Y is given by
$$ f(x,y)= \begin{cases}
2(x^2-xy) && \text{ for } 0 <x<1,-x<y<x\\
0 && \text{elsewhere}\\
\end{cases} $$
and asking to find P(x+y<0.5)
as shown in the graph boundaries show be like this
graph
$$\int\limits_{0}^{1/2}\int\limits_{0}^{\frac{1}{2}-y}cxy^2(1-x)(1-y^2)\, dxdy = 1$$
but I am confuse because y depends on x, so it shouldn't be like this.
any explanation ?
thank you
$$ \Bbb P (X+Y>3)=\Bbb P (Y>3-X)=1-\Bbb P(Y<3-X) \\=1-\int_0^{3}\int_0^{3-y}e^{-x-y}dx~dy=4e^{-3} $$
We are looking for the region to the right of the triangle enclosed between the x axis, y axis and the line, which should be the same as the converse of the region to the left, im not sure why your answer is like that, unless there is some other piece of information that is missing from the question, or maybe I've made a mistake.
Best Answer
You want to integrate over the intersection of $0<x<1$, $-x<y<x$, and $x+y<1/2$.
Because $y$ may be negative, then $x$ may be greater than $1/2$. Indeed when $x=1$, then $-1<y<-1/2$ is in the interval.
So the outer integration is over all of $x\in[0;1]$ and the inner is over $y\in[-x;\min\{x,1/2-x\}]$.
$$\int_0^1\int_{-x}^{\min\{x,1/2-x\}} 2(x^2-xy) \,\mathrm dy\,\mathrm dx$$
You can split this into a sum of two integrals, one over $x\in[0;1/2)$ and the second over $x\in[1/2;1]$.