Let $(X,Y)$ be a point chosen at random from the triangle $\{x,y:0\leq x\leq y\leq 1\}$. $f_{X,Y}(x,y)=2$ if $(x,y)$ is in the triangle, and it is 0 otherwise. Find the probability density function for $X$.
What confuses me about this problem is understanding how $x$ and $y$ make up the triangle. If I'm understanding this correctly, then the biggest triangle we can make has vertices $(0,1),(1,1),$ and $(0,0)$. If this is the case then the probability should be 1. As $x$ increases, $y$ can at least be $x$ which means that $y$ is dependent on $x$. From this though I'm not sure where to go from here.
Best Answer
You're right, that's the support of your random (multivariate) variable.
Huh... what? What you know is that "total" probability is $1$, i.e. $\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f_{X,Y}(x,y) dx dy=1$. Now, because the density here is constant, denoting by $S$ the support region and by $A_S$ its area we get $\int_{S} 2 dx dy= 2 \, A_S= 1$ and this is indeed true, because the area of the triange is $\frac12$. Then, it's all right.
Yes. In fact, if you have a bounded support that it's not a rectangle, (or a cartesian product of rectangles) then the variables are dependent.
You have the joint density $f_{X,Y}$. To get the single variable ("marginal") density, you sum (integrate) over the other variable ("marginalize") :
$$ f_X(x)=\int f_{X,Y}(x,y) dy $$
Because the density is constant, and the support is known, what remains is just to get the integrations limits right, i.e. which is the range for the integrating variable ($y$) for each fixed $x$. Can you go on from here?