Suppose the point (X, Y ) is chosen randomly from the region bounded by the lines x = 0, y = 1, and y.. I don't know how to do this problem but I tried to do it two ways, both of which I have no idea if they are correct or not. All I know is the answer is f(x,y)=2. Can someone help me out? Thanks!



 A: *

*I think your bounds are wrong.
Mathematically, since you that the points are distributed "uniformly", the height of the volume is flat. So, let $f(x,y) = h$. this is what we are solving for. Using calculus, we have
$$\int_0^1\int_x^1f(x,y)\,dydx = \int_0^1\int_x^1 h\,dydx = 1.$$
Notice that your bounds are wrong. After solving, we find that $h  =2$.

*I don't know what you did for the bottom one. But geometrically, we know the volume of the wedge is one. The area of triangle is 
$$\frac{1}{2}\times 1\times 1 = \frac{1}{2}$$
To get the volume, we need to multiply this by some height $h$, and we know it is equal to $1$, hence
$$\frac{1}{2}h = 1\implies h = 2.$$
So intuitively, we know that the joint density over this region is $f(x,y) = 2$.
A: The fact that the point $(X,Y)$ is chosen uniformly from the triangular region $T$ means (by definition) that the joint probability density function $f_{X,Y}(x,y)$ of the coordinates $X,Y$ is constant inside that region
$$
f_{X,Y}(x,y)=C.
$$ 
How to determine this constant? Imposing normalization, i.e.
$$
\iint_T dx dy\ f_{X,Y}(x,y) =1\Rightarrow C\times (\text{area of }T) =1.
$$
The area of the triangle is given by $1/2\times 1\times 1=1/2$, so the constant $C=2$.

Your second approach to the problem is not really understandable (it is not clear what $x$ is supposed to represent. On top of this, $x^2=2$ clearly does not imply $x=2$!). Your first approach starts from the rather bad notation
$$
\int_0^x\int_0^1 \frac{1}{2}bh\ dy dx\ .
$$
First, you should never use the same symbol ($x$) for an integration variable, and a limit of integration. Second, when you have a double integral, one of which has one (or both limits of integration) depending on the other variable, you shoud be very clear about which integral should be performed first [in your way of writing it, the outer integral depends on $x$, so it seems the final result of the integration should still depend on $x$!]. Third, if $b$ and $h$ are meant to represent base and height of the triangular region, then you don't need any integration at all (you are already computing the area of that region!). If you really want to use integrals to compute the area of that triangle (it is a longer way, but may be instructive) you should write
$$
\int_T dx dy=\int_0^1 dx \int_0^x dy=\int_0^1 dx\ x=\frac{x^2}{2}\Big|_0^1=\frac{1}{2}
$$
as before.
