This is the region:
The area should be $\frac32$.
For $x \in (0,1)$,
$$f_X(x) = \int_0^{2-x} \frac23 \, dy$$
For $y \in (0,2)$,
$$f_Y(y) = \int_0^{\min(1,2-y)} \frac23 \, dx$$
If I'm understanding this correctly, then the biggest triangle we can make has vertices $(0,1),(1,1),$ and $(0,0)$.
You're right, that's the support of your random (multivariate) variable.
If this is the case then the probability should be 1.
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.
As $x$ increases, $y$ can at least be $x$ which means that $y$ is dependent on $x$.
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?
Best Answer
The condition that the two number chosen randomly from unit interval$(0, 1)$ is same as the conditions $0\lt x. y \lt 1$ . Let the two numbers be $x, y$ Then the probability is that $|x-y| \gt 0. 5$Draw the graph of the two functions and use geometric probability to get answer.
You need to just divide the sum of areas of two corner triangles with the square. It is easy to see probability is $0. 25$