Question

Let f(x,y) =3/2 \(x^2\leq y \leq 1\) \(0\leq x\leq 1\)
a)find \(P(0\leq X\leq 1/2)\)
b) find \(P (1/2 \leq Y \leq1\)
c)Find \(P(X\geq 1/2,Y\geq 1/2)\)
Please, help

Answer 1

I got
\[f_X(x)= \dfrac{3}{2}(1-x^2)~~~0\leq x\leq 1\]
\[f_Y(y) =\dfrac{3}{2}\sqrt{y}~~~0\leq y\leq 1\]
and after testing, I have they are dependent.
However, I don't know how to do for a, b, c
@Zarkon

Answer 2

a) \[\int_0^{1/2}f_X(x) dx=\int_0^{1/2}(3/2)(1-x^2)dx\] right?
the same with b, right?

Answer 3

for c) it is

need confirm a, b only

Answer 4

Yup a) and b) look good :)
Although the integral bounds for b) will be 1/2 to 1, but I'm pretty sure you knew that ;)

Answer 5

@kirbykirby but I don't know why we have to take
\[\int_{1/2}^{1}\int_{1/2}^{1/\sqrt2}(3/2) dxdy + \int_{1/2}^{1}\int_{1/\sqrt 2}^\sqrt y (3/2) dx dy\]
I understand the second integral, but the first one. Is it not that it is a rectangular and we just take width x height as \((1/\sqrt 2)-1/2)(1/2)\) ??
why do we have to have that integral?

Answer 6

this is regarding c) right?

Answer 7

yes

Answer 8

ah well it seems that they have broken up the integral into two regions . Maybe the diagram here is more clear

However, it is a uniformly distributed region, so indeed, you can just take the geometric area .. but you must divide that area by the total area of the 2-D region covered by your joint density function