Hi guys,
Im given X, Y, Z are independent uniform random variables on the interval (0,1).
The question is find P(x<y<z).
Im thinking its a triple integration of f(x,y,z) dxdydz ? not too sure of the bounds though..
thanks for your time
Hi guys,
Im given X, Y, Z are independent uniform random variables on the interval (0,1).
The question is find P(x<y<z).
Im thinking its a triple integration of f(x,y,z) dxdydz ? not too sure of the bounds though..
thanks for your time
No need to integrate. You can solve this by counting, since
$\displaystyle
(X < Y < Z) \cup (X < Z < Y) \cup \cdots \cup (Z < Y < X)
$
is the whole space (except for a set of measure 0) and each of those events is disjoint and equally likely. Hence, the probability is 1/6.
If you did want to integrate, which I wouldn't recommend, the integral would be of this form:
$\displaystyle \displaystyle \int_0 ^ 1 \int_0 ^ z \int_0 ^ y \ dx \ dy \ dz$
That one I integrate, although if you apply the same idea I initially used it makes things a bit easier. First, I would calculate 1 - P(XY > Z) and note that
$\displaystyle
(0 < Z < XY < X < Y) \cup (0 < Z < XY < Y < X)
$
is the equivalent to (XY > Z) (except on a set of probability 0); moreover we have a disjoint union of equally likely sets, so it suffices to calculate
$\displaystyle
1 - 2 \cdot \int_0 ^ 1 \int_0 ^ x \int_0 ^ {xy} \ dz \ dy \ dx.
$
Another approach would be to take the -log of both sides. You end up needing to find P(Gamma(2, 1) > Exponential(1)) with this method, which is easy enough. There may be an easier way that I'm not seeing that lets you get this immediately.
No typo, that is what it is. Clearly $\displaystyle XY < Z$ or $\displaystyle XY \ge Z$ correct? I can make the inequality strict because equality happens on a set of probability 0; it doesn't matter at all, but I made the inequalities strict so that I could get a disjoint union.
Both the methods I listed give the same answer, which leads me to believe I'm not making a stupid mistake.