X and Y are independent uniform variables distributed between 0 and 1.
Z = XY
How to calculate the distribution function of Z?
Hello,
Their joint pdf, that is to say the pdf of (X,Y) is the product of their pdf, since they're independent.
It equals 1, for 0<x<1 and 0<y<1.
Then, use the Jacobian transformation that transforms (X,Y) into (XY,Y) (see here : http://www.mathhelpforum.com/math-he...ion-r-v-s.html for more information) to get the joint pdf of (XY,Y)
Then integrate with respect to the second variable to get the pdf of XY
how would i do it if i want to take the integral
from 0 to 1 for dy and 0 to z/x for dx
In other words, if I have the
P(X <=1, Y <= u/x)
= double integral of 0 to 1 dx and 0 to u/x dy
how to solve this integral?
I solved that a few months ago on this board.
You may have to look through a few pages to find it.
The actual problem was to find the density of
$\displaystyle W=(XY)^Z$
where X,Y and Z are iid U(0,1) rvs.
It turns out that W is U(0,1) too.
On the way to solving that I found the density of XY.
The cdf of Z is
$\displaystyle F(z) = \Pr(Z < z) = \Pr(XY < z) = \Pr \left(Y < \frac{z}{X} \right)$
where $\displaystyle 0 \leq z \leq 1$
$\displaystyle = \int_{x = 0}^{x = z} \int_{y = 0}^{y = 1} \, dy \, dx + \int_{x = z}^{x = 1} \int_{y = 0}^{y = z/x} \, dy \, dx $.
The pdf of Z is $\displaystyle f(z) = \frac{dF}{dz}$
I got $\displaystyle f_Z(z)=-\ln z$ on 0<z<1.
And that is a valid density and I think that's what I got the last time.
$\displaystyle f_{XY}(x,y)=1$ on 0<x<1, 0<y<1.
Let $\displaystyle Z=XY$ and $\displaystyle W=X$.
The inverse relationship is $\displaystyle W=W$ and $\displaystyle Y=Z/W$.
So, $\displaystyle f_{ZW}(z,w)=1/w$ on 0<w<1, 0<z<w.
And $\displaystyle f_{Z}(z)=\int_z^1 dw/w=-\ln z$ on 0<z<1.
NEXT, let $\displaystyle Q=Z^U$ wher U is U(0,1), indep of X and Y, and it turns out Q is U(0,1).