# Thread: distribution function of Z = XY

1. ## distribution function of Z = XY

X and Y are independent uniform variables distributed between 0 and 1.

Z = XY

How to calculate the distribution function of Z?

2. 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

3. 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?

4. 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

$W=(XY)^Z$

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.

5. Originally Posted by vcarter3
X and Y are independent uniform variables distributed between 0 and 1.

Z = XY

How to calculate the distribution function of Z?
Originally Posted by vcarter3
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?

The cdf of Z is

$F(z) = \Pr(Z < z) = \Pr(XY < z) = \Pr \left(Y < \frac{z}{X} \right)$

where $0 \leq z \leq 1$

$= \int_{x = 0}^{x = z} \int_{y = 0}^{y = 1} \, dy \, dx + \int_{x = z}^{x = 1} \int_{y = 0}^{y = z/x} \, dy \, dx$.

The pdf of Z is $f(z) = \frac{dF}{dz}$

6. Originally Posted by mr fantastic
The cdf of Z is

$F(z) = \Pr(Z < z) = \Pr(XY < z) = \Pr \left(Y < \frac{z}{X} \right)$

where $0 \leq z \leq 1$

$= \int_{x = 0}^{x = z} \int_{y = 0}^{y = 1} \, dy \, dx + \int_{x = z}^{x = 1} \int_{y = 0}^{y = z/x} \, dy \, dx$.

The pdf of Z is $f(z) = \frac{dF}{dz}$
I'm having trouble understanding the limits.

7. I got $f_Z(z)=-\ln z$ on 0<z<1.

And that is a valid density and I think that's what I got the last time.

$f_{XY}(x,y)=1$ on 0<x<1, 0<y<1.

Let $Z=XY$ and $W=X$.

The inverse relationship is $W=W$ and $Y=Z/W$.

So, $f_{ZW}(z,w)=1/w$ on 0<w<1, 0<z<w.

And $f_{Z}(z)=\int_z^1 dw/w=-\ln z$ on 0<z<1.

NEXT, let $Q=Z^U$ wher U is U(0,1), indep of X and Y, and it turns out Q is U(0,1).

8. Originally Posted by Random Variable
I'm having trouble understanding the limits.
Draw the square defined by x = 0, x = 1, y = 0 and y = 1. Draw the part of the curve y = z/x that lies inside the square. Integrate over the region of the square that's below this curve.

The result is consistent with baldeagle's calculation.

9. Originally Posted by mr fantastic
The result is consistent with boldeagle's calculation.
If that's the case, then you must be correct.

10. Originally Posted by mr fantastic
The cdf of Z is

$F(z) = \Pr(Z < z) = \Pr(XY < z) = \Pr \left(Y < \frac{z}{X} \right)$

where $0 \leq z \leq 1$

$= \int_{x = 0}^{x = z} \int_{y = 0}^{y = 1} \, dy \, dx + \int_{x = z}^{x = 1} \int_{y = 0}^{y = z/x} \, dy \, dx$.

The pdf of Z is $f(z) = \frac{dF}{dz}$
How did you know to choose z as the point to break the integral into two?

11. Originally Posted by matheagle
I got $f_Z(z)=-\ln z$ on 0<z<1.

And that is a valid density and I think that's what I got the last time.

$f_{XY}(x,y)=1$ on 0<x<1, 0<y<1.

Let $Z=XY$ and $W=X$.

The inverse relationship is $W=W$ and $Y=Z/W$.

So, $f_{ZW}(z,w)=1/w$ on 0<w<1, 0<z<w.

And $f_{Z}(z)=\int_z^1 dw/w=-\ln z$ on 0<z<1.

NEXT, let $Q=Z^U$ wher U is U(0,1), indep of X and Y, and it turns out Q is U(0,1).
Some questions:
1) You mean X = W for the inverse yes?
2) Why is w the upper bound for z in your joint density function f(z,w)?
3) How did you get z and 1 for the limits of integration for fz

12. Originally Posted by vcarter3
How did you know to choose z as the point to break the integral into two?
What part of post #8 don't you understand?