# Independence of Random Variables

1. Dec 2, 2017

### showzen

1. The problem statement, all variables and given/known data
Given $f_{X,Y}(x,y)=2e^{-x}e^{-y}\ ;\ 0<x<y\ ;\ y>0$,
The following theorem given in my book (Larsen and Marx) doesn't appear to hold.

2. Relevant equations
Definition
$X$ and $Y$ are independent if for every interval $A$ and $B$, $P(X\in A \land Y\in B) = P(X\in A)P(Y\in B)$.
Theorem
$X$ and $Y$ are independent iff $f_{X,Y}(x,y)=g(x)h(y)$.
If so, there is a constant $k$ such that $f_X(x)=kg(x)$ and $f_Y(y)=(1/k)h(y)$.

3. The attempt at a solution
Consider $g(x)=2e^{-x}$ and $h(y)=e^{-y}$. Then, $f_{X,Y}(x,y)=g(x)h(y)$, therefore theorem indicates that $X$ and $Y$ are independent.
The constant $k$ is
$k=\int_0^\infty h(y)dy=1$
$k=\int_0^y g(x)dx = 2(1-e^{-y})$
There is a contradiction in the value of k and it is not constant.

Am I missing something, or is the theorem incomplete in that it is lacking details on the intervals that the random variables are defined on?

2. Dec 2, 2017

### MathematicalPhysicist

The theorem is correct, you should have: $f_X(x) = \int_x^\infty f_{X,Y}(x,y)dy$ and $f_Y(y) = \int_0^y f_{X,Y}(x,y)dx$.

3. Dec 2, 2017

### Ray Vickson

You are definitely missing something: your "product" formula for $f_{X,y}(x,y)$ holds only over the region $0 < x < y$. (Presumably, $f_{X,Y}$ is zero outside that region, but the question does not say that.) Anyway, a bivariate density for a pair of independent random variables would need to be non-zero on the whole of the bivariate space $0 < x,y < \infty$.

To see explicitely that $X,Y$ are NOT independent, compute the marginal densities
$$f_X(x) = \int_{-\infty}^{\infty} f_{X,Y}(x,y) \, dy = \int_x^{\infty} e^{-x} e^{-y} \, dy$$
and
$$f_Y(y) = \int_{-\infty}^{\infty} f_{X,Y}(x,y) \, dx =\int_0^y e^{-x} e^{-y} \, dx .$$
Finish the calculations and then check if $f_X(x) f_Y(y)$ agrees with your $f_{X,Y} (x,y)$

Last edited: Dec 2, 2017
4. Dec 2, 2017

### showzen

$f_X(x) = 2\int_x^\infty e^{-x}e^{-y}dy = 2e^{-2x}$
$f_Y(y) = 2\int_0^y e^{-x}e^{-y}dx = 2e^{-y}(1-e^{-y})$
$f_X(x)f_Y(y) = 4e^{-2x}e^{-y}(1-e^{-y}) \neq f_{X,Y}(x,y)$
So we can show explicitly that they are not independent...
But we can separate $f_{X,Y}(x,y)$ into a product of $g(x)$ and $h(y)$ which implies independence by the theorem?

5. Dec 2, 2017

### Ray Vickson

No, no, no! Absolutely not! Your $f_{X,Y}(x,y)$ is NOT a product of some $g(x)$ and some $h(y)$ over the whole quarter plane $0 < x, y < \infty$. It is a product, but only over the restricted region $0 < x < y < \infty$, and that fact makes $X$ and $Y$ very definitely not independent. In fact, you should do what I suggested in my first response, which is to compute the marginals $f_X(x)$ and $f_Y(y)$. You can use these to compute $P(X \in A)$ and $P(Y \in B)$ for $A = \{ 0 < x < 1 \}$ and $B= \{ 0 < y < 2 \}$. Now compute $P(X \in A) \cdot P(Y \in B)$. Finally, compute $P(X \in A \; \& \; Y \in B)$ by integrating $f_{X,Y}(x,y)$ over the appropriate 2-dimensional region. Do you get different answers? (I do.)

6. Dec 2, 2017

### showzen

Yes, the answers are different. Can $X$ and $Y$ only be independent if they are defined over a rectangle?

7. Dec 2, 2017

### Ray Vickson

Basically, yes, and for obvious reasons: you cannot make $g(x) h(y)$ come out at zero over a part of the $(x,y)$ plane where both $g(x) \neq 0$ and $h(y) \neq 0$.