Joint density h(z1,z2) can be factored out as g(z1)q(z2)
then z1 and z2 are indepent without finding marginals.
Why is that?
Read page 11: http://www.colorado.edu/Economics/mo...intdensity.pdf
Well, I'm not sure whether the OP meant:
1) Suppose you can write the joint density function of $\displaystyle Z_{1}$ and $\displaystyle Z_{2}$, $\displaystyle H(z_{1},z_{2})$ as the product $\displaystyle g(z_{1})q(z_{2})$. Show $\displaystyle Z_{1}$ and $\displaystyle Z_{2}$ are independent.
or 2) Suppose you can write the joint density function of $\displaystyle Z_{1}$ and $\displaystyle Z_{2}$, $\displaystyle H(z_{1},z_{2})$ as the product $\displaystyle g(z_{1})q(z_{2})$ where $\displaystyle g$ and $\displaystyle q$ are PDF's of $\displaystyle Z_{1}$ and $\displaystyle Z_{2}$ respectively. Show $\displaystyle Z_{1}$ and $\displaystyle Z_{2}$ are independent.
#2 is answered by that white paper (from my alma mater!), but #1 is not. Even if the OP meant #2, I like #1 better! It seems like it should be true, but I haven't been able to work out the proof.
I've known this for 20 years. But I have no idea where the proof is.
This is for continuous rvs.
If the joint density factors into two function, that consist solely of the two variables AND the region is a rectangle.
(That's important. And it can be an infinite rectangle.)
Then the two rvs are independent.
That's why I knew...
f(x,y)=c on $\displaystyle 0<x<y<1$ meant that X and Y were dependent.
That's a triangle and not a rectangle.
A rectangles means a<x<b and c<y<d, hence there is no relationship between x and y.
Likewise $\displaystyle f(x,y)=e^{-x-y}$ on x>0, y>0 means that they are independent and one can see that these are iid exponential (1) rvs.
I was about to go to bed, but I decided to prove it myself.
Say $\displaystyle f(x,y)=g(x)h(y)$, where g and h may not be the marginals. Also a<x<b and c<y<d is our support.
Then by integration we have
$\displaystyle f_X(x)=\int_c^d g(x)h(y)dy=g(x)(H(d)-H(c))$
and
$\displaystyle f_Y(y)=\int_a^b g(x)h(y)dx=h(y)(G(b)-G(a))$.
Thus
$\displaystyle f(x,y)=g(x)h(y)={f_X(x)\over H(d)-H(c)}\cdot {f_Y(y)\over G(b)-G(a)}$.
Or
$\displaystyle f(x,y)=K_1f_X(x)\cdot K_2f_Y(y)$, where
$\displaystyle K_1={1\over H(d)-H(c)}$ and $\displaystyle K_2={1\over G(b)-G(a)}$.
Now take the double integral to show that $\displaystyle K_1\cdot K_2=1$, i.e.
$\displaystyle 1=\int_a^b\int_c^d f(x,y)dy dx=K_1\int_a^bf_X(x)dx \cdot K_2\int_c^df_Y(y)dy=K_1K_2$
leaving us with $\displaystyle f(x,y)=f_X(x)f_Y(y)$.
I haven't seen this theorem in any book in 20 years.
I asked a colleague about it and he said it's in Hogg's book.
I just found it on page 77 of the THIRD edition.
The proof is exactly what I did two days ago.
Hello,
This is what I remember... It involves Fubini's theorem for integrals... And I'm sorry, it involves bits of measure theory ^^'
Assume that g and q are pdf, that is their integral is 1.
Let $\displaystyle \tilde{f}$ be a measurable, positive and bounded function.
$\displaystyle \mathbb{E}(\tilde{f}(Z_1))={\color{red}\int \tilde{f}(z_1,z_2) ~ P_{(Z_1,Z_2)}(dz_1,dz_2)}=\int_{\mathbb{R}^2} \tilde{f}(z_1,z_2) ~ h(z_1,z_2) ~dz_1dz_2$ (by the "transfer formula") ----- you can skip the red part if you don't want to hear of measure.
now define f such that $\displaystyle f(x)=\tilde{f}(x,y)$
$\displaystyle \mathbb{E}(f(Z_1))=\int_{\mathbb{R}^2} f(z_1) h(z_1,z_2) ~dz_1dz_2$
since a density is integrable and that f is measurable, then we can use Fubini's theorem :
$\displaystyle \mathbb{E}(f(Z_1))=\int_{\mathbb{R}} f(z_1) \left(\int_{\mathbb{R}} h(z_1,z_2) ~dz_2\right) ~dz_1$
So again, using the "transfer formula", we get that the pdf of $\displaystyle Z_1$ is $\displaystyle \int_{\mathbb{R}} h(z_1,z_2) ~dz_2=\int_{\mathbb{R}} g(z_1)q(z_2) ~dz_2=g(z_1) \int_{\mathbb{R}} q(z_2) ~dz_2=g(z_1)$
Similarly, we can get the pdf of $\displaystyle Z_2$ equal to $\displaystyle q(z_2)$
and so we have the joint probability function which is equal to the product of the two marginal pdf.
Thus $\displaystyle Z_1$ and $\displaystyle Z_2$ are independent.
Does it look correct and understandable ?
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The "transfer formula" says that if X has a pdf h, and if f is a measurable bounded function, then :
$\displaystyle \mathbb{E}(f(X))=\int f(x)h(x) ~dx$
(the reciprocal is true, that is if you find an expression of the expectation like this, then h is the pdf of X)
Matheagle, but that function doesn't factor if you write the PDF over $\displaystyle \mathbb{R}^2$ without a "support clause".
$\displaystyle f(x,y) = 2\chi_{(0<x<y<1)}(x,y)=2\chi_{(0<x<1\wedge x<y<1)}(x,y)$
Where $\displaystyle \chi_{(A)}(x)=\begin{cases}
1 & \textrm{if }x\in A \\
0 & \textrm{if }x\notin A\end{cases}$
Your point is well taken though. You have to be careful when you look at a joint PDF to decide whether it factors or not.
Also, in your proof, you don't say so, but I assume that the support doesn't have to be finite. It looks to me like the proof works in that case as well so long as $\displaystyle \lim_{x\rightarrow \pm \infty}G(x),H(x)$ are finite.
Correct, that's why you need a rectangle region.
But my undergrads don't like these indicator functions.
I just let them examine the joint density and the region.
This way they can tell in a second whether the rvs are indep or not.
What's suprising is that this basic theorem is not mentioned in
Walpole, Wackerly....
Yet the a basic version of the Central Limit Theorem is proved.
And moreover these books waste your time asking to check if the rvs are
indep by finding the marginals and that's just silly.
And in Hogg's book, he uses the infinities instead of a,b,c,d.
My a,b,c,d need not be finite.
I thought about the cases of G and H going to infinity as insert our bounds,
but since this is a finite measure I don't think thats possible.
This is why I looked at Hogg's book.
I was wondering about those cases, but our densities are well behaved functions.