Results 1 to 9 of 9

Math Help - Can anyone explain this?

  1. #1
    Junior Member
    Joined
    Nov 2008
    Posts
    53

    Can anyone explain this?

    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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by ninano1205 View Post
    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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jul 2008
    Posts
    138
    Well, I'm not sure whether the OP meant:

    1) Suppose you can write the joint density function of Z_{1} and Z_{2}, H(z_{1},z_{2}) as the product g(z_{1})q(z_{2}). Show Z_{1} and Z_{2} are independent.

    or 2) Suppose you can write the joint density function of Z_{1} and Z_{2}, H(z_{1},z_{2}) as the product g(z_{1})q(z_{2}) where g and q are PDF's of Z_{1} and Z_{2} respectively. Show Z_{1} and 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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    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 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 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.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    I was about to go to bed, but I decided to prove it myself.
    Say 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
    f_X(x)=\int_c^d g(x)h(y)dy=g(x)(H(d)-H(c))
    and
    f_Y(y)=\int_a^b g(x)h(y)dx=h(y)(G(b)-G(a)).
    Thus
    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

    f(x,y)=K_1f_X(x)\cdot K_2f_Y(y), where

    K_1={1\over H(d)-H(c)} and K_2={1\over G(b)-G(a)}.

    Now take the double integral to show that K_1\cdot K_2=1, i.e.

    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 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.
    Last edited by matheagle; March 11th 2009 at 05:43 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    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 \tilde{f} be a measurable, positive and bounded function.

    \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 f(x)=\tilde{f}(x,y)


    \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 :
    \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 Z_1 is \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 Z_2 equal to q(z_2)

    and so we have the joint probability function which is equal to the product of the two marginal pdf.
    Thus Z_1 and Z_2 are independent.


    Does it look correct and understandable ?

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
    The "transfer formula" says that if X has a pdf h, and if f is a measurable bounded function, then :
    \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)
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    If f(x,y) factors and you're not on a rectangle.
    Then the variables are dependent.
    For example let f(x,y)=2 on 0<x<y<1.
    The support must be a rectangle.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Jul 2008
    Posts
    138
    Matheagle, but that function doesn't factor if you write the PDF over \mathbb{R}^2 without a "support clause".

     f(x,y) = 2\chi_{(0<x<y<1)}(x,y)=2\chi_{(0<x<1\wedge x<y<1)}(x,y)

    Where \chi_{(A)}(x)=\begin{cases}<br />
1 & \textrm{if }x\in A \\<br />
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 \lim_{x\rightarrow \pm \infty}G(x),H(x) are finite.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Explain this!
    Posted in the Math Challenge Problems Forum
    Replies: 4
    Last Post: December 30th 2010, 03:41 PM
  2. How can i explain???
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: June 6th 2010, 04:37 PM
  3. Can someone explain this to me?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: August 29th 2009, 08:11 AM
  4. Can someone explain how I do this one?
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: July 15th 2008, 08:22 PM
  5. Can you explain to me how to do this?
    Posted in the Algebra Forum
    Replies: 1
    Last Post: June 8th 2008, 03:55 AM

Search Tags


/mathhelpforum @mathhelpforum