Transformation of multivariate distribution

D

duvin103time

New member
Joined
Jun 14, 2006
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9
K=(k1,
k2,
k3)

P=(p11,p12,p13,
p21,p22,p23,
. . .
. . .
. . .
pn1,pn2,pn3)

Q = P*K

Random vector K have a distribution f(K). P is a constant matrix.
What is the distribution of Q?

I tried Jacobian. I pad zeros to make K n-dim vector, and pad random submatrix to make P a nxn matrix. But I cannot get a unique distribution of Q, because of the random submatrix padded to P.

Many thanks
 
This question threw me until I realized what you are trying to compute does not exist.

Your use of the word distribution with the notation f(K) leaves me guessing what is required here. By distribution do you mean the probability distribution or measure, the cumulative distribution function, or the probability density?

In my experience, the most common use of the word distribution refers to the probability distribution, also known as probability measure. Let \(\displaystyle \L \mu\) be the (probability) distribution for the random vector \(\displaystyle K.\) Then the distribution for the random vector \(\displaystyle Q(K)\) is \(\displaystyle \mu \circ Q^{-1} .\) This means for \(\displaystyle Q(K)\) the probability of an event \(\displaystyle A\) is \(\displaystyle \mu( Q^{-1} (A))\) where \(\displaystyle Q^{-1}(A)\) is the inverse image. The distribution \(\displaystyle \mu \circ Q^{-1}\) is well-defined since \(\displaystyle Q\) is a measurable function. So \(\displaystyle \mu \circ Q^{-1}\) is a correct answer to the question of what is the distribution of \(\displaystyle Q\).

It appears that you are trying to compute the density for \(\displaystyle Q .\) But there can be no density for \(\displaystyle Q .\) This is because the image \(\displaystyle S\) of \(\displaystyle Q\) is a subspace of \(\displaystyle R^n\) of at most dimension \(\displaystyle 3 < n\), so \(\displaystyle S\) has measure 0 in \(\displaystyle R^n.\) Therefore there can be no function \(\displaystyle g\) such that \(\displaystyle \int_S g =1\), which is what a density has to satisfy. But is the question asking for the density? It would be possible to define a density on the subspace \(\displaystyle S\) using the concept of Hausdorff measure. Back to the second paragaph: what is really required here?
 
duvin103time said:
I need probability density.

Considering the Dirac's Delta Function, I thought the density does exist.
I don't know whether the Dirac's Delta Function is allowed. The generalized function has measure 0, but its integral is 1.
see http://mathworld.wolfram.com/DeltaFunction.html
Click to expand...
Your attempt to calculate a density using the standard change-of-variable formula with Jacobian says to me that you want to calculate a Lebesgue measurable function on \(\displaystyle R^n\) with real values. As you indicate by using the words "generalized function," the Dirac Delta function is not a Lebesgue measurable function.

Is this problem for a class? Have you been given theoretical tools to deal with the problem? As you've found, the standard change-of-variable formula cannot handle the problem.
 
duvin103time said:
"The Dirac's Delta Function is not a Lebesgue measurable function, so the standard change-of-variable formula cannot handle the problem."
Is that right?

Please, see http://mathforum.org/kb/thread.jspa?thr ... &tstart=15
,and find the fault in that discussion. I am confused.
Click to expand...
You changed my statements by adding the "so." I did not link those two statements.

On the Dirac Delta function not being a Lebesgue measureable function, see http://en.wikipedia.org/wiki/Dirac_delta_function. The standard change-of-variable formula does not handle change of dimensions. Those statements were correct as far as they go.

But I see the possibilities in your solution

f(inv(P3)*q3)/|det(P3)|*delta(qn-Pn*inv(P3)*q3)

provided in that other thread. I don't know enough about the theory behind the delta function to know whether this solution is rigorous, but I see what you did. Essentially you computed a density on \(\displaystyle R^3\) assuming the first 3 rows of P form an invertible matrix. But to extend this density to vectors \(\displaystyle q \in R^n\), you have to check that \(\displaystyle q\) is in the image of \(\displaystyle Q.\) You express this by using the delta function.

Can you rigorously defend this use of the delta function? Denoting your density by \(\displaystyle g\), can you show that for any measurable set \(\displaystyle A \subset R^n\) that \(\displaystyle \int_A g = \int_{Q^{-1}(A)} f\)?

The use of the delta function may be a good notational device, but I'll make two points for better understanding. First, the density is not a Lebesgue measurable function. That is OK, you were not requiring that. Second, if you actually use the density to compute probabilities, you will end up using ordinary integration on \(\displaystyle R^3\) and checking that the vectors are in the image of \(\displaystyle Q\) as I described.

The issue for me was understanding what you were looking for. As I pointed out and someone in the other thread pointed out, your notation and use of terms was confusing.

This was an interesting question. Thanks for posting it. Good luck!
 
You wrote:
Can you rigorously defend this use of the delta function?
My answer is:
I cannot.
:)

I am an electronics engineer. And my math knowledge is limited. Thus my terms may be confusing. I wish I could do it better next time. :wink:
Discussions with you make the analysis more rigorous. Thanks a lot!
 
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