I Derive Convolution Expression for Z_PDF(z)

[rabbed]

Hello again, a bit late (as usual after doing some thinking).

Does this make sense?

The line is (x(t), y(t)) = (t, z-t)

Z_PDF(z) = integral wrt t from -inf to inf of X_PDF(x(t)) * Y_PDF(y(t)) * sqrt( x'(t)^2 + y'(t)^2 ) * dt / sqrt(det( [ z'(x(t)) z'(y(t)) ] * [ z'(x(t)) z'(y(t)) ]^T ))

The line integral divided by the square root of the squared determinant of J where J is the gradient [ z'(x(t)) z'(y(t)) ]
(the square root which is also the length element of z divided by the area element of x and y at a point)
It would be nice to convert the line integral into a double integral with dx and dy multiplying area elements into length elements of z, but maybe that's not possible..

 

[Stephen Tashi]

The line is (x(t), y(t)) = (t, z-t)

Z_PDF(z) = integral wrt t from -inf to inf of X_PDF(x(t)) * Y_PDF(y(t)) * sqrt( x'(t)^2 + y'(t)^2 ) * dt / sqrt(det( [ z'(x(t)) z'(y(t)) ] * [ z'(x(t)) z'(y(t)) ]^T ))

The line integral divided by the square root of the squared determinant of J where J is the gradient [ z'(x(t)) z'(y(t)) ]
(the square root which is also the length element of z divided by the area element of x and y at a point)
It would be nice to convert the line integral into a double integral with dx and dy multiplying area elements into length elements of z, but maybe that's not possible..

I don't see any definition for z(t).

Does you forumula work for the case where X and Y are each uniformly distributed on [0,1] ?

 

[rabbed]

I don't see any definition for z(t).

Right, but z(t) isn't used anywhere. z = f(x,y) = x + y, does it need to be a function of t?

Does you forumula work for the case where X and Y are each uniformly distributed on [0,1] ?

Since both square roots become sqrt(2) we end up with the convolution formula, so it should be ok?

 

[rabbed]

By the way,
since (x(t), y(t)) = (t, z-t)
is
integral wrt t from -inf to inf of X_PDF(x(t)) * Y_PDF(y(t)) * sqrt( x'(t)^2 + y'(t)^2 ) * dt
equal to
integral wrt x from -inf to inf of X_PDF(x) * Y_PDF(z-x) * sqrt( (dx/dx)^2 + (dy/dx)^2 ) * dx?

In that case:
Z_PDF(z) = integral wrt x from -inf to inf of X_PDF(x) * Y_PDF(z-x) * sqrt( (dx/dx)^2 + (dy/dx)^2 ) * dx / sqrt(det( [ (dz/dx) (dz/dy) ] * [ (dz/dx) (dz/dy) ]^T ))

 

[Stephen Tashi]

Right, but z(t) isn't used anywhere. z = f(x,y) = x + y, does it need to be a function of t?

If z is a function of the two variables (x,y) then what do you mean by z' ?

 

[rabbed]

If z is a function of the two variables (x,y) then what do you mean by z' ?

depends on what variable it's derivated with respect to
z = x + y
dz/dx = 1 (z'(x) = 1)
dz/dy = 1 (z'(y) = 1)

 

[Stephen Tashi]

depends on what variable it's derivated with respect to
z = x + y
dz/dx = 1 (z'(x) = 1)
dz/dy = 1 (z'(y) = 1)

You should be using the notation for partial derivatives. (The Insight: https://www.physicsforums.com/insights/partial-differentiation-without-tears/ is relevant to the question "Is z a function of t ?")

 

[rabbed]

You should be using the notation for partial derivatives. (The Insight: https://www.physicsforums.com/insights/partial-differentiation-without-tears/ is relevant to the question "Is z a function of t ?")

Yep, I know the other notation is a bit flawed, but it has it's uses. (but tell me if my logic was wrong somewhere!)

So it this better?

By the way,
since (x(t), y(t)) = (t, z-t)
is
integral wrt t from -inf to inf of X_PDF(x(t)) * Y_PDF(y(t)) * sqrt( x'(t)^2 + y'(t)^2 ) * dt
equal to
integral wrt x from -inf to inf of X_PDF(x) * Y_PDF(z-x) * sqrt( (dx/dx)^2 + (dy/dx)^2 ) * dx?

In that case:
Z_PDF(z) = integral wrt x from -inf to inf of X_PDF(x) * Y_PDF(z-x) * sqrt( (dx/dx)^2 + (dy/dx)^2 ) * dx / sqrt(det( [ (dz/dx) (dz/dy) ] * [ (dz/dx) (dz/dy) ]^T ))

 

[Stephen Tashi]

So it this better?

The problem is that your notation can't be interpreted. For example, what function is "dy/dx" ?

A mathematical function has a domain and a co-domain. What is the domain of "dy/dx" and what is it's co-domain?

 

[rabbed]

Since z = x + y,
y = z - x
and
dy/dx = -1?

 

[Stephen Tashi]

As far as I can tell, you aren't presenting any logical arguments. You are conjecturing various formulas and asking for criticism of them. That's a permissible approach in the early stages of an investigation, but you should follow-up a conjecture by testing it with some simple examples instead of relying on my comments. it's ok to make conjectures by resorting to "magic" - such as writing down symbols like "dx/dx" without asking what they symbolize. But you should proceed to working specific examples that force you to make specific interpretations. (I'm about to get busy for a few days with the jobs of being executor of an estate, so I'm not going to have time to criticize a hundred different conjectures.) - In fact I just got a phone call and I must leave right now.

 

[rabbed]

I think the essence here is that for multiple variables/dimensions we need to use line/surface integrals
and that it starts to make more sense to use the joint PDF of the source RV's.
XY_PDF(x,y)*sqrt(dx^2 + dy^2) = X_PDF(x)*|dx| * Y_PDF(y)*sqrt(1 + (dy/dx)^2)

Hm, maybe regular multidimensional integration comes in when considering inequalities, like Z < X+Y.

I'll keep exploring. Thanks for all help!