• We’re currently investigating an issue related to the forum theme and styling that is impacting page layout and visual formatting. The problem has been identified, and we are actively working on a resolution. There is no impact to user data or functionality, this is strictly a front-end display issue. We’ll post an update once the fix has been deployed. Thanks for your patience while we get this sorted.

Easy elementary school level math problem.

S

StatsManD

Member
LOL I said the whole question wrong. I tried to do it from memory. I am typing up the correct question now.

The real question is much harder:

Let h(x,x') = Integral(p(x|y)*p(y|x')dy.

Let p0(x) be a random approximation of p(x). (p0(x) can be any function)
Let pi+1(x) be integral(h(x,x')*pi(x')dx'

x' is a dummy argument.

Provided the joint density p(x,y) over the domain X,Y.

Prove that sequence {pi(x)} converges monotonically to p(x).


🙂

So maybe it isn't grade school level math.

So maybe I am lying about it being for my cousin too.
 
Originally posted by: StatsManD
No this isn't elementary algebra. I lied about this. This is high level college math. Maybe even graduate level.
Click to expand...

This post is very similiiar to some other easy maths i saw yesterday... verrrrry suspicious me thinks.
 
Not only are you not capable of doing the problem, but you are too much of a coward to be truthful about asking for help.
 
It seems like you are leaving out a few crucial details, but I'll give you my thoughts:

You have an iterative integral operator, so my first inclination was to somehow use the contraction mapping theorem to get existence of a limit. I am guessing these are probablilities so the kernal h(x,x') will always be less than one. (Using also that you are integrating over a probability space.) Using this fact, you can show (Assuming your space of functions you are considering satisfies the requirements of the contraction mapping theorem.) that the mapping pi to pi+1 is a contraction and hence converges.

To show it converges to p, show that it satisifes the integral equation. p(x)=int(h(x,x')p(x')dx' By the contraction mapping theorem, this limit is unique so it's the only solution.

Finally, when you say converges monotonically, I will assume pointwise monotonic convergence. For this it seems easy to show that pi+1(x)<=pi(x) for all x, again since the supremum of h(x,x') is strictly less than one.

Hope this helps.
 
Originally posted by: Unitary
It seems like you are leaving out a few crucial details, but I'll give you my thoughts:

You have an iterative integral operator, so my first inclination was to somehow use the contraction mapping theorem to get existence of a limit. I am guessing these are probablilities so the kernal h(x,x') will always be less than one. (Using also that you are integrating over a probability space.) Using this fact, you can show (Assuming your space of functions you are considering satisfies the requirements of the contraction mapping theorem.) that the mapping pi to pi+1 is a contraction and hence converges.

To show it converges to p, show that it satisifes the integral equation. p(x)=int(h(x,x')p(x')dx' By the contraction mapping theorem, this limit is unique so it's the only solution.

Finally, when you say converges monotonically, I will assume pointwise monotonic convergence. For this it seems easy to show that pi+1(x)<=pi(x) for all x, again since the supremum of h(x,x') is strictly less than one.

Hope this helps.
Click to expand...

Best first post ever?
 
Originally posted by: SearchMaster
Originally posted by: Unitary
It seems like you are leaving out a few crucial details, but I'll give you my thoughts:

You have an iterative integral operator, so my first inclination was to somehow use the contraction mapping theorem to get existence of a limit. I am guessing these are probablilities so the kernal h(x,x') will always be less than one. (Using also that you are integrating over a probability space.) Using this fact, you can show (Assuming your space of functions you are considering satisfies the requirements of the contraction mapping theorem.) that the mapping pi to pi+1 is a contraction and hence converges.

To show it converges to p, show that it satisifes the integral equation. p(x)=int(h(x,x')p(x')dx' By the contraction mapping theorem, this limit is unique so it's the only solution.

Finally, when you say converges monotonically, I will assume pointwise monotonic convergence. For this it seems easy to show that pi+1(x)<=pi(x) for all x, again since the supremum of h(x,x') is strictly less than one.

Hope this helps.
Click to expand...

Best first post ever?
Click to expand...

If he was in the Highly Technicial forum. In OT, you deserve to get flamed for stuff like this, which is what most people are approppriately doing. 😀
 
Yea, I didn't write up the whole problem. The whole problem is a more than a page long, full of equations. I just typed up what I thought was necessary.
 
Originally posted by: SearchMaster
Originally posted by: Unitary
It seems like you are leaving out a few crucial details, but I'll give you my thoughts:

You have an iterative integral operator, so my first inclination was to somehow use the contraction mapping theorem to get existence of a limit. I am guessing these are probablilities so the kernal h(x,x') will always be less than one. (Using also that you are integrating over a probability space.) Using this fact, you can show (Assuming your space of functions you are considering satisfies the requirements of the contraction mapping theorem.) that the mapping pi to pi+1 is a contraction and hence converges.

To show it converges to p, show that it satisifes the integral equation. p(x)=int(h(x,x')p(x')dx' By the contraction mapping theorem, this limit is unique so it's the only solution.

Finally, when you say converges monotonically, I will assume pointwise monotonic convergence. For this it seems easy to show that pi+1(x)<=pi(x) for all x, again since the supremum of h(x,x') is strictly less than one.

Hope this helps.
Click to expand...

Best first post ever?
Click to expand...

Unitary for elite
 
You are quickly headed down the path of bannage I can tell, but I've learned to appreciate you for your entertainment value. I gaurentee you'll be loved and adored here.
 
You must log in or register to reply here.

TRENDING THREADS

Back
Top