Easy elementary school level math problem.

[StatsManD]

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.

 

[KLin]

How about you tell your "cousin" to do his own damn homework.

 

[her209]

Wow... really? This is Elementary Algebra...

 

[StatsManD]

No this isn't elementary algebra. I lied about this. This is high level college math. Maybe even graduate level.

 

[blahblah99]

That's 2nd grade math, you should be able to do that while you're sleeping.

 

[fishmonger12]

Originally posted by: StatsManD
No this isn't elementary algebra. I lied about this. This is high level college math. Maybe even graduate level.

This post is very similiiar to some other easy maths i saw yesterday... verrrrry suspicious me thinks.

 

[StatsManD]

Fixed the problem. I had to lie about it being easy, to get more people to try and do it.

 

[Leros]

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.

 

[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.

 

[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.

Best first post ever?

 

[Fullmetal Chocobo]

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.

Best first post ever?

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.

 

[George P Burdell]

Too elementary for me. Yawn.

 

[StatsManD]

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.

 

[snoopdoug1]

Originally posted by: George P Burdell
Too elementary for me. Yawn.

 

[slsmnaz]

Maybe coming here to do your homework is what caused this to happen??

 

[Pepsi90919]

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.

Best first post ever?

Unitary for elite

 

[MisterJackson]

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.