Math Help Please

[DVK916]

Given two random variables X and Y, let Y have a non probabilistic measure with infinite cumulative density, and let X|Y have a probability measure with cumulative density of 1. Prove that X must also have a non probabilistic measure with infinite cumulative density.

Essentially prove this.

Given Integral(fy(y)dy)=infiniti
Integral(fx|y(x|y)dx)=1

Then

Integral(fx(x)dx)= infiniti.

 

[imported_Tick]

Um.... can you write it out with better symbols? How bout write it in Scientific Notebook or something and upload the screen shot to pics.bbzzdd?

 

[DVK916]

http://tinypic.com/4bokyrp.jpg

Here is a better representation of the question.

 

[hypn0tik]

Use the definition of a conditional pdf and work form there.

 

[hypn0tik]

It would help if you knew a thing or two about joint pdf's...

 

[DVK916]

I tried, I can't get it to work out.

 

[BoomerD]

huh? Whatdhe say?

/me is pretty math challenged...I do ok with basic elementary algebra...sometimes, but after that, it might as well be Chinese...School is gonna be fun the next couple of semesters...Math in each one...

 

[hypn0tik]

Originally posted by: DVK916
I tried, I can't get it to work out.

Show me your attempt

 

[Jeff7]

I find your definition of "low level math" disturbing.
:Q

I've had Calc I and II, and this is like nothing I've ever seen before.

 

[hypn0tik]

Originally posted by: Jeff7
I find your definition of "low level math" disturbing.
:Q

I've had Calc I and II, and this is like nothing I've ever seen before.

OP is just trying to show off. Don't be alarmed. Probability and random variables aren't trivial as demonstrated by the OP.

 

[potoba]

yeah, this is one of those probability concepts I remember using it in my quantum class. We called it dirac integral or sth...

 

[DVK916]

This is what I have

Integral(fx(x)dx)=integral(integral(f(x|y)*f(y)dy)dx)

Now this integral isn't seperatable since f(x|y) is a function of both x and y.

So I can't just say it equals 1*infiniti which is wrong.

 

[DVK916]

This was on my Final on my Advanced Bayesian Probability course. It is a high level course with Calculus probability theory as a prequirement.

 

[DVK916]

I also want the proof for the discrete case. I have trying to do this problem since our final when I couldn't do it. Got a B in the class though. Still sucks.

 

[imported_Tick]

Originally posted by: DVK916
This was on my Final on my Advanced Bayesian Probability course. It is a high level course with Calculus probability theory as a prequirement.

SO WHY DID YOU CALL IT BASIC? If I can't figure it out, and I know Diff EQ. and 3 semesters of Calc, IT'S NOT BASIC!!

 

[DVK916]

Originally posted by: Tick

Originally posted by: DVK916
This was on my Final on my Advanced Bayesian Probability course. It is a high level course with Calculus probability theory as a prequirement.

SO WHY DID YOU CALL IT BASIC? If I can't figure it out, and I know Diff EQ. and 3 semesters of Calc, IT'S NOT BASIC!!

Well I thought by calling it basic would get more people to look at it, calling it advance might scare people away.

 

[hypn0tik]

Integral(f_Y(y) dy) = infinity from c to d
Integral(f_X(x|Y=y) dx) = 1 from a to b

You know that f_X(x|Y=y) = f_XY(x,y)/f_Y(y))

Substitute above

Integral(f_X(x|Y=y) dx) = 1 --> Integral(f_XY(x,y)/f_Y(y)) dx) = 1 [with limits a to b]
Since f_Y(y) is independent of x, you can take it out of the integral and put it on the other side.

Integral(f_XY(x,y) dx) = f_Y(y) [with limits a to b]

Now, take the integral with respect to y of both sides [with limits c to d]

See picture here

 

[DVK916]

OMG, it is so simple, stupid me couldn't figure it out.

 

[DVK916]

Oh and thanks

 

[hypn0tik]

No problem

 

[Kromis]

Originally posted by: Jeff7
I find your definition of "low level math" disturbing.
:Q

I've had Calc I and II, and this is like nothing I've ever seen before.