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Math help, simple algebra problem (kind of)

D

DVK916

Banned
True of False:

If the laplace transformation of a density function does not exsist, then its moments do not exsist.

If true, prove it.
If false, give a counter example.


ok, so it isn't algebra.
 
If i remember correctly....the moment is the integral of the density function times the nth moment. So it would be true, it won't exist.... (you can construct the density function from the moments)

take this with a grain if salt b/c probabillty is not my strength.
 
What the heck do they mean by a moment; I've just spent the last 9 months having it shoved in my face that it's the torque.
 
Originally posted by: DVK916
True of False:

If the laplace transformation of a density function does not exsist, then its moments do not exsist.

If true, prove it.
If false, give a counter example.


ok, so it isn't algebra.
Click to expand...

Too bad those double chins don't increase your spelling skills.
 
Originally posted by: TruePaige
..Yes...Yes...nobody taught me that in Algebra.

At least not while I was watching. 0.o
Click to expand...

Its squeezed in between factoring and compleating the square.
 
Originally posted by: DVK916
I am almost positive it is true, but I can't figure out how to prove it.
Click to expand...

I think to do this is that you have to prove that if you have a valid correlation function, you have a valid moment. When you take the fourier of the correlation (or is it auto correlation???) you get the Power Spectral density function. Since Fourier is a one to one function, it implies that the function it intergrated be a 1 to 1 function. If you try to take the fourier of an invalid correlation function (which in turn means it has a invalid density function....remember that the correlation involves the density function) you will get an unreal and negative transform. If you have an invalid desnity function, you cannot have a valid moment.

I think that is how it worked, but we did that stuff with fourier, not laplace (but the only difference is the integration limits anyways)....man I hated that class....
 
Will you ever learn to do your own homework, or do you plan to have random internet people do your work for you throughout your life?
 
Originally posted by: zinfamous
Will you ever learn to do your own homework, or do you plan to have random internet people do your work for you throughout your life?
Click to expand...


Do your own damn homework! :|
 
my pathetic answer which I jsut made up whilst downing some Mountain due:

The laplace transformation will not change the problem but simply provide you with a different domain for analysis.

If the laplace transform does not exist, the function, in it's original domain, will yeild only a probabilty for your variable of 0 and nothing more, therefore the function will have no deviations from 0, and therefore there will be no moments but 0.

 
Originally posted by: Goosemaster
my pathetic answer which I jsut made up whilst downing some Mountain due:

The laplace transformation will not change the problem but simply provide you with a different domain for analysis.

If the laplace transform does not exist, the function, in it's original domain, will yeild only a probabilty for your variable of 0 and nothing more, therefore the function will have no deviations from 0, and therefore there will be no moments but 0.
Click to expand...

I call bull**** :laugh:
 
Originally posted by: Goosemaster
Originally posted by: Goosemaster
my pathetic answer which I jsut made up whilst downing some Mountain due:

The laplace transformation will not change the problem but simply provide you with a different domain for analysis.

If the laplace transform does not exist, the function, in it's original domain, will yeild only a probabilty for your variable of 0 and nothing more, therefore the function will have no deviations from 0, and therefore there will be no moments but 0.
Click to expand...

I call bull**** :laugh:
Click to expand...

sounded like good bull😉

 
Originally posted by: Gibson486
Originally posted by: Goosemaster
Originally posted by: Goosemaster
my pathetic answer which I jsut made up whilst downing some Mountain due:

The laplace transformation will not change the problem but simply provide you with a different domain for analysis.

If the laplace transform does not exist, the function, in it's original domain, will yeild only a probabilty for your variable of 0 and nothing more, therefore the function will have no deviations from 0, and therefore there will be no moments but 0.
Click to expand...

I call bull**** :laugh:
Click to expand...

sounded like good bull😉
Click to expand...

my knowledge is limited..I jsut tried my best to formualte an answer with what little I know🙁

 
True. If the Laplace transform exists, you can calculate the moments of the pdf as the derivatives.

Let X be a nonnegative random variable and F(s) be the Laplace transform of the pdf. Then

F'(s)=d/ds E[e^(-sX)]=E[-Xe^(-sX)]
...
F^(n)(s)=d^n/ds^n E[e^(-sX)]=E[(-X)^n e^(-sX)]

Evaluating at s=0 yields

E[X]=-F'(0)
E[X^2]=+F''(0)
...
E[X^n]=(-1)^n F^(n)(0)

Edit: This isn't an algebra problem...
 
Originally posted by: randumb
True. If the Laplace transform exists, you can calculate the moments of the pdf function as the derivatives.

Let X be a nonnegative random variable and F(s) be the Laplace transform of the pdf function. Then

F'(s)=d/ds E[e^(-sX)]=E[-Xe^(-sX)]
...
F^(n)(s)=d^n/ds^n E[e^(-sX)]=E[(-X)^n e^(-sX)]

Evaluating at s=0 yields

E[X]=-F'(0)
E[X^2]=+F''(0)
...
E[X^n]=(-1)^n F^(n)(0)
Click to expand...

Why would you feed that troll? DVK916's got more chins than # of brain lobes.
 
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