Math help, simple algebra problem (kind of)

[Goosemaster]

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)

mother of god....

 

[randumb]

Originally posted by: Parasitic

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)

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

So that he hopefully stops posting threads about statistics every day.

 

[Goosemaster]

Originally posted by: randumb

Originally posted by: Parasitic

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)

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

So that he hopefully stops posting threads about statistics every day.

So I'm guessing I was REALLY REALLY off🙁

 

[Synomenon]

Yes, DVK please stop posting all of your homework online everyday.

 

[zinfamous]

Originally posted by: randumb

Originally posted by: Parasitic

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)

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

So that he hopefully stops posting threads about statistics every day.

Right...it only encourages him. He clearly hasn't learned a damn thing form his classes, b/c he simply tries to get ATOT to do all of his work for him. If you keep helping him out, he'll keep asking. Just ignore the BS, whether or not you want to flex your stats pene and post your knowledge 😉

 

[jman19]

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.

Do your own homework you tool.

 

[StormRider]

Originally posted by: randumb
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...

Unfortunately, this isn't what the OP was asking to prove. Logically, he was trying to prove:

If ~P then ~Q

What you just proved (I didn't check if it was correct) was

If P then Q

The two logic statements are not equivalent.

However, the following statement is equivalent to the first statement

If Q then P

 

[AnyMal]

Originally posted by: zinfamous

Originally posted by: randumb

Originally posted by: Parasitic

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)

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

So that he hopefully stops posting threads about statistics every day.

Right...it only encourages him. He clearly hasn't learned a damn thing form his classes, b/c he simply tries to get ATOT to do all of his work for him. If you keep helping him out, he'll keep asking. Just ignore the BS, whether or not you want to flex your stats pene and post your knowledge 😉

Where's LoKe when you need him? Oh wait......