A Calculating the statistical properties of the given PDF

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To calculate the characteristic function, mean, and variance from a given probability density function (PDF) p(x), standard formulas can be applied, such as the characteristic function φ(t) = ∫ e^(itx)f(x)dx and moments m_k = ∫ x^kf(x)dx. A uniform distribution example was discussed, where p(x) = 1/2b, leading to confusion about the correct form of the characteristic function. The conversation highlighted the importance of understanding the relationship between the characteristic function and its implications for calculating moments and variance. Ultimately, the discussion emphasized the need for clarity in notation and the application of Fourier transform properties in statistical calculations.
Arman777
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For instance if we are given only a PDF in the form of ##p(x)##, how can one calculate the characteristic function, the mean, and the variance of these PDF's ?

Any site or explanation will be enough for me
 
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There are standard formulas given pdf ##f(x)##. Char. fcn. ##\phi(t)=\int_R e^{itx}f(x)dx##, moments ##m_k=\int_R x^kf(x)dx##, variance ##=m_2-m_1^2##.
 
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Arman777 said:
Any site or explanation
A textbook on statistics ?
 
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Okay I understand it. Thanks for the help
 
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Okay It seems that I did not or at least I thought I was. Let me take a uniform distribution in the form of ##p(x) = 1/2b##. The characteristic function is given by

$$p(k) = \frac{ie^{-ikx}}{2ak}$$

From here I want to calculate the mean and the variance as I have said before. I want to use this equation

1618137328629.png


so I get

$$ln(\frac{ie^{-ikx}}{2ak}) = ik<x>_c - \frac{k<x^2>_c}{2}$$

$$ln(\frac{i}{2ak})-ikx = ik<x>_c - \frac{k<x^2>_c}{2}$$

but from here I don't know what to do..
 
Arman777 said:
Okay It seems that I did not or at least I thought I was.
...
as I have said before.
Sorry I missed that :wink: -- can't make much sense of the first one and don't believe the second

Interesting thing, this FT aspect of a pdf. Not much in ordinary textbooks, I grant you.I will switch to 'shut up and learn mode' after these comments:
Arman777 said:
The characteristic function is given by
$$p(k) = \frac{ie^{-ikx}}{2ak}$$
This can't be right:
##p(x) = 1/2b## suggests a uniform distribution from ##-b## to ##+b##
So I would expect (using ##\phi(t)##, not ##p(k)## which is confusing)
mathman said:
$$\phi(t)=\int_R e^{itx}f(x)dx$$
something that depends on ##k## and ##b##, but not on ##x## ! (Lazy me: ##\displaystyle{\sin bt\over bt} ##, which I now 'all of a sudden' recognize and remember :cool: -- from the FT world)

By the same token the ##\int x^n p(x)## goodies you want to derive from ##\ \phi(k)\ ## should be convolutions in the ##t## domain (right ?)

##\ ##
 
BvU said:
Not much in ordinary textbooks, I grant you.
Our textbook is really hard to understand since its not for starters..
BvU said:
p(x)=1/2b suggests a uniform distribution from −b to
Yes my mistake, sorry about that. But in our book the notation is p(k) so I cannot use ##\phi(t)##
 
BvU said:
By the same token the ∫xnp(x) goodies you want to derive from ϕ(k) should be convolutions in the t domain (right ?)
I don't know what this means
 
A property of Fourier transforms is that the transform of a product is a convolution vice versa.

##\ ##
 
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Okay, this time I really solved the problem. Thanks for the help.
 
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