Why we are using the (characteristics function) CHF for evaluating the (probability density function) PDF of any random variable, why not directly evaluate PDF for random variable..
Thanks @George Stoica ,
but if i have random variable x, so can compute PDF directly. but are the advantage by using the CHF, canyou describe in detail...
Dear Atul,
One of the known applications of the characteristic function is the determination of the relationships between the statistical moments for one or more random variables.
It is not necessary to use CHF . For example if you know the distribution function (DF) then its derivative is the PDF.
In rider to what they said above, having random variable only will not be enough to compute the pdf unless if you know how it distributed, in that case you can use that information to derive what you wants
Sincerely Auwal Hamisu
Dear Kumar,
If a random variable admits a density function, then the characteristic function is its dual, in the sense that each of them is a Fourier transform of the other. Thus it provides the basis of an alternative route to analytical results compared with working directly with probability density functions or cumulative distribution functions.
Note however that the characteristic function of a distribution always exists, even when the probability density function or moment-generating function do not.
Sincerely,
Elsawah
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