Probability Density Function in Terms of Moments by sparkunder20

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									             Probability Density Function in Terms of Moments
                                              Sahand Rabbani


Problem Statement
In this article, we attempt to express the probability density function f (x) of a random variable X in terms
of the moments E [X n ], n = {0, 1, 2, . . .}, of the distribution.


Solution
Consider the nth moment of the distribution f (x):
                                                         ∞
                                          E [X n ] =           xn f (x)dx.
                                                        −∞

Now consider the Taylor Series expansion of ex :
                                                            xn
                                                         ∞
                                                 ex =          .
                                                        n=0
                                                            n!

Thus, the expansion of e−j2πxs is
                                                       ∞                n
                                                           (−j2πxs)
                                         e−j2πxs =                  .
                                                       n=0
                                                               n!
Now, we consider the following infinite series:
                            (−j2πs)n                       (−j2πs)n
                         ∞                             ∞                     ∞
                                     E [X n ] =                                   xn f (x)dx
                        n=0
                               n!                      n=0
                                                              n!             −∞

                                                                     (−j2πxs)n
                                                         ∞       ∞
                                                 =                                   f (x) dx
                                                        −∞       n=0
                                                                         n!
                                                         ∞
                                                 =             e−j2πxs f (x) dx
                                                        −∞
                                                 = F (s),
where F (s) is the Fourier transform of f (x). We now have an expression for the Fourier transform of a
distribution in terms of its moments:
                                                               (−j2πs)n
                                                             ∞
                                F (s) = F{f (x)}(s) =                   E [X n ] .
                                                           n=0
                                                                  n!
We recover the probability density function by taking the inverse Fourier transform of this expression:
                                                     (−j2πs)n
                                                 ∞
                             f (x)   = F −1                   E [X n ] (x)
                                                 n=0
                                                        n!

                                                            (−j2πs)n
                                          ∞                ∞
                                     =        ej2πxs                 E [X n ] ds.
                                         −∞             n=0
                                                               n!


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S. Rabbani                                                         Probability Density Function in Terms of Moments


Example
To verify this result, we apply the formula to the standard normal distribution. The moments of the standard
normal distribution are given by
                                                    0           n odd
                                        E [X n ] =   −n/2 n!
                                                    2    (n/2)! n even.
Applying the formula derived above gives
                                                                                       
                                    ∞
                                                              (−j2πs)n             n!
                       f (x) =          ej2πxs                            2−n/2         ds
                                   −∞                 n>0
                                                                 n!              (n/2)!
                                                     n even

                                                         (−j2πs)2m 2−m
                                    ∞                 ∞
                              =         ej2πxs                                  ds
                                   −∞                m=0
                                                              m!

                                                         (−2π 2 s2 )m
                                    ∞                 ∞
                              =         ej2πxs                             ds
                                   −∞                m=0
                                                            m!
                                    ∞
                                                       2 2
                              =         ej2πxs e−2π       s
                                                              ds
                                   −∞
                                              2 2
                                                 s
                              = F −1 e−2π              (x)
                                   1   2
                              =   √ e−x /2 .
                                   2π
This is in fact the probability density function of the standard normal random variable.




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