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Suppose I have given a finite number of grid values belonging to a probability density function. Moreover, I have the associated values of the density support. For instance:

support density value
0.06    0.07
-0.04   0.11
-0.02   0.52
0.00    1.56
0.02    7.87
0.04    19.18
0.06    13.66
0.08    3.40
0.10    0.98
0.13    0.33
0.15    0.14
0.17    0.07
0.19    0.00
0.22    0.43
0.24    0.01

Does anyone know the formula to calculate the first four moments of the distribution?

I would appreciate any help. Many thanks in advance!

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    $\begingroup$ You could interpolate the density and then numerically integrate this function? You know, $E[X]=\int_\mathbb{R} xf_X(x)dx$ etc... $\endgroup$
    – Alex
    Commented Jul 4, 2020 at 13:15

2 Answers 2

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The key is:

$$ \mathbf{E}[X^k] = \sum_{i=1}^n x_i^k p(x_i) $$

($X$ discrete variable, $x_i$ realizations, and $p(x_i)$ realization probabilities)

See this link for further details.

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  • $\begingroup$ Great, this helps a lot! thanks! $\endgroup$
    – Walter
    Commented Jul 4, 2020 at 13:27
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    $\begingroup$ @Walter Glad to help. Unless you are waiting for other answers, please consider marking this answer as accepted (marking procedure available here: stackoverflow.com/help/someone-answers) $\endgroup$
    – ir7
    Commented Jul 4, 2020 at 13:35
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    $\begingroup$ As the densities do not yet sum to one, please also consider proper normalisation, no? $\endgroup$ Commented Jul 5, 2020 at 6:29
  • $\begingroup$ @Kermittfrog Good point. If they don’t add up to one, one needs to normalize them. See, for example, math.stackexchange.com/questions/278418/… $\endgroup$
    – ir7
    Commented Jul 5, 2020 at 16:45
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Just to add, you did not mention which kind of momement. These calculated by formula in ir7 are called general moments. However, there are also:

  • Central moments defined as $E[X-EX]^k$
  • Standardized moments defined as $E\big[\frac{X-EX}{\sigma(X)}\big]^k$,

where $EX$ is first general moment (expected value) and $\sigma(X)$ is second general moment (standard deviation).

Note that for $k=1$ the central moment and standardized moment are always 0. For $k=2$, the standardized moment is always 1.

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