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Mathematics | Reimann Zeta Distribution Model

  • Last Updated : 19 Feb, 2021

Introduction :

Suppose an event can occur several times within a given unit of time. When the total number of occurrences of the event is unknown, we can think of it as a random variable. When a random variable X takes on values on discrete time interval from 1 to infinity, one choice of a probability density is the Reimann Zeta distribution whose probability density function is given by as follows.

f(x) = \frac{1}{\zeta(\alpha+1)}x^{-(\alpha+1)}

Above expression will be applicable only when given below condition will follow.

 x = 1,2,3,....    . 
 f(x) = 0, Otherwise

Where, \alpha     is the parameter and \zeta(\alpha+1)     is the value of the zeta function, defined by as follows.

\zeta(y) = 1+(\frac{1}{2})^y+ (\frac{1}{3})^y+(\frac{1}{4})^y + ...... +  (\frac{1}{n})^y = \Sigma_{k=1}^{\infty} k^{-y}

The random variable X following Reimann Zeta Distribution is represented as follows.

X ~ RIE(\alpha    )

Expected Value :

The Expected Value of the Reimann Zeta distribution can be found by summing up products of Values with their respective probabilities as follows.

\mu = E(X) = \Sigma_{x=-\infty}^{\infty} x.f(x)
\mu = \Sigma_{x=1}^{\infty} x.\frac{1}{\zeta(\alpha+1)}.x^{-(\alpha+1)}
\mu = \frac{1}{\zeta(\alpha+1)}\Sigma_{x=1}^{\infty} x^{-\alpha}

Using the property \zeta(y) = \Sigma_{k=1}^{\infty} k^{-y}    , we get the following expression as follows.

\mu = \frac{\zeta(\alpha)}{\zeta(\alpha+1)}

Variance and Standard Deviation : 

The Variance of the Riemann Zeta distribution can be found using the Variance Formula as follows.

σ^2 = E( X − μ )^2 = E( X^2 ) − μ^2
E(X^2) = \Sigma^{\infty}_{x=-\infty} x^2.f(x) 
E(X^2) = \frac{1}{\zeta(\alpha+1)} \Sigma_{x=1}^{\infty} x^2.x^{-(\alpha+1)}
E(X^2) = \frac{1}{\zeta(\alpha+1)} \Sigma_{x=1}^{\infty} x^{1-\alpha}

Using the property \zeta(y) = \Sigma_{k=1}^{\infty} k^{-y}    , we get the following expression as follows.

E(X^2) = \frac{\zeta(\alpha-1)}{\zeta(\alpha+1)}
So, Var(X) = E(X^2) - \mu^2
Var (X) = (\frac{\zeta(\alpha-1)}{\zeta(\alpha+1)})^2 - (\frac{\zeta(\alpha)}{\zeta(\alpha+1)})^2
Var(X) = \sigma^2 = \frac{[\zeta(\alpha-1)]^2 - [\zeta(\alpha)]^2}{[\zeta(\alpha+1)]^2}

Standard Deviation is given by as follows.

\sigma = \frac{1}{\zeta(\alpha+1)} \sqrt{[\zeta(\alpha-1)]^2 - [\zeta(\alpha)]^2}

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