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A discrete probability distribution has the following property

$$ P(X \ge x) = {x^{- E(x)}} $$

where $x$ is a positive natural number and $E(X)$ is the finite expected value of the distribution. This is a power law distribution where the exponent turns out to be the expected value of the distribution.

Question: I am trying to find is there if an interpretation of this distribution i.e. what does it signify when the exponent happens to be the expected value. Is there anything that is significant or non trivial about such a distribution.

Edit: Glen has given a numerical answer which is correct. Actually prior to posting this question, I had done the same calculation using the zeta function and obtained the same unique answer. However, what I was looking for in my question is the interpretation of this distribution, may be something like a physical interpretation if at all it exits. To see where I am coming form, I observed this distribution in economic data. So while zeta function gives a numerical fit, I wanted to see if I can explain the distribution in the layman (economic) sense.

Note: I have asked the same question at MSE

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    $\begingroup$ With a discrete distribution on positive integers you have an example of unbounded Zipf's law with shape parameter $1.83377265$. If instead this had been a continuous distribution restricted to values of $1$ or more, you would have an example of a Pareto distribution with shape parameter $\phi \approx 1.6180339887$. Both are power law distributions, and that is probably the closest you get to an interpretation unless you are a golden ratio fan $\endgroup$
    – Henry
    Oct 2, 2016 at 17:14
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    $\begingroup$ @Henry with the Pareto distribution with scale-parameter equal to 1 $$P(X \geq x) = x^{-\alpha}$$ we have $E(X) = \frac{\alpha}{\alpha-1}$ which leads to $\alpha=2$ $\endgroup$ Feb 14, 2020 at 16:45
  • $\begingroup$ @SextusEmpiricus - you seem to be correct, though it would still be a power law distribution $\endgroup$
    – Henry
    Feb 14, 2020 at 19:14

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Consider the expression

$$P(X \ge k) = {k^{- \beta}},\;\; \beta>0,\;\; k \in \{1,2,3,...\}$$

We have

$$P(X < k) = 1- k^{- \beta}$$

Then

$$P(X=1) = P(X<2) = 1- 2^{- \beta}$$ $$P(X=2) = P(X<3) - P(X=1)= 1- 3^{- \beta} - 1 + 2^{- \beta} = 2^{- \beta} - 3^{- \beta} $$

etc, which gives us the probability mass function as

$$P(X=k) = k^{- \beta} - (k+1)^{- \beta}$$ and the distribution function

$$P(X\leq k) = 1- (k+1)^{- \beta}$$

This is a proper distribution "family" for all permissible values of $\beta$. Its expected value is

$$E(X)= \sum_{k=1}^{\infty} k\cdot \big(k^{- \beta} - (k+1)^{- \beta}\big)$$

and one can infer that it is decreasing in $\beta$. Graphs of the pmf for three distinct values of $\beta$ and for values of the variable up to $10$ are

enter image description here

As $\beta$ increases, probability mass concentrates closer to the origin, and so the expected value gets lower. We see that all pmf's follow a decaying path reminiscent of the geometric distribution (which is the discrete counterpart of the Exponential distribution), but our generic distribution falls faster than the Geometric (not in the picture).

The last $\beta$ value is the one that makes ~ $E(X) = \beta$ and as shown elsewhere it satisfies $E(X) = \zeta[E(X)]$ , the zeta function. The value $\zeta(3/2)$ is used in physics, and the value $\zeta (2)$ in the Basel problem. We are somewhere in between. Graphically, the case $E(X) = \beta$ does not appear to posses any special characteristic.

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