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I would like to generate random numbers $X$'s from a desired distribution whose properties should meet the following requirements:

  1. $X \in [0, \infty) $

  2. The mean of the r.v. is around 1, i.e., $\mathbb{E}[X] \approx 1$

  3. The distribution shows "long tail". "Long tail" in the sense that satisfies the typical description: https://en.wikipedia.org/wiki/Long_tail

  4. To be more quantitative, let's say at least $P(X \gt 5) = 0.1$

  5. Although it's possible to combine multiple distributions to achieve the above goals, I was looking for a single distribution that can be written as a compact, closed-form probability density function.

In words, I was looking for a distribution whose mode or mean is around 1 and has fat tail that extends to large values. Can you suggest a distribution that satisfies these properties?

(It's possible that this is naive and no such distribution exists.)

The closest candidate came to my mind is $\chi^2$ distribution, which is controlled by the parameter $k$. However, it's either the mean is too high or the tail probability is too low. Below is an example of $\chi^2(k=3)$. Ideally I would like to move the mean to 1, and make the tail "fatter".

enter image description here

A use case would be to use this distribution as a random number generator, such that the mean of the generated numbers is around 1 while being able to generate large numbers.


Just wanted to point out that, although what @stans suggested to choose log-normal distribution with $\mu = -\sigma^2/2$ satisfies the requirement of $\mathbb{E}[X] = 1$, it doesn't create enough tail probability.

In fact, in order to satisfy the mean=1 condition, $\mu$ needs to be shifted to the very left so that the tail probability $P(X>5)$ gets squeezed smaller. Doing a grid search in the range $\sigma \in [1, 8]$, it seems that the largest tail probability happens around $\sigma=1.79$, at which $P(x>5) \approx 0.036$

enter image description here

Python code to generate log-normal distribution and the corresponding $P(X>5)$:

import numpy as np
import scipy.stats

sigma = 1
mu = -0.5 * sigma**2

s = sigma  
scale = np.exp(mu)

tail_prob = 1.0 - scipy.stats.lognorm(s=s, scale=scale).cdf(5)
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    $\begingroup$ You could scale your chi-square -- choose a smaller d.f. say 1, and then scale it up to get the mean right; equivalently, take a gamma density with rate=shape (which makes the mean 1), and you then vary the shape parameter to give whatever tail property you need (lower shape = heavier) $\endgroup$ – Glen_b -Reinstate Monica Jan 22 '18 at 22:49
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    $\begingroup$ Actually that won't work for the specific proportion you asked for, sorry; the biggest proportion above 5 you can get with a gamma density whose mean is 1 is about 0.0586; this happens when the shape is just above 0.1053. You can get a heavier tail (in the more usual senses) by making the shape smaller, but that will reduce the proportion above 5. $\endgroup$ – Glen_b -Reinstate Monica Jan 22 '18 at 23:35
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    $\begingroup$ This question is starting to look too open-ended. To see why, consider that the distribution that assigns probability $9/10$ to $0$ and probability $1/10$ to $10$ has a mean exactly $1$ and exactly satisfies your constraint--but otherwise looks almost nothing like your plots. Could you focus this question to include (a) what you really mean by "long-tail" or "fat tail" and (b) what additional criteria you are thinking of that would narrow the scope of possible answers to something reasonable? $\endgroup$ – whuber Jan 25 '18 at 21:30
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    $\begingroup$ @DJ Those don't seem relevant to the question, since neither has an expectation at all. How were you thinking of applying them? $\endgroup$ – whuber Jan 25 '18 at 21:56
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    $\begingroup$ Re the edit to clarify "long tailed": this implies you could take literally any long-tailed distribution $G$ with a finite expectation, scale and shift it as appropriate to put a tenth of its probability to the right of $5$ with a mean of $10$, and mix it with an atom at $0$ to produce what you want. In other words, the set of answers to your question is no smaller than the set of all finite-expectation long-tailed distributions! That's not enough information to recommend a procedure to generate random numbers. Perhaps you could tell us what you're hoping to model with $X$? $\endgroup$ – whuber Jan 26 '18 at 17:46
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Log-normal for the right choice of $\mu$ and $\sigma$. In other words, if $X$ ~ $\rm{LN}(\mu,\sigma^2)$ then

$ 1 = \rm{E}[X] = \exp\{\mu + \sigma^2/2\}\ \ \ \ <=>\ \ \ \mu = -\sigma^2/2. $

Parameter $\sigma$ means "tail fatness" and can be set arbitrarily high.

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  • $\begingroup$ @stans Yours can't work for the specific proportion requested for the same reason my gamma (in comments) didn't -- while you do make the tail heavier (in the more usual senses) by making $\sigma$ higher, you cannot achieve even $0.0364$ in the tail beyond $5$ with a mean-1 lognormal, let alone anything as big as $0.1$. The biggest tail proportion beyond $5$ occurs when $\sigma$ is just below $1.7941226$. In fact to get so large a proportion beyond $5$ you'll probably need a fairly light-tailed distribution. $\endgroup$ – Glen_b -Reinstate Monica Jan 22 '18 at 23:45
  • $\begingroup$ @Glen_b Thank you for the calculation. I guess, your and my suggestions can be made heavier in the tails by considering mixtures of respective distributions. Mixing over the scale parameter may push us far enough. $\endgroup$ – stans - Reinstate Monica Jan 23 '18 at 5:33
  • $\begingroup$ If the aim is to get a high proportion beyond a particular value, it may be better(as I suggest above) to look at lighter tailed distributions, not heavier-tailed. Mixtures will work either way (i.e. can generate both lighter and heavier tailed distributions). ... I think your answer is still the kind of thing the OP is after -- it seems like the specific proportion beyond 5 was somewhat arbitrary. $\endgroup$ – Glen_b -Reinstate Monica Jan 23 '18 at 5:54
  • $\begingroup$ @Glen_b thanks for your comments! I think what stans suggested did meet my goal. For example, a log-Normal with \mu=-2.0 and \sigma=2.0 with result in a distribution that has E(X)=1 and Prob(X > 5) = 0.17. $\endgroup$ – cwl Jan 23 '18 at 15:55
  • $\begingroup$ @cwl That's odd. I get about 0.0356 when I do it in R: > pnorm(log(5),-2,2,lower.tail=FALSE) $\:\qquad$ $\:$ [1] 0.03555934 ... or if we work directly with the lognormal: plnorm(5,-2,2,lower.tail=FALSE), which gives the same output. $\endgroup$ – Glen_b -Reinstate Monica Jan 23 '18 at 22:09
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How about a random variable which takes values 1/2p with probability p and 1/(2-2p) with probability 1-p? The mean is always 1. The min is 0, but the max is arbitrary.

If a condition is needed so that the range in unbounded, create a mixture where this variable is the shift for an Exponential 1 random variable.

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  • $\begingroup$ That's not a long-tailed distribution in any standard sense of the word. Objections to this solution have already been expressed in comments to the question. $\endgroup$ – whuber Jan 26 '18 at 18:47
  • $\begingroup$ @whuber It is a rather lengthy discussion, and nonetheless the numerous edits have not precluded this as a viable answer. I do note you previously posited this kind of distribution, perhaps as a strawman. I think we are in agreement that there is a lack of a rigorous definition here. $\endgroup$ – AdamO Jan 26 '18 at 19:00
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    $\begingroup$ The Wikipedia reference included with the latest edit, although it does not provide a technical definition, makes it reasonably clear that bounded distributions won't suffice. I have pointed out that despite the requirements of the question, the answers are still completely open-ended and I have (therefore) asked the OP to try to focus the question more. $\endgroup$ – whuber Jan 26 '18 at 19:02

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