I'm looking for a continuous distribution which I can parameterize such that

  1. The expected value is roughly zero
  2. The expected maximum given $x$ draws from that distribution is only very weakly increasing in $x$
  3. $Prob(\max $ of $ x $ draws $ > 0)$ is small for "large" $x$, say around $x\in (30, 100)$
  4. (nice to have): The expected maximum of $x$ variables drawn from the distribution has a nice closed-form solution
  5. (nice to have): The distribution has not too many parameters and is not a very peculiar one that is only known to probability fetishists.

Phenomenon I am trying to capture a phenomenon which I can here describe as a rigged lottery with continuous outcomes. Most lottery tickets are of (varying) negative value. However, as people take part in the lottery, the expected value must be non-negative. Second, there are lottery insiders which, instead of the unconditional value, use insider information to attain the maximum of their draws. These are observed to purchase many tickets. Properties 2 and 3 control that these insiders indeed purchase many tickets (and the value that they expect from that).

Now I try to back out the lottery distribution consistent with this behavior.

I was starting off with Frechet (also because it has convenient properties regarding the maximum), but it doesn't allow enough freedom. I manage to calibrate it such that 1. and 2. hold, but then 3. breaks very quickly.

I looked in the related distributions but couldn't find anything. In general, how does one find (search) distributions when having so specific demands? In specific here, is there any?

  • 3
    $\begingroup$ These demands actually aren't very specific at all: they scarcely narrow your options. There are too many possible solutions. It's more effective to start with what you know about the phenomenon you are studying, then choose a family of distributions that is rich enough to capture the likely behavior you will encounter but otherwise is as simple as possible. Some additional characteristics to consider are whether you need to vary the scale or shape; what both tails will behave like; and what degree of asymmetry you want to permit. $\endgroup$ – whuber Apr 29 '16 at 16:53
  • $\begingroup$ Re the new criterion (4): that's always a wise thing to ask for! Since the CDF of the maximum of $n$ iid values drawn from a distribution $F$ is $F^n$, and its derivative is $nF^{n-1}F^\prime$, you will need a distribution for which $F^n$ has a "nice closed form" for all $n$, suggesting you rewrite $$F(x)=e^{G(x)}.$$ Now we're back to where we started ... . If you can't speak to the left tail, the need for scaling, symmetry, or other useful characteristics, maybe you could say something about the phenomenon you are trying to model? $\endgroup$ – whuber Apr 29 '16 at 17:28
  • 5
    $\begingroup$ I was going to suggest a constant value of zero. $\endgroup$ – eric_kernfeld Apr 29 '16 at 17:32
  • $\begingroup$ @whuber So I need a distribution with a fat tail below zero because of 3.. However, this should be counter-effected by some mass with large and positive values, in order to get the expected value right (1.). The derivative of the pdf at and above 0 is then somewhat constrained by 2. $\endgroup$ – FooBar Apr 29 '16 at 17:49
  • $\begingroup$ You haven't yet said whether these distributions should even be continuous! Things like this, which may be obvious to you because you know what you're trying to model, become significant issues for anyone trying to give a thorough, well-reasoned answer. That's why it can be such an advantage to describe your real problem. $\endgroup$ – whuber Apr 29 '16 at 18:36

How about $y \tilde{} \mathscr{N}(-a, fa)$ for small $a>0$ and $f>1$?

  1. E$[y]$ = $-a$.
  2. E[max of $x$ draws] = $-a + 2fa\sqrt{\ln x}$ = $a[2f\sqrt{\ln x}-1]$.
  3. Probability of maximum > 0 is similarly increasing in x, and controlled by $f$. It is something like $f_Z(z) = x F_Y(y)^{x-1}f_Y(y)$ where Y is your variable ($\mathscr{N}(-a, fa)$), $Z$ is the max of $x$ draws, $F$ denotes cdf, and $f$ denotes pdf. You can work it out.
  4. Property 2 has a closed form solution.
  • $\begingroup$ The tentative nature of this answer--which in fact satisfies the first three criteria--nicely emphasizes how overly broad the question still is. $\endgroup$ – whuber Apr 29 '16 at 17:36
  • $\begingroup$ Re 4: You do not have a closed form solution. That expression is only an approximation. $\endgroup$ – whuber Apr 29 '16 at 18:34

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