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The variance of a binomial distribution is not adjustable. If you know its mean and the number of trials, the variance is also determined. I need a distribution that has an additional degree of freedom, where the variance and the mean are independent.

In other words, I need a discrete probability distribution with the following properties:

  1. Its support is the integers $0, 1, 2, \dots, N$, where $N$ is a non-negative integer (the number of trials, in analogy to the binomial distribution).
  2. The distribution is symmetric around $N/2$, which is also the mean.
  3. The variance is a free parameter that I can adjust (within a feasible range, see comments).
  4. For suitable parameters, the variance should go to zero if $N$ is even, or to $1/4$ if $N$ is odd (the variance of drawing one of $(N-1)/2$ or $(N+1)/2$ at random).

The binomial distribution with $p=1/2$ fulfills 1. and 2., but fails at 3 and 4, because the variance is a function of $p$ and $N$. Is there any other discrete probability distribution that meets these requirements?

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    $\begingroup$ The binomial distribution doesn't satisfy 2 unless p=1/2. The terminology "variance and mean are independent" makes no sense. I think you mean that that given that the mean is specified (say N/2) the variance is a single parameter that is not a function of the mean of the distribution. $\endgroup$ – Michael R. Chernick Oct 8 '17 at 22:22
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    $\begingroup$ It must be the case that you are looking for a distribution that doesn't involve independent trials with a constant p because then the binomial is the only possibility. $\endgroup$ – Michael R. Chernick Oct 8 '17 at 22:29
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    $\begingroup$ @MichaelChernick In the binomial, the variance is determined from the mean in the sense that given the mean $=pN$, you can deduce $p$ and from there the variance. By "independent" I meant precisely what you said. $\endgroup$ – becko Oct 8 '17 at 22:58
  • $\begingroup$ @MichaelChernick Of course, the discrete distribution I am looking for doesn't even have to be related to trials of any sort. $\endgroup$ – becko Oct 8 '17 at 22:59
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    $\begingroup$ I don't think there can be such a distribution without more constraints, as the support + the symmetry implies a strict maximum on the variance that is a function of the mean, namely, $N^2/4$ (with the distribution having $p(0) = p(N) = 1/2$, $p(\text{anything else}) = 0$. $\endgroup$ – jbowman Oct 8 '17 at 23:43
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You can probably just use a compound binomial distribution where $p$ follows some other symmetric distribution with mean = 0.5. The beta-binomial distribution comes to mind. You just have to make sure your $\alpha$ and $\beta$ are equal so $p$ has reflection symmetry about the mean. You can increase and decrease them together to adjust the variance and leave the mean unchanged, as you wish according to the compound variance formula.

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  • $\begingroup$ Oh, almost. I realize I need the variance to go to zero for a suitable choice of parameters. I have added this requirement to the question. The beta-binomial does not satisfy it. $\endgroup$ – becko Oct 11 '17 at 14:52
  • $\begingroup$ I am sorry for changing the question. $\endgroup$ – becko Oct 11 '17 at 15:15

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