2
$\begingroup$

Suppose I know that for a discrete, non-negative r.v. $X$ that $X | X \geq 1$ has $\mu = 3.3$ while when $X \geq 0$ has $\mu = 2.1$. That is, the subset of the population that already has a value $1$ or more has a proportionally higher mean. I would like to assume that $X$ is more likely to evaluate to some value $x \leq 5$ so that the distributions of $X$ and $X | X \geq\ 1$ could roughly be modeled by a gamma distribution. Is there a way that I can estimate what the standard deviation for both distributions? What is the most information that can be gained from these small pieces?

$\endgroup$
7
  • $\begingroup$ If you are getting lots of zeros in your variable and it's a count, I would try a ZIP (zero-inflated poisson) model first before rolling my own. $\endgroup$
    – Neal Fultz
    Commented Apr 17, 2014 at 17:23
  • 3
    $\begingroup$ Why don't you estimate the standard deviation directly from data? $\endgroup$
    – Aksakal
    Commented Apr 17, 2014 at 17:49
  • $\begingroup$ @Aksakal Well, $\mu$ isn't really an average but the per unit value of some variable assuming it is uniformly distributed among units. Because of the way the data is setup going through and getting the individual frequencies I would need to calculate directly would take quite a while. Since I only need a rough estimate for purposes here I was wondering if there was a more expedient solution. $\endgroup$
    – 114
    Commented Apr 17, 2014 at 18:28
  • $\begingroup$ @114, this is confusing. Are you trying to estimate $\mu$ then? Is the mean known? What is the objective here? $\endgroup$
    – Aksakal
    Commented Apr 17, 2014 at 18:42
  • $\begingroup$ @Aksakal Assume that the means above have been estimated in a nice way, but the data is not available. What is the most (if anything) that can be done with this information and does that include a method for estimating standard deviation? $\endgroup$
    – 114
    Commented Apr 17, 2014 at 19:00

1 Answer 1

1
$\begingroup$

You can derive the parameters for a very wide range of two-parameter discrete distributions with this information. If you have $p(x=0|\theta_1, \theta_2)$ and $\mathbb{E}(x|\theta_1, \theta_2)$ then finding $(\theta_1, \theta_2)$ is a two-dimensional root-finding problem.

In your case, $p(x=0)$ can be found from your two expectations by noting that $\mathbb{E}(x) = (1-p(x=0))\mathbb{E}(x|x\geq 1)$. You could work directly with the two expectations, but it's often easier to work with $p(x=0)$ than with $\mathbb{E}(x|x\geq 1)$.

For example, assume the data follows a Negative Binomial distribution parameterized so that the mean is $r(1-p)/p$ and variance is $r(1-p)/p^2$, $p(x=0) = p^r$. From your information, $p(x=0) = 0.3636...$. A small amount of programming comes up with $p = 0.5525, r=1.705$ as parameters that reproduce your information to a pretty high degree of accuracy. From the parameters, of course, you can calculate any quantity of interest.

$\endgroup$
1
  • $\begingroup$ Thanks! This is exactly the sort of thing I was looking for. $\endgroup$
    – 114
    Commented Apr 17, 2014 at 20:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.