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Let $F(x| \alpha, \beta)$ denote the cumulative density function of a probability distribution. Let $[a,b]$ and $[c,d]$ be two disjoint subsets of the support of $F$. Suppose that $F(b) - F(a) = p$ and $F(d) - F(c) = q$, for some non zero $p$ and $q$ such that $p + q < 1$. Can you always uniquely determine $\alpha$ and $\beta$? Otherwise, under what conditions/probability distributions can you determine the parameters?

For examples:

1) $F$ is the normal distribution, and $F(0) = 0.4$ and $F(1) - F(0) = 0.3$. Can you uniquely determine the mean and standard deviation?

2) Consider the skew normal distribution, with probability density function $f(x| \xi, \omega, \alpha) = 2\phi(\frac{x-\xi}{\omega})\Phi(\alpha \frac{x-\xi}{\omega})$. If we fix the location $\xi$, can we uniquely determine the scale $\omega$ and skew $\alpha$ under conditions set out above?

Alex
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  • sorry, i slightly abused the word 'disjoint' as clearly in example one the subsets $(-\infty, 0]$ and $[0, 1]$ are not disjoint. Please pardon the sloppiness in the definition here. – Alex Jan 18 '16 at 03:24
  • Have you ever seen (or sketched for yourself) a normal density curve? If you have, and have paid any attention to the details, the datum $F(0)=0.5$ is wiggling its eyebrows and winking an eye, and pointing out with suggestive movements of its elbow and shoulder and thumb, the value of the mean to you. – Dilip Sarwate Jan 18 '16 at 07:39
  • Oh yes, you are right, that is a bad example, I will change it. – Alex Jan 18 '16 at 08:15
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    This note is relevant: http://www.johndcook.com/quantiles_parameters.pdf . It shows how to solve this problem for the normal, log normal, Cauchy, Weibull, gamma, and inverse gamma probability distributions. – Alex Jan 18 '16 at 22:43
  • These questions are related: http://stats.stackexchange.com/questions/30236/is-there-a-method-to-estimate-distribution-parameters-given-only-quantiles, http://stats.stackexchange.com/questions/6022/estimating-a-distribution-based-on-three-percentiles – Alex Jan 18 '16 at 22:46
  • Your definition 2 doesn't work as it stands, since the way you define it there, $f$ is the same for any $\mu,\sigma$. – Glen_b Jan 19 '16 at 04:19
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    thanks @Glen_b, fixed now. Anyway, by the linked paper, I suppose the answer to 2) is also yes, since the skew normal distribution forms a location scale family. – Alex Jan 19 '16 at 04:25
  • This question appears to be too general to admit any complete and useful answer. Once it is rephrased rigorously, it asks under what conditions a function $f:\mathbb{R}^k\to\mathbb{R}^2$ is one-to-one. The function in question maps the parameters $\theta\in\mathbb{R}^k$ to the ordered pair $(F([a,b]),F([c,d]))\in\mathbb{R}^2$. In statistical practice $f$ is continuous and obviously its image lies in $[0,1]\times[0,1]$, but these conditions are too weak to do anything with. – whuber Jan 19 '16 at 14:23
  • Hi @whuber. Appreciate you taking the effort to phrase the question rigorously and in generality. In my case I would like $k = 2$ (i.e. 2 parameters). The question arose as a parallel to solving a system of linear equations: given two pieces of consistent (probability) information, is possible to uniquely determine 2 unknown parameters. I hope that those of you more knowledgeable about exotic distributions can come up with a counter example if you feel this statement is not true. Alternatively, statements like, this is true for all unimodal, continuous distributions are also very good... – Alex Jan 19 '16 at 22:13
  • ... and at least John Cook (as linked) believes this is useful enough to produce an answer given specific distributions. – Alex Jan 19 '16 at 22:14
  • Nobody's denying the potential utility. I would just like to suggest that you consider narrowing your question. It doesn't take an "exotic" family of distributions to produce a counterexample: the procedure can fail for just about any family, depending on how $a,b,c,d$ are chosen. Generically, though, it will succeed for any one- or two-parameter family and will necessarily fail for any family with more than two parameters. – whuber Jan 19 '16 at 23:38
  • it would be really helpful if you could provide such a counter example as I am finding it very difficult to imagine that this result is not true, at least for a unimodal distribution. – Alex Jan 19 '16 at 23:54

1 Answers1

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For your first example, $$F(0) = \Phi\left(\frac{0-\mu}{\sigma}\right) = \Phi\left(\frac{-\mu}{\sigma}\right) = 0.4$$ and so $$-\frac{\mu}{\sigma}= \Phi^{-1}(0.4) \approx -0.2534\tag{1}$$ while $$F(1) = 0.7 = \Phi\left(\frac{1-\mu}{\sigma}\right)$$ and so $$\frac{1-\mu}{\sigma}= \Phi^{-1}(0.7) \approx 0.5243\tag{2}$$ and you should be able to calculate the values of $\mu$ and $\sigma$ from $(1)$ and $(2)$.

Dilip Sarwate
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  • yes, this forms a system of linear equations and the initial conditions forces it to be a consistent system. – Alex Jan 19 '16 at 22:09