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Is it possible to easily build a Skewed Normal Distribution with these 3 parameters?

  1. -Mean (or median)
  2. 99.7-th Percentile for data to the left of the mean (median)
  3. 99.7-th Percentile for data to the right of the mean (median)

In other words, since it is practical to build the classical Normal distribution having an idea of its location and scale (which is immediately identified with the 99.7th percentile), can I do the same with the skewed Normal distribution by defining a left and a right scale?

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I take the question to ask if you can estimate the three parameters of a skew normal distribution from the median and the two quantiles you give, the 0.3% and 99.7% percentiles (why did you choose those two?). The cumulative distribution function of a random variable with a skew normal distribution $\mathcal{SN}(\xi,\omega,\alpha)$ can be written (see Skew Normal Distribution) as $$ \Phi(\frac{x-\xi}{\omega}) - 2T(\frac{x-\xi}{\omega},\alpha) $$ where $T$ is Owen's $T$ function.

Using this we can set up a system of three equations with three unknowns, which we can try to solve numerically to get estimates of the three parameters. Denote by $q_1, q_2=M,q_3$ the three percentiles 0.3%,50%,99.7% estimated from the data. Then the equations are $$ \Phi(\frac{q_i-\xi}{\omega}) -2T(\frac{q_i-\xi}{\omega},\alpha)=\begin{cases} 0.003 ~\text{if}~ i=1 \\ 0.5 ~~~~~\text{if}~ i=2 \\ 0.997 ~\text{if}~ i=3 \end{cases} $$ which will need to be solved numerically. The same approach can clearly be used with other quantiles!

In R, the T.Owen function is included in the sn package of A. Azzalini, which also includes the cdf (cumulative distribution function) psn which could be used directly.

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    $\begingroup$ Thanks for the complete answer. The reason why I chose the 0.3 and 99.7 percentile is that I want to approximate a skewed distribution that gives values in the interval [M-a,M+b], where the probability decreases towards the interval boundaries, and it is acceptable to have values outside the boundaries with low probability (0.003 in this case). $\endgroup$
    – Vit Ill
    Feb 17, 2017 at 8:57

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