[Math] How to find which percentile is a value in a skewed normal distribution

statistics

I have a skewed normal distribution for which I know the average, standard deviation, skewness & kurtosis (which is different from zero).

Given a number $X,$ how can estimate which percentile corresponds to that value? (I'm ok with getting an approximate value of this percentile.)

I used z-score tables in the past (before having skewed distributions), but they seem to apply only to non-skewed distributions.

Thanks for your help.

Best Answer

The Wiki page Skew normal distribution provides the information to estimate the parameters using the sample mean ($\bar{x}$), standard deviation ($s$), and skewness ($\hat{\gamma}$). The 3 parameters to be estimated are $\mu$, $\sigma$, and $\alpha$.

If $|\hat{\gamma}|<1$, then $\hat{\alpha}$ is found in two steps:

$$\delta =\sqrt{\frac{\pi \left| \hat{\gamma} \right| ^{2/3}}{2 \left(\left| \hat{\gamma} \right| ^{2/3}+\left(\frac{4-\pi }{2}\right)^{2/3}\right)}}$$

$$\hat{\alpha} = \text{sgn}(\hat{\gamma})\sqrt{\frac{\delta }{1-\delta ^2}}$$

Otherwise $\hat{\alpha}$ is the solution to

$$\hat{\gamma} =\frac{\sqrt{2} (4-\pi ) \hat{\alpha} ^3}{\left((\pi -2) \hat{\alpha} ^2+\pi \right)^{3/2}}$$

which needs to be performed numerically. Then $\hat{\mu}$ and $\hat{\sigma}$ are

$$\hat{\sigma} =\frac{s}{\sqrt{1-\frac{2 \hat{\alpha} ^2}{\pi \left(\hat{\alpha} ^2+1\right)}}}$$

$$\hat{\mu} =\bar{x}-\frac{\sqrt{\frac{2}{\pi }} \hat{\alpha} \hat{\sigma} }{\sqrt{\hat{\alpha} ^2+1}}$$

Now armed with the parameter estimates, then one can estimate the cumulative distribution function:

$$Pr(X \le x)=\Phi\left(\frac{x\, -\hat{\mu} }{\hat{\sigma} }\right)-2 T\left(\frac{x\, -\hat{\mu} }{\hat{\sigma} },\hat{\alpha} \right)$$

where $T$ is the Owen's T function: $T(x,a)=\frac{\int_0^a \frac{\exp \left(-\left(\left(t^2+1\right) x^2\right)\right)}{2 \left(t^2+1\right)} \, dt}{2 \pi }$ .

Here is an implementation using Mathematica:

I know code should be given as text but in this case because it is unlikely that you have Mathematica (and it would look much messier as text), it should be instructive as to the process.

To estimate the percentage of the distribution no larger than a specified value you'll need to use the cumulative distribution function (CDF) described on the Wiki page. Using Mathematica for values of $X$ being 7.5 and 5.2:

If you have access to the statistical package R, then the sn package will perform these calculations.

Related Solutions

[Math] Mean of a portion of a normal distribution

Suppose $X$ is normal with mean $\mu$ and standard deviation $\sigma$. Then $Z=\frac{X-\mu}{\sigma}$ is normal with mean $0$ and standard deviation $1$, and $X=\sigma Z + \mu$. Then

$$E[X \mid X \in [a,b]]=E[\sigma Z + \mu \mid X \in [a,b]] \\ = \mu + \sigma E[Z \mid X \in [a,b]] \\ = \mu + \sigma E \left [Z \left | Z \in \left [ \frac{a-\mu}{\sigma},\frac{b-\mu}{\sigma} \right ] \right. \right ] \\ = \mu + \sigma \frac{E \left [ Z 1_{[\frac{a-\mu}{\sigma},\frac{b-\mu}{\sigma}]}\right ]}{P \left ( Z \in \left [\frac{a-\mu}{\sigma},\frac{b-\mu}{\sigma} \right ] \right )} \\ = \mu + \sigma \frac{\int_{\frac{a-\mu}{\sigma}}^{\frac{b-\mu}{\sigma}} x e^{-\frac{x^2}{2}} dx}{\int_{\frac{a-\mu}{\sigma}}^{\frac{b-\mu}{\sigma}} e^{-\frac{x^2}{2}} dx}.$$

The numerator can be calculated in terms of elementary functions using the FTC, since $-x e^{-x^2/2}$ is the derivative of $e^{-x^2/2}$.

The denominator is an integral of the standard normal density, which cannot be calculated in terms of elementary functions, but can be easily evaluated using software, using functions generally called either "erf" (for "error function") or "normcdf".

$$= \mu + \sigma \frac{ e^{-\frac{\left ( a-\mu\right ) ^2}{2\sigma^2}} - e^{-\frac{\left ( b-\mu\right ) ^2}{2\sigma^2}} }{ \sqrt{\frac{\pi}{2}}\left ( \text{erf}\left ( \frac{b-\mu}{\sqrt{2}\sigma}\right ) - \text{erf}\left ( \frac{a-\mu}{\sqrt{2}\sigma}\right ) \right ) }$$

[Math] Calculate Percentile of Skewed Dataset

The straightforward way is to use the definition of percentile (which differs a bit from text to text and software to software) and count observations. This works for data from any distribution. (Differences in definitions do not matter much for large samples.) Roughly speaking, the 90th percentile is a value below which one finds not more than 90% of the observations, and above which one finds not more than 10% of them.

Here are a few examples of percentiles for a couple of datasets in R, one normal (symmetrical) and one exponential (right-skewed). Notice that percentiles of small samples do not necessarily match percentiles of the populations from which they are sampled. (The method you have been using for normal data seems to conflate the two kinds of percentiles.) In the data displays below the numbers in brackets give the index of the first observation in the that row.

x = round(sort(rnorm(50, 100, 15)), 1);  x  # generate 50 obs from Norm mean=100, SD=15
x
 [1]  61.1  69.4  71.1  73.0  73.9  77.5  78.0  78.0  79.0  81.5
[11]  83.4  85.9  86.5  87.8  87.9  88.0  88.8  90.0  90.7  91.3
[21]  92.0  93.0  95.3  97.9  97.9  99.2  99.2 100.0 100.2 101.0
[31] 102.9 103.4 104.3 104.6 105.4 107.2 108.5 108.6 109.5 109.6
[41] 111.3 111.5 111.9 118.0 119.5 119.6 119.6 119.9 121.5 128.4
quantile(x, .9)  # 90th percentile
   90% 
119.51 
quantile(x, .7)  # 70th percentile
   70% 
105.94 
qnorm(c(.9, .7), 100, 15)  # 90th and 70th percentiles of POPULATION
[1] 119.2233 107.8660

x = round(sort(rexp(60, rate=1/50)), 1);  x  # generate 60 obs from EXP mean=50
x
 [1]   0.0   0.4   0.5   0.9   0.9   1.3   2.4   3.8   4.4   6.1
[11]   7.5   7.9   8.0   9.0   9.9  11.0  11.5  13.3  15.4  16.4
[21]  19.6  20.3  25.1  25.4  28.0  28.8  29.3  29.5  31.1  32.0
[31]  32.1  34.2  37.2  40.6  42.0  42.0  49.5  52.2  55.6  56.8
[41]  59.6  64.3  73.9  74.7  78.7  87.9  90.1  95.4  97.2 105.2
[51] 110.3 113.8 114.6 172.5 187.0 188.2 188.8 207.9 259.7 265.3
quantile(x, .9)
   90% 
173.95 
quantile(x, .7)
  70% 
67.18 
qexp(c(.9, .7), rate=1/50)  # 90th and 70th percentiles of POPULATION
115.12925  60.19864

Best Answer

Related Solutions

[Math] Mean of a portion of a normal distribution

[Math] Calculate Percentile of Skewed Dataset

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