Diff for "FAQ/gamma" - CBU statistics Wiki
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Differences between revisions 9 and 11 (spanning 2 versions)
 ⇤ ← Revision 9 as of 2008-02-12 17:32:51 → Size: 1099 Editor: PeterWatson Comment: ← Revision 11 as of 2008-02-13 10:47:46 → ⇥ Size: 681 Editor: PeterWatson Comment: Deletions are marked like this. Additions are marked like this. Line 5: Line 5: The Gamma distribution [http://www.aiaccess.net/English/Glossaries/GlosMod/e_gm_gamma_distri.htm: may be produced by summing exponential distributions.] There are two parameters, n and $$\lambda$$. n is the number of summed exponentials and $$\lambda$$ is the exponential rate. When $$\lambda$$ is very small compared to n a negative skewed version of the gamma distribution results with no upper limit. The Weibull distribution is negatively skewed and may be generated [http://www.taygeta.com/random/weibull.xml using random variables which are uniform on the interval [0,1]] Line 7: Line 7: The below produces two exponential random variables (n=2) with a very small $$\lambda$$ (=1/10000) in SPSS and sums them. One resulting simulated data set produced using this macro had a skew of -1.03. The below produces an open ended negatively skewed weibull distribution with parameters, 2 and 20. It has a median of$$2^text{-0.05}((ln(2)^text{0.05}) = approx. 0.95.$$ Line 10: Line 15: define !gamma ( !pos !tokens(1)                /!pos !tokens(1)).!do !i=!1 !to !2 !by 1.compute !concat(a,!i)=-(10000)*ln(rv.uniform(0,1)*10000).!doend.!enddefine.!gamma 1 2.exe.compute sum=-(a1+a2). compute wrv= (-(1/2)*(ln(1-rv.uniform(0,1))))**(0.05).

# How do I produce random variables which follow a negatively skewed distribution?

Most distributions such as the exponential and log-Normal distributions are positively skewed with the mode of the distribution occurring for lower values.

The Weibull distribution is negatively skewed and may be generated [http://www.taygeta.com/random/weibull.xml using random variables which are uniform on the interval [0,1]]

The below produces an open ended negatively skewed weibull distribution with parameters, 2 and 20. It has a median of

$$2text{-0.05}((ln(2)text{0.05}) = approx. 0.95.$$

compute wrv= (-(1/2)*(ln(1-rv.uniform(0,1))))**(0.05).
exe.

None: FAQ/gamma (last edited 2013-03-08 10:17:36 by localhost)