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| 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).] |
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| 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 $$2^text{-0.05}ln(2)^text{0.05}$$ = 0.95. which is of form $$A^text{-1/B}ln(2)^text{1/B}$$ where A and B are the two parameters of the Weibull distribution. (See [http://www.weibull.com/AccelTestWeb/weibull_distribution.htm here for formulae).] |
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| 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. |
compute alpha=2. compute beta=20. |
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| !gamma 1 2. exe. compute sum=0. exe. compute sum=-(a1+a2). |
compute rvw=((-1/alpha)*(ln(1-rv.uniform(0,1))))**(1/beta). compute med=((alpha)**(-1/beta))*(ln(2)**(1/beta)). |
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
$$2text{-0.05}ln(2)text{0.05}$$ = 0.95.
which is of form
$$Atext{-1/B}ln(2)text{1/B}$$
where A and B are the two parameters of the Weibull distribution. (See [http://www.weibull.com/AccelTestWeb/weibull_distribution.htm here for formulae).]
compute alpha=2. compute beta=20. compute rvw=((-1/alpha)*(ln(1-rv.uniform(0,1))))**(1/beta). compute med=((alpha)**(-1/beta))*(ln(2)**(1/beta)). exe.