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I am implementing an exponentially distributed random number generator (RNG) based on George Marsaglia's Ziggurat algorithm. I previously used the algorithm to create a normally distributed RNG. By the nature of the algorithm, the standard deviation is 1 and the mean is zero. In order to modify the curve, all one needs to do is multiply by the desired standard deviation and/or add the desired mean.

However, it doesn't appear to be quite as simple in the case of an exponential distribution, where I want to vary lambda. Yes, simply multiplying my X values from the RNG by some constant modifies its Y-intercept, but the X-intercept remains the same.

So my question to you is how can I modify the lambda of a set of exponentially distributed values whose original lambda is equal to 1? Thanks

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If the random variable $X$ is standard exponential with density $\mathrm e^{-x}\mathbf 1_{x\geqslant0}$ and if $\lambda$ is positive then the random variable $Y=X/\lambda$ is exponential with parameter $\lambda$ and density $\lambda\mathrm e^{-\lambda y}\mathbf 1_{y\geqslant0}$.

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  • $\begingroup$ so just divide x by lambda? $\endgroup$ – audiFanatic Jul 8 '13 at 21:29

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