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If $X$ is uniformly distributed on $(0,1)$, then the random variable $ \lambda(-\ln(1-X))^{1/k}\ $, is Weibull distributed with parameters $k$ and $\lambda$.

With this, I can get random numbers distributed weibull from a uniform random number generator.

but if I have a translated Weibull distribution

$ f(x, k, \lambda, \theta) = \frac{k}{\lambda} (\frac{x-\theta}{\lambda})^{k-1} \exp^{-(\frac{x-\theta}{\lambda})^k} $

how is the transformation to generate random numbers distributed like a translated weibull from a generator of uniform random numbers ?

EDIT

the question it's not about the density function, it about of the transformation. How I can generate random numbers distributed like a translated weibull ? (using a uniform random number generator)

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  • $\begingroup$ It seems that you are saying you know how to generate values of $X-\theta$ and you wish to have a set of values of $X$ instead. In other words, given any realization $x_i-\theta$ you need a formula for $x_i$ in terms of $x_i-\theta$ and $\theta$. Is this correct? $\endgroup$
    – whuber ♦
    Commented Feb 18, 2014 at 19:06
  • $\begingroup$ @whuber I can generate random numbers with an uniform distribution, and with the transformation ($ \lambda(-\ln(1-X))^{1/k}\ $) exposed in the question, therefore I can change the distribution of the numbers to a Weibull distribution ( $[0, \inf)$). how I can change the distribution of the randon numbers uniformely distributed to random numbers distributed like a translated weibull distribution ( $[\theta, \inf)$ ). $\endgroup$
    – Gabriel
    Commented Feb 18, 2014 at 19:50
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    $\begingroup$ Have you noticed that to convert $x-\theta$ to $x$ you only need to add $\theta$? $\endgroup$
    – whuber ♦
    Commented Feb 18, 2014 at 19:52
  • $\begingroup$ @whuber , the question it's not about the density function, it about of the transformation. How I can generate random numbers distributed like a translated weibull ? (using a uniform random number generator). $\endgroup$
    – Gabriel
    Commented Feb 18, 2014 at 20:01
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    $\begingroup$ To expand on @whuber's comment / answer, let $y = x - \theta$. $y$ then has a standard Weibull distribution; you can see this by substituting $y$ for $x-\theta$ in your formula for $f(\dots)$. You can generate $y$ as you describe above, then get back to $x$ by adding $\theta$. $\endgroup$
    – jbowman
    Commented Feb 18, 2014 at 20:11

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What the commenters are saying is this:

If $T$ is your translated Weibull, then you can generate variates $t_i$ from the translated Weibull with the following: $$t_i = \theta + \lambda(-\ln(1-x_i))^{1/k}\ ,$$ where the $x_i$ are your uniform random numbers.

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