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I understand that if i want a sample from an exponential distribution left truncated at a, i can just take a sample from a regular exponential distribution and add the value of a to every single observation. This makes intuitive sense, since the pdf of such a truncated distribution with scale 1 would just be: $f(x) = e^{-(x-a)}$. I imagine it is quite easy, but how would i mathematically proof, that the sample would indeed equate to a sample directly drawn from the truncated distribution? Besides, does this method work for any continuous distribution?

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To prove this, just show that if $Z$ is a usual exponential random variable (with parameter $1$) and if $X = a+Z$, then $f_X(x) = e^{-(x-a)}$ (for $x \ge a$ and $0$ otherwise). You can show this using the formula for the PDF of a transformation of a random variable, or by finding the CDF of $X$ and then differentiating it.

(I'm assuming you already know that the truncated distribution has PDF $e^{-(x-a)}$ (for $x\ge a$ and $0$ otherwise), but if you don't, to show this, you can compute $P(Z \le x\mid Z > a)$, and then differentiate with respect to $x$ to get the PDF of the truncated distribution.)

And no, this sort of thing will not work for all distributions.

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