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Asymetric double exponential distribution $X \sim AL(p,\theta_1, \theta_2)$ has the following form of density function: $$ f(x) = p \theta_1 e^{−\theta_1x} \mathbb{1}_{x<0} + (1 − p) \theta_2 e^{\theta_2x} \mathbb{1}_{x\geq0} $$ I know the Inverse CDF method of simulation random variable, but in this case I think it is not possible, because CDF can not be presented in one simple form, so my question is:

How can I simulate this distribution?

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Well, it should be fairly obvious, since $$\Pr[X \ge 0] = \int_{x=0}^\infty f_X(x) \, dx = 1-p,$$ and $$\Pr[X < 0] = p.$$ Thus to simulate realizations of $X$, all you need to do is simulate a $\operatorname{Bernoulli}(p)$ variable, and based on the value of this variable, simulate either an $-\operatorname{Exponential}(\theta_1)$ or $\operatorname{Exponential}(\theta_2)$ variable.

Incidentally, you have the signs reversed if $\theta_1, \theta_2 > 0$. Your present parametrization is valid only if $\theta_1, \theta_2 < 0$.

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  • $\begingroup$ Thank you, I made a mistake, because my density function is not positive. It should be equal: $ f(x) = p \theta_1 e^{-\theta_1 x} \mathbb{1}_{x \geq 0} + (1-p) \theta_2 e^{\theta_2 x} \mathbb{1}_{x<0} $. My question now is: If I will simulate number which is less than $p$, then I have to simulate random variable with density $f(x) = p \theta_1 e^{-\theta_1 x}$ or $f(x) = \theta_1 e^{-\theta_1 x} $? $\endgroup$ – MathMen Jun 14 '17 at 17:29
  • $\begingroup$ @JonasAl-Hadad You simulate the respective random variable attached to the corresponding Bernoulli probability. So if for example $p = 0.3$, $\theta_1 = 5$, $\theta_2 = 9$, and you generate a $Y \sim \operatorname{Uniform}(0,1)$ variable and find that it is $y = 0.2 < p$, then you will generate a $\operatorname{Exponential}(\theta_1 = 5)$ variable and keep the sign positive. If you generated $y = 0.8 > p$, then you would generate $\operatorname{Exponential}(\theta_2 = 9)$ and flip the sign. $\endgroup$ – heropup Jun 14 '17 at 18:01

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