How do you find the CDF of a double exponential distribution, I'm having problems with removing the absolute x
Double exponential has same probability distribution for +ve & -ve sides( I.e. symmetry) So first take CDF of -ve side of distribution. Then, 0.5 + CDF of +ve side of distribution
In April 2011 the CDF for the double exponential distribution was given as 0.5exp(lambda_x) for x<0 and 0.5(1-exp(-lambda_x)) + 0.5 for x >=0 How was this obtained?
You're given: \[g(x)= \frac{1}{2}\lambda e^{-\lambda\mid x\mid }\] So for \[x\leqslant 0\] \[g(x)=\frac{1}{2}\lambda e^{\lambda x}\] We need to integrate to find the CDF: So for \[ x\leqslant 0\] \[G(x)=\int_{-\infty }^{x}\frac{1}{2}\lambda e^{\lambda t}dt = \frac{1}{2}e^ {\lambda x}\] Now for \[x\geq 0\] \[G(x)=P(x\leq 0)+\int_{0}^{x}\frac{1}{2}\lambda e^{-\lambda t}dt=\frac{1}{2}+\left [ -\frac{1}{2} e^{-\lambda t}\right ]_{0}^{x}=1-\frac{1}{2}e^{-\lambda x}\]
Katherine, can we use the alternate process mentioned in the CMP core reading given below? Question: Generate a random variate X from the double exponential distribution with density function: Solution: The density f is symmetric about 0; we can therefore generate a variate Y having the same distribution as |X| and set X = + Y or X = - Y with equal probability. Now the density of |X| : Algorithm: The simulated numbers obtained are different in this process than the process you mentioned where we take F(x) values for x<0 and x>0. Are both methods allowed?
