Question

Obtain the inverse function of the cdf of the Laplace pdf, given by \(f(x)=\) \((1 / 2) e^{-|x|}\), for \(-\infty

Short answer

The inverse of the CDF of the Laplace distribution is given by \(F^{-1}(u) = log(2u + 1)\) for \(u < 1 / 2\) and \(F^{-1}(u) = -log(2 - 2u)\) for \(u >= 1 / 2\). The R function that generates a random sample from this distribution is given by: \n\n```R \ninverse_laplace_cdf <- function(n){ \n u <- runif(n) \n samples <- ifelse(u < 0.5, log(2*u + 1), -log(2 - 2*u)) \n return(samples) \n} ```

Step-by-step solution

Derive the Cumulative Distribution Function (CDF)

To find the inverse of the Cumulative Distribution Function (CDF), it's first necessary to derive the CDF from the given Probability Density Function (PDF). Given a Laplace PDF \(f(x) = (1 / 2) e^{-|x|}\), the CDF can be calculated by integrating the PDF from \(-\infty\) to \(x\): For \(x < 0\), we get: \(F(x) = \int_{-\infty}^{x} f(t) dt = \int_{-\infty}^{x} (1 / 2) e^{t} dt = e^x - 1 / 2\) For \(x >= 0\), we get: \(F(x) = \int_{-\infty}^{x} f(t) dt = 1/2 + \int_{0}^{x} (1 / 2) e^{-t} dt = 1 - e^{-x} / 2\)

Derive the Inverse of Cumulative Distribution Function (CDF)

To find the inverse of the CDF, denoted as \(F^{-1}(u)\), set \(F(x) = u\) and solve for \(x\). Given \(F(x) = u\), For \(u < 1 / 2, e^{x} = 2u + 1\) gives \(F^{-1}(u) = log(2u + 1)\) For \(u >= 1 / 2, e^{-x} = 2 - 2u\) gives \(F^{-1}(u) = -log(2 - 2u)\)

Write an R function using the Inverse CDF

We now write an R function that uses the inverse transform method to generate random variables that follow the given distribution: \n\n```R \ninverse_laplace_cdf <- function(n){ \n u <- runif(n) \n samples <- ifelse(u < 0.5, log(2*u + 1), -log(2 - 2*u)) \n return(samples) \n} ``` \n\n In the R function above, the `runif(n)` generates n number of uniforms random variables `u`. The `ifelse()` function is used to check whether each `u` is less than 0.5 in order to implement the different solutions for \(F^{-1}(u)\) we derived in Step 2. The generated variable `samples` then gives us a sample of size n from the desired Laplace distribution.