When do we call a function a cumulative distribution function (cdf)?

From several statistics and probability theory courses you might know that any Function is a cumulative distribution function (cdf) if: - \(F : \mathbb R → [0, 1]\)

• \(F\) is increasing monotonously

• \(F\) is right continuous

• \(F: \mathbb{R} \rightarrow[0,1] \text { with }\lim _{x \rightarrow-\infty} F(x)=0 \text { and } \lim _{x \rightarrow \infty} F(x)=1\)

Inverse transform method and proof of correctness

Let \(F\) be a continuous and strictly increasing \(c d f\) on \(\mathbb{R}\), we can define its inverse \(F^{-1}:[0,1] \rightarrow \mathbb{R}\). Let \(U \sim \mathcal{U}[0,1]\) then \(X=F^{-1}(U)\) has cdf \(F\). Quick Proof / Verification: \[ \begin{aligned} \mathbb{P}(X \leq x) &=\mathbb{P}\left(F^{-1}(U) \leq x\right) \\ &=\mathbb{P}(U \leq F(x)) \\ &=F(x) \end{aligned} \]

So now let \(F\) be a cdf on \(\mathbb{R}\) and define its generalized inverse \(F^{-1}:[0,1] \rightarrow \mathbb{R}\) \[ F^{-1}(u)=\inf \{x \in \mathbb{R} ; F(x) \geq u\} \] Let \(U \sim \mathcal{U}[0,1]\) then \(X=F^{-1}(U)\) has cdf \(F\).

Example: Weibull Distribution

So now that we understood the math laid out before we can apply it. Let’s try to find the inverse of a weibull distribution function which is basically a more general case of the exponential function as for \(\alpha=1\) it is the exponential distribution. Weibull distribution. Let \(\alpha, \lambda>0\) then the Weibull cdf is given by \[ u= F(x)=1-\exp \left(-\lambda x^{\alpha}\right), x \geq 0 \]

Now let’s start solving for \(u\): \[ \begin{aligned} u &=F(x) \Leftrightarrow \log (1-u)=-\lambda x^{\alpha} \\ & \Leftrightarrow x=\left(-\frac{\log (1-u)}{\lambda}\right)^{1 / \alpha} \end{aligned} \] Because of \((1-U) \sim \mathcal{U}[0,1]\) when \(U \sim \mathcal{U}[0,1]\) we can now use this to define our random variable that follows a weibull distribution. \[ X=\left(-\frac{\log U}{\lambda}\right)^{1 / \alpha} . \]

This formula basically is the mathematical way of describing what we did before. It assigns every value on the y-Axis a value on the x-Axis. Hence, it returns our wanted random variable which follows a weibull distribution. Cool!

So we can check if this is actually true (of course it is because it’s mathematically correct lol).

Implementation in R / verification

weibull_simulation <- function(n, alpha= 1, rate= 0.5){
  ((-log((1-runif(n)))/rate))^(1/alpha)
}

plot(weibull_simulation(100, alpha = 1, rate= 3), main= "Weibull Simulation", ylab= "Values")

hist(weibull_simulation(20000, alpha = 1, rate = 3), probability = TRUE, n=100, main= "Histogram of the distribution", ylab= "Density", xlab = "Values")
lines(seq(0,5,length=20000),dexp(seq(0,5,length=20000),rate = 3), col= "red")
legend("topright",legend= "Density of \nexponential distribution",col = "red", fill= "red", cex= 0.6)

And indeed for a large nit follows an exponential distribution like it should.

Of course this also works for other parameters:

hist(weibull_simulation(200000, alpha = 2, rate = 1), probability = TRUE, n=100, main= "Histogram of the distribution", ylab= "Density", xlab = "Values")
lines(seq(0,5,length=20000),dweibull(seq(0,5,length=20000), shape = 2, scale = 1), col= "red")
legend("topright",legend= "Density of \nweibull distribution",col = "red", fill= "red", cex= 0.6)