A quantile function Q is defined in terms of its distribution function F as:
$Q(p)=inf\{ x\in R:p \le F(x)\},p\in(0,1)$
But i don't understand very well how it works exactly. Suppose we are managing a cdf for Normal Distribution N(0,1)
$F(x) = \frac{1}{2} \left[ 1 + erf \left( \frac{x-\mu}{\sigma \sqrt{2}} \right) \right]$
If i want to know the probability for Q(1) i have to find the minimum $x \in R$ such that $p \le F(x)$ where in this case:
$$\text{find the minimum x s.t. } \frac{1}{2} \left[ 1 + erf \left( \frac{x-0}{1 \sqrt{2}} \right) \right] \ge 1$$
$$\text{for }x = 0.841 \text{ we have } \frac{1}{2} \left[ 1 + erf \left( \frac{0.841}{ \sqrt{2}} \right) \right] = 1$$
Hence Q(1) for N(0,1) is 0.841.
Suppose now we want to use the Weibull Distribution with $k = 1$ and $\lambda = 1$, the cdf is $F(x) = 1 – e^{-(x/\lambda)^{k}}$
From an external quantile applet (
Quantile applet
) i know that Q(1) = 0.632
but i'm not able to compute this result:
$$\text{find the minimum x s.t. } 1 – e^{-(x/\lambda)^{k}} \ge 1$$
$$1 – e^{-(0.632/1)^{1}}$$
$$ = 0.468472 \neq 1$$
Where i'm wrong?
Best Answer
Mathematically, Quantiles are simple the inverse of the CDF. As such the input $p$ is bounded by $(0,1)$ as $p$ is the probability that the event has occured. In order to understand this, one must understand the CDF. The CDF takes in a value and calculates the probability that the event occurs for any value before. When dealing with distributions such as Normal or Weibull, the CDF only equals $1$ at infinity, because there is a chance that the event occurs at near infinte values for the Random Variable. If you want to do the quantile of Weibull, you can take the CDF equation and solve for $X$ in terms of $P$. The result, for Weibull, should be $X = \lambda((-\log(1-P))^{1/k})$.