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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?

  • 0
    Q(1) for the normal should be infinity. There does not exist any x such that $1 \le F(x)$ for the normal distribution. Same thing for the Weibull. I think the applet you are using is using percentages rather than probabilities, i.e. Q(1) for them corresponds to Q(.01) for you.2011-07-13
  • 0
    F is a distribution, and therefore always in [0, 1]. Q(1) is the smallest value that it a.s. never exceeded. In the case of the normal distribution Q(1)=$\infty$2011-07-13
  • 0
    Ok thank you, the applet i used has a wrong data representation and leads me to confuse "quantile" with "x".2011-07-13

5 Answers 5