Does the expression $e^\frac{-1}{x^a}$ match any known distribution function?

Question

I have the following expression:

$$\exp\left(-\frac{1}{x^a}\right)\,,$$

where $a$ is a positive constant.

As far as I can see, this is a valid distribution function if we define it to equal $0$ if $x\le0$. Does this match any known cumulative distribution function? What's throwing me off is the $x^a$ in the denominator of the exponent.

Answer 1

For $a>0$ and $x\geq0$, $$ c(x) = e^{-x^{-a}} $$ is an increasing function such that $c(0)=0$, $c(x)>0$ for $x>0$ and $c(+\infty) = 1$. Therefore, it can be interpreted as a CDF, and the relative PDF is $$ p(x) = \frac{dc}{dx} = a e^{-x^{-a}} x^{-1 - a} $$ This may resemble the Weibull distribution, but it's not. It is the Fréchet distribution, also known as inverse Weibull distribution. See also: https://stats.stackexchange.com/q/99389.