I am usually pretty confident when it gets to normal distribution functions but I keep asking myself one question.

Given the following function: $f(x)= 1/ e^{-x^2}$ where $e$ is Euler's number, is this also a normal distribution function?? When drawing it, it has all the characteristics of a Gaussian curve, but the thing that keeps me from getting to a definite answer is that the above-mentioned function does not follow the formula for a normal function, which is: enter image description here

How would you answer this question? Is $1/e^{-x^2}$ just an exponential function, whereas the normal function is a special type of exponential function??

Thanks a lot in advance.

Kind regards, Helena

  • 4
    $\begingroup$ Are you sure the the graph of the function $f(x) = 1/e^{-x^2} = e^{x^2}$ "has all the characteristics of a Gaussian curve"? $\endgroup$ Apr 1, 2019 at 18:48
  • $\begingroup$ The answer to the question as asked has been covered, but I wondered if it's possible you didn't quite ask what you intended; if your question had instead been about $e^{-x^2}$ the answer would be "not unless you multiply by the correct factor to make it integrate to 1". $\endgroup$
    – Glen_b
    Apr 1, 2019 at 22:16

2 Answers 2


No, it isn't.

When it doubt, plot. The black line is your function $\frac{1}{\exp(-x^2)}$, the red line is the standard normal density. They are about as different as they can be.

non-normal distribution

xx <- seq(-2,2,by=.01)

\begin{align} & \frac 1 {e^{-x^2}} = e^{x^2} \\[10pt] = {} & {\large e^{-\frac 1 2\cdot\frac{(x-\mu)^2}{\sigma^2}}} \text{ only if } \mu=0 \text{ and } \sigma^2 = -1. \end{align} But $\sigma^2$ cannot be $-1$ unless $\sigma$ is imaginary.

Moreover, note that $\displaystyle \int_{-\infty}^\infty e^{x^2}\,dx = +\infty. $


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