# Is the following function a normal distribution?

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: 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??

Kind regards, Helena

• 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"? – Alecos Papadopoulos Apr 1 '19 at 18:48
• 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". – Glen_b -Reinstate Monica Apr 1 '19 at 22:16

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. xx <- seq(-2,2,by=.01)
plot(xx,1/exp(-xx^2),type="l",ylim=c(0,5),xlab="",ylab="")
lines(xx,dnorm(xx),col="red")


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