0
$\begingroup$

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

$\endgroup$
  • 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"? $\endgroup$ – Alecos Papadopoulos Apr 1 '19 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 -Reinstate Monica Apr 1 '19 at 22:16
1
$\begingroup$

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)
plot(xx,1/exp(-xx^2),type="l",ylim=c(0,5),xlab="",ylab="")
lines(xx,dnorm(xx),col="red")
| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

\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. $

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.