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Is the first derivative of

$Pr(x)=\frac{e^{\beta_0+\beta_i x_i}}{1+e^{\beta_0+\beta_i x_i}}$

a Gaussian function?

ECII
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1 Answers1

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The short answer is NO. (Differentiating cannot introduce an $x^2$ term in the exponential.)

Longer answer: Out of curiosity I just differentiated the (almost trivial) case $\beta_0 = 0$, $\beta_1 = 1$, because this claim was made in another question here recently.

The derivative is $e^x/ (1 + e^x)^2 $. Note that this is symmetric about zero. Thus, if it is equal to some Gaussian $\frac{1}{\sigma \sqrt{2 \pi}} e^{ (x - \mu)^2/2\sigma^2 }$ then the mean is necessarily $\mu = 0$. Moreover, for the functions to agree when $x = 0$, we must have $\sigma \sqrt{2 \pi} = 4$.

Thus, the only possible Gaussian would have $\mu = 0$, $\sigma = 4/\sqrt{2 \pi}$. This is plotted below. The differing behaviour is clear. (Admittedly they are a better match than I imagined though).

enter image description here

curve( dnorm(x, sd= 4/sqrt(2 * pi)), xlim=c(-7,7), col = "red" )
curve( exp(x)/ ( (1 + exp(x))^2 ), add=T )
P.Windridge
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  • Although you are correct, the graphical comparison is unconvincing. How do we know that some other choice of Gaussian parameters won't exactly reproduce the logistic curve? (One way to prove this is to look at the tail behaviors, such as plotting the hazard functions: the logistic hazard is bounded whereas a Gaussian hazard is not.) – whuber Apr 17 '15 at 16:07
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    @whuber- it is easy to see from the formula that the logistic curve is symmetric about zero. Thus the mean of any candidate Gaussian must be zero. Also both curves must take the same value at $x=0$. This fixes the s.d. $\sigma = 4/\sqrt{2\pi}$. The Gaussian only has these two parameters!! – P.Windridge Apr 17 '15 at 16:25
  • I.e. the Gaussian curve plotted above is the only remotely plausible candidate. – P.Windridge Apr 17 '15 at 16:33
  • I mean "remotely plausible", if we couldn't already see from the formulae that the tails are different (as I noted in the preliminary comment). – P.Windridge Apr 17 '15 at 16:38
  • Re your first comment: That's the kind of argument that ought to go into your answer, rather than in a comment. – whuber Apr 17 '15 at 17:24
  • @whuber - done. – P.Windridge Apr 17 '15 at 22:06