# Proof of expectation of mean of normal distribution is 0

I am reading a book in which the proof of Expectation of normal is 0 is proved. I couldn't figure out how the right hand integral of equation 6.1.2 is less than infinity and how c1 is exactly chosen? Any help is appreciated

• Where is this from ? It is in fact not true that $$\int_{-\infty}^{\infty}e^{-c_1 x} <\infty$$ You'd need to replace $x$ by $|x|$ for it to be true. Oct 2, 2023 at 11:43
• Can you explain your statement
– Jay
Oct 2, 2023 at 11:51
• If $c_1>0$, then $e^{-c_1 x}\to \infty$ as $x\to-\infty$, so the integral $\int_{-\infty}^0 e^{-c_1 x} dx$ is trivially divergent. Oct 2, 2023 at 11:54
• I found a pdf version of this textbook in which (6.1.2) is correctly rendered: in particular, it shows that "$e^{-c_1 x}$" is intended to be "$e^{-c_1|x|}$" (a function of the absolute value of $x$).
– whuber
Nov 25, 2023 at 15:57

In one (pdf) copy of this passage the assertion instead is

There exists $$c_1\gt 0$$ such that $$f(x) = \max\{|x|,x^2\}e^{-x^2/2}\le c_1 e^{-c_1 |x|}$$ for all $$x\in \mathbb R.$$

Here is a screen shot:

In this version, "$$-c_1 x$$" is "$$-c_1|x|.$$" Nevertheless, the assertion is false.

A good way to study arguments like this is to plot the functions involved. Here are the graphs of $$|x|e^{-x^2/2}$$ (dotted orange), $$x^2 e^{-x^2/2}$$ (dashed green), and their maximum (solid blue).

What the text intends to do is bound the first two curves above by a curve whose area is easy to find and is finite. That cannot be done using curves of the form $$c \exp(-c |x|).$$ The best any curve of this form can do at the points where $$f$$ is maximized is shown in orange here:

It's not even close to covering the graph of $$f.$$

The problem is easily fixed by considering an even simpler comparison curve, such as $$c \exp(-|x|)$$ for some value of $$c$$ to be found. A little Calculus will show that $$c=4$$ (barely) works and any larger value of $$c$$ therefore works with room to spare:

The area under the dashed orange (bi-exponential) curve is $$8.$$ Consequently, the areas under the graphs of $$|x| \exp(-x^2/2)$$ and $$x^2\exp(-x^2/2)$$ must each be less than $$8.$$ Because $$8$$ is finite, the point is made.