The expected value of a function $g$ of a random variable $X$ is defined as the Stieljes integral:
$$E[g(x)] = \int_{-\infty}^\infty g(x) \,dF(x),$$
where $F(x)$ is the cumulative distribution function of $X$. In practice, we also have:
$$E[g(x)] = \int_{-\infty}^\infty g(x) f(x) \,dx,$$
where $f(x)$ is the density function. I am working with the function $g(x) = a \, \text{sech}^2(a \, x)$. From an engineer's point of view, this function resembles a Dirac delta (multiplied by 2) for large enough $a$. Consequently, I tried to calculate $E[\delta(x)]$ rather than $E[a \, \text{sech}^2(a \, x)]$ for very large $a$. On the one hand, I got:
$$E[\delta(x)] = \int_{-\infty}^\infty \delta(x) f(x)\, dx = f(0).$$
On the other hand, using the first formula,
$$E[\delta(x)] = \int_{-\infty}^\infty \delta(x)\, dF(x).$$
Since $\delta$ is equal to zero almost everywhere, the set $\{x:\delta(x) \neq 0\} = \{ 0 \}$ is countable and hence has a Lebesgue measure of zero. I have read that if the Lebesgue measure is zero, the integral is zero even if the value of the function is infinite. Hence, it follows that
$$E[\delta(x)] = 0.$$
Obviously, there is a flaw in my calculations. Actually I am familiar with Riemann integration and have little understanding of Lebesgue integration.
P.S.: Computer simulations give one result or the other depending on the ratio between $a$ and the sample size.
Best Answer
The dirac delta is what is called a generalised function.
Basically, the dirac delta is defined to work as a useful exception to the usual rule of a discrete atom having zero Lebesgue measure.
As an exception, naturally it doesn't follow the rule. So, if it is useful to do so then we might extend the notion of expectation to say:
$$\mathsf E(\delta(X-c)) = f(c)$$