Question:
If $X$ is a random vairable and $\delta$ is the Dirac delta, is $\delta(X)$ meaningful? Useful?
Motivation:
In this other question, we have $X\sim Exp(p)$, and it is asked whether there exists a function $g:\Bbb R\to \Bbb R$, independent of $p$, such that $\Bbb Eg(X) = p$. The comments revealed that this is not possible.
However, it is possible to approximate such an $g$ by approximating a "one-sided" Dirac delta. For example, with $g_\alpha = \alpha I_{[0,\frac1\alpha]}$ for $\alpha>0$, where $I_{A}$ is an indicator function, we have in fact $$ \lim_{\alpha\to\infty}\Bbb Eg_\alpha(X) = \lim_{\alpha\to\infty}\Bbb E\left[\alpha I_{[0,\frac1\alpha]}(X)\right] = p. $$ Knowing this, I am now tempted to simply say that $g=\delta$ somehow answers the problem (since $g_\alpha$ has the properties of $\delta$ in the limit). In other words, $\delta(X)$ should be some sort of "random variable distribution(?)" with an expected value of $p$.
Caveat: The limit is not really $\delta$, because I don't think $\int_0^\infty\delta(x)f(x)\ dx$ is well-defined in general; or maybe it is $\frac{f(0)}{2}$? I would like something like $\delta^+$ that satisfies $\int_0^\infty\delta^+(x)f(x)\ dx = f(0)$, but I'm out of my depth here. Anyway, that's not the point of my question, so please just imagine I had chosen a better example...
I somehow suspect $\delta(X)$ makes sense if you "intend to" use it in an integral, but naturally $\delta(X)$ will not be a random variable in the usual sense (right?). For example, I might argue that $\Bbb E \delta(X)$ is meaningful, because we can write $$ \Bbb E \delta(X) \overset?= \int_\Bbb R \delta(x)f_X(x)\ dx = f_X(0). $$
I should mention that my knowledge of the Dirac delta is rudimentary. I know how it can be used in integrals, and that it can be seen as a measure or as a distribution. I understand the measure perspective, but to be honest, I don't know the formalities of distribution theory. I think this question naturally points to a distribution perspective, because of the family of functions, and this is why I am on shaky ground here.
Standard random variables have their sample space be numbers, but random variables are much more general than that. You can have random functions, random sets, and in this case, random functionals. A functional is given as an inner product with a function, and outputs a number. In this case, the random functional can have an inner product with a function, and the output is a random number.
The problem here is that $\delta(X)$ is not a function of x, which a delta function usually is. So you would have here $$\langle \delta(X), f(x) \rangle=\int_\mathbb{R} \delta(X) f(x) dx = ???$$ if $X=0$. It would be $= 0$ otherwise.
What is a far more interesting object is $\delta(x-X)$. The math here would yield
$$\langle \delta(x-X), f(x) \rangle=\int_\mathbb{R} \delta(x-X) f(x) dx = f(X)$$ so evaluating $f$ at a random value. There is a rich amount of math that can be done with this object.