Simple mean or kernel estimator?

Question

Let $X$ be a continuous random variable with support $\mathcal{X}$ and density $f(x)$. Suppose I'm interested in constructing a consistent estimator of $E(X)$ using $n$ i.i.d. observations $(X_1,..., X_i, ...,X_n)$ with $X_i\sim X$.

As a first simple thought, I would consider $$ \hat{\mu}_n=\frac{1}{n} \sum_{i=1}^n X_i $$ which is a consistent estimator of $E(X)$ as $n$ goes to infinity, under some conditions.

Suppose I want to complicate my life and take the definition of $E(X)$ which is $$ E(X)=\int_{x\in \mathcal{X}} x f(x)dx $$ At this point, I could consider a kernel estimator for $f(x)$ (let me denote it by $\hat{f}_n(x)$) which is consistent for $f(x)$ under certain assumptions. In turn, I could compute $$ \tilde{\mu}_n=\int_{x\in \mathcal{X}} x \hat{f}_n(x)dx $$ which is a consistent estimator of $E(X)$ as $n$ goes to infinity, under some conditions.

If the above is correct, it seems to me that $\hat{\mu}_n$ and $\tilde{\mu}_n$ achieve the same objective, although under different (more or less stringent) set of conditions. Is this true?

Answer 1

They do achieve the same objective, and we can make the comparison even clearer. Let $\hat F_n$ give the empirical CDF and let $\hat f_n$ be the corresponding pmf that puts a mass of $1/n$ on each observed value (which are almost surely unique here). Then the first moment of $\hat F_n$ is $$ \int x \,\text d \hat F_n = \sum_{x \in \{x_1,\dots,x_n\}} x \hat f_n(x) = \bar x_n \to_p \int x \,\text d F = \text E[X] $$ so even $\hat \mu_n$ can be viewed this way.

I think this is a consequence of how we can think of moments as being functionals of probability measures via $\nu \mapsto \int x \,\text d\nu$ so it is less surprising that we can approximate a moment of $P$ (the probability measure with CDF $F$) by using moments of estimators of $P$.