Central limit theorem for the function of an iid random variable

Question

Given an iid random variable $X$, instead of the distribution $\sqrt{n}(n^{-1}\sum{X_{i}}-E[X])$ which is the result that the central limit theorem provides , I am interested in the distribution of $\sqrt{n}(n^{-1}\sum{h(X_{i})}-E[h(X)])$. How is this proven? In other words, if $X$ is i.i.d., does that imply that $h(X)$ is i.i.d as well?

Answer 1

Is the new variable IID ?

Define $Y = h(X)$, you ask whether $\mathbf{Y} = \{Y_i\}_{i=1}^n \stackrel{\mathrm{i.i.d}}{\sim} p_Y$ if we have that $\mathbf{X} = \{X_i\}_{i=1}^n \stackrel{\mathrm{i.i.d}}{\sim} p_X$.

Let's tackle a "simple" case, in which $h$ is invertible (and thus $h^{-1}$ exists). The CDF of $Y$ is $$F_Y(y) = P[Y\leq y] = P[\,h(X) \leq y\,] = P[\,X\leq h^{-1}(y)\,] = F_X(h^{-1}(y))$$ Because $\{X_i\}_{i=1}^n$ is i.i.d, we can factor the joint CDF $$ F_\mathbf{X}(x_1, \dots x_n) = P[\,X_1\leq x_1, \dots, X_n \leq x_n\,] = \prod_{i=1}^nF_{X_i}(x_i) = \prod_{i=1}^nF_{X}(x_i)$$ But then, using the same trick as before, we have that the joint Y-CDF can factor: \begin{align*} F_\mathbf{Y}(y_1, \dots y_n) &= P[\,Y_1\leq y_1, \dots, Y_n \leq y_n\,]\\ & = P[\,h(X_1)\leq y_1, \dots, h(X_n) \leq y_n\,]\\ & = P[\,X_1\leq h^{-1}(y_1), \dots, X_n \leq h^{-1}(y_n)\,]\\ & \\ &= \prod_{i=1}^nF_{X_i}(h^{-1}(y_i))= \prod_{i=1}^nF_{X}(h^{-1}(y_i))\\ &= \prod_{i=1}^nF_{Y}(y_i) \end{align*} So your joint $Y$-CDF factors, thus $\mathbf{Y}$ is i.i.d !

Is the new variable integrable ?

A more problematic condition is the requirement that $\mathrm{E}[Y]$ and $\mathrm{E}[Y^2]$ exist.