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?

  • 1
    $\begingroup$ If the random variables $X_i$ are i.i.d. then the random variables $h(X_i)$ must also be i.i.d, but that doesn't tell you whether the CLT applies, since the CLT has requirements beyond the random variables being i.i.d. - they must also have finite mean and variance. $\endgroup$
    – fblundun
    Commented Feb 28, 2021 at 13:09
  • $\begingroup$ Yes, assuming finite second moments for the new random variable $h(X)$. Hence, we can apply the central limit theorem to $h(X)$? Can anyone give a little more explanation about why $h(X)$ is i.i.d. as well? $\endgroup$
    – shem
    Commented Feb 28, 2021 at 13:12
  • 2
    $\begingroup$ Would delta method be something you are looking for? $\endgroup$
    – B.Liu
    Commented Feb 28, 2021 at 13:50
  • $\begingroup$ I thought about the delta method, but I don't see how it applies as I'm explicitly interested in the expression $\sqrt{n}(n^{-1}\sum{h(X_{i})}-E[h(X)])$. $\endgroup$
    – shem
    Commented Feb 28, 2021 at 13:57
  • $\begingroup$ The delta method answers the question about the distribution of $\sqrt{n}(h(S_n) - h(\mu))$, where $S_n = \sum_{i=1}^nX_i/n$. The question is about $\sum_{i=1}^n h(X_i)/n$, not $h(\sum_{i=1}^nX_i/n)$ $\endgroup$ Commented Feb 28, 2021 at 13:57

1 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.

  • $\begingroup$ Suppose I assume my function $h(.)$ is continuous. Do you think the continuous mapping theorem would work here? Thanks for your answer! $\endgroup$
    – shem
    Commented Feb 28, 2021 at 15:08

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