Consider a random variable $X$ with values in $\left[0,\infty\right)$ such that $E\left[X\right]=\infty$. Given $M > 0$ I want to estimate the expected value of $X$ truncated at $M$. That is I want to estimate $\mu_M = E\left[X|X \le M\right]$. I observe a sequence $X_1, X_2, \ldots, X_n$, and then discard values greater than $M$, then compute the mean, essentially: $$ \hat{\mu} = \frac{\sum_{1\le i \le n} I_{X_i \le M} X_i}{\sum_{1\le i \le n} I_{X_i \le M}}. $$ Is $\hat{\mu}$ governed by a central limit theorem? Will it converge to $\mu$? How is the asymptotic variance of $\hat{\mu}$ affected by $n$ and $\mathcal{P}\left(X \le M\right)$?

  • $\begingroup$ Well, your truncated variables is restricted to $[0,M]$ so clearly have a finite variance, so the usual CLT applies. $\endgroup$ Commented Dec 3, 2020 at 23:46
  • $\begingroup$ So perhaps the convergence is $\sqrt{n \mathcal{P}\left(X \le M\right)} \left(\hat{\mu} - \mu_M\right) \sigma_M^{-1} \to \mathcal{N}\left(0,1\right)$? $\endgroup$ Commented Dec 4, 2020 at 0:12

1 Answer 1


Since you have truncated to a finite support, the variance of the truncated random variable must be finite, and so you can apply the classical central limit theorem to the parts that make up your estimator. So yes, the asymptotic behaviour of the numerator and denominator in your expression are both governed by the central limit theorem. As to convergence, the law-of-large numbers also applies, so yes, it will converge. (Note that the LLN holds even in some cases with infinite variance.)

In this case the situation is complicated by the fact that the numerator and denominator in your expression are correlated random variables that are both asymptotically normal. Below I will show you their asymptotic normal distributions useing the CLT.

Suppose we denote the "effective sample size" and "effective sample total" respectively by:

$$n_\text{eff}(\mathbf{x}_n) = \sum_{i=1}^n \mathbb{I}(x_i \leqslant M) \quad \quad \quad T(\mathbf{x}_n) = \sum_{i=1}^n x_i \cdot\mathbb{I}(x_i \leqslant M).$$

Let $\lambda = \mathbb{P}(X_i \leqslant M)$ denote the probability that an observation is with the trucation range, and define the conditional mean $\mu_M = \mathbb{E}(X_i|X_i \leqslant M)$ and conditional variance $\sigma_M^2 = \mathbb{V}(X_i|X_i \leqslant M)$. To facilitate our analysis we take $Y_i = \mathbb{I}(X_i \leqslant M)$ and we note that:

$$\mathbb{E}(X_i Y_i | Y_i) = Y_i \cdot \mu_M \quad \quad \quad \mathbb{V}(X_i Y_i | Y_i) = Y_i \cdot \sigma_M^2.$$

Thus, applying the law of iterated expectation and the law of iterated variance we have:

$$\begin{align} \mathbb{E}(T(\mathbf{X}_n)) &= \sum_{i=1}^n \mathbb{E}(X_i Y_i) \\[6pt] &= \sum_{i=1}^n \mathbb{E}(\mathbb{E}(X_i Y_i|Y_i)) \\[6pt] &= \sum_{i=1}^n \mathbb{E}(Y_i \cdot \mu_M) \\[6pt] &= \sum_{i=1}^n \lambda \mu_M \\[12pt] &= n \lambda \mu_M, \\[12pt] \mathbb{V}(T(\mathbf{X}_n)) &= \sum_{i=1}^n \mathbb{V}(X_i Y_i) \\[6pt] &= \sum_{i=1}^n \Big[ \mathbb{E}(\mathbb{V}(X_i Y_i | Y_i)) + \mathbb{V}(\mathbb{E}(X_i Y_i | Y_i)) \Big] \\[6pt] &= \sum_{i=1}^n \Big[ \mathbb{E}(Y_i \cdot \sigma_M^2) + \mathbb{V}(Y_i \cdot \mu_M) \Big] \\[6pt] &= \sum_{i=1}^n \Big[ \lambda \sigma_M^2 + \lambda (1-\lambda) \mu_M^2 \Big] \\[12pt] &= n \lambda (\sigma_M^2 + (1-\lambda) \mu_M^2). \\[12pt] \end{align}$$

Consequently, we have:

$$\mathbb{E} \bigg( \frac{T(\mathbf{X}_n)}{n \lambda } \bigg) = \mu_M \quad \quad \quad \mathbb{V} \bigg( \frac{T(\mathbf{X}_n)}{n \lambda} \bigg) = \frac{\sigma_M^2 + (1-\lambda) \mu_M^2}{n \lambda}.$$

We can then form the standardised pivotal quantity and apply the central limit theorem to get:

$$\sqrt{n \lambda} \cdot \frac{T(\mathbf{X}_n)/n \lambda - \mu_M}{\sqrt{\sigma_M^2 + (1-\lambda) \mu_M^2}} \overset{\text{Approx}}{\sim} \text{N}(0,1).$$

Undertaking similar calculations gives:

$$\mathbb{E} \bigg( \frac{n_\text{eff}(\mathbf{X}_n)}{n \lambda } \bigg) = 1 \quad \quad \quad \mathbb{V} \bigg( \frac{n_\text{eff}(\mathbf{X}_n)}{n \lambda} \bigg) = \frac{1-\lambda}{n \lambda},$$

with the standardised pivotal quantity:

$$\sqrt{n \lambda} \cdot \frac{n_\text{eff}(\mathbf{X}_n)/n \lambda - 1}{\sqrt{1-\lambda}} \overset{\text{Approx}}{\sim} \text{N}(0,1).$$

This gives the asymptotic distributions of the scaled numerator and denominator for $\hat{\mu}$. The usual thing to do next would be to approximate the ratio by a T-distribution using an appropriate degrees-of-freedom. One could go further by examining the correlation between these quantities and taking account of this in looking at the behaviour of the ratio. In any case, the above shows how the CLT applies here.

  • $\begingroup$ This is really great. Would it be appropriate to apply the multivariate CLT to the vector of $n_{eff}$ and $T\left(X_n\right)$ then use the delta method to get the asymptotics of $\hat{\mu}$? I think I can figure out the covariance of those two quantities. $\endgroup$ Commented Dec 4, 2020 at 19:17
  • $\begingroup$ Yes, that would be reasonable. It might be worth testing your results with simulations to see how good the resulting distributional approximation is, but in theory, that sounds okay. $\endgroup$
    – Ben
    Commented Dec 4, 2020 at 22:25

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