# Is the average of n independent Laplace random variables a Gaussian distribution?

Does the average $$\frac{\sum^n_i X_i}{n}$$ converge to a normal when $$n \to \infty$$. Here $$X_i$$ are independently distributed Laplace samples, with zero mean, and different standard deviation $$\sigma_i$$.

I know this can be applied from the general Central Limit Theorem However, I am not sure if the Laplace distribution satisfies Lyapunov condition?

• The Lyapunov condition seems to hold (do you want the calculations?), but the Lyapunov CLT says something about $\sum\frac{X_i}{\sigma_i}$, not $\sum\frac{X_i}{n}$. The first will converge to a standard normal, but the second may not (if $(\sigma_i)_i$ grows fast enough). Typo, or intended? If the latter, the Lyapunov CLT may not be what you are looking for. Jul 28, 2020 at 8:28
• Update: it looks like the Lyapunov condition does not hold in general, either, at least not for all possible sequences $(\sigma_i^2)$. Jul 28, 2020 at 9:50
• Incidentally, The Laplace Distribution and Generalizations by Kotz, Kozubowski & Podgórski looks helpful. Jul 29, 2020 at 7:18

# TL;DR

You cannot use either the Lyapunov or the Lindeberg CLT to say anything about the convergence in distribution of $$\frac{1}{s_n}\sum_{i=1}^n X_i$$ (where $$s_n^2=\sum_{i=1}^n\sigma_i^2$$) without additional conditions on the sequence of variances $$(\sigma_i^2)$$.

Neither CLT would say anything about $$\frac{1}{n}\sum_{i=1}^n X_i$$. If the sequence of variances $$(\sigma_i^2)$$ grows fast enough, I strongly doubt that this average converges to anything reasonable.

Assume that $$X_i\sim\text{Laplace}(0,b_i)$$ for a parameter $$b_i>0$$. Then $$\sigma_i^2=2b_i^2$$. As above, let

$$s_n^2=\sum_{i=1}^n\sigma_i^2=2\sum_{i=1}^n b_i^2.$$

The key property we need is that $$|X_i|\sim\text{Exp}\big(\frac{1}{b_i}\big)$$. This allows us to easily calculate the expectations we need in the Lyapunov or Lindeberg CLTs.

The condition for the Lyapunov CLT is that there is some $$\delta>0$$ such that

$$\lim_{n\to\infty}\frac{1}{s_n^{2+\delta}} \sum_{i=1}^nE\big(|X_i|^{2+\delta}\big) =0.$$

We have $$E\big(|X_i|^{2+\delta}\big) = b_i^{2+\delta}\Gamma(\delta+3),$$ so $$\frac{1}{s_n^{2+\delta}} \sum_{i=1}^nE\big(|X_i|^{2+\delta}\big) = \frac{\sum_{i=1}^n b_i^{2+\delta}\Gamma(\delta+3)}{\big(\sum_{i=1}^n 2b_i^2\big)^\frac{2+\delta}{2}} = \frac{\Gamma(\delta+3)}{2^\frac{2+\delta}{2}}\frac{\sum_{i=1}^n b_i^{2+\delta}}{\big(\sum_{i=1}^n b_i^2\big)^\frac{2+\delta}{2}}.$$ So the condition is that there is some $$\delta>0$$ such that $$\frac{\sum_{i=1}^n b_i^{2+\delta}}{\big(\sum_{i=1}^n b_i^2\big)^\frac{2+\delta}{2}} \to 0.$$ However, this does not hold in general. Consider $$b_i=\frac{1}{i}$$. Recall that $$\sum_{i=1}^\infty\frac{1}{i^2}=\frac{\pi^2}{6}$$. So the denominator in the fraction goes to $$\frac{\pi^{2+\delta}}{\sqrt{6}^{2+\delta}}$$, whereas the numerator has some other finite but nonzero limit. So the condition that the fraction goes to zero does not hold for this choice of $$(b_i)$$.

The condition for the Lindeberg CLT is that for all $$\epsilon>0$$, $$\lim_{n\to\infty}\frac{1}{s_n^2}\sum_{i=1}^nE\big(X_i^21_{|X_i|>\epsilon s_n}\big) = 0.$$ The expectation here is just a moment of a left-truncated exponential distribution. We have $$E\big(X_i^21_{|X_i|>k}\big) = \int_k^\infty \frac{x^2}{b_i}e^{-\frac{x}{b_i}}\,dx = e^{-\frac{k}{b_i}}(2b_i^2+2b_ik+k^2).$$ So the Lindeberg condition is that $$\frac{1}{s_n^2}\sum_{i=1}^nE\big(X_i^21_{|X_i|>\epsilon s_n}\big) = \sum_{i=1}^n e^{-\frac{\epsilon s_n}{b_i}}\frac{2b_i^2+2b_i\epsilon s_n+\epsilon^2 s_n^2}{s_n^2} \to 0.$$ But that again does not hold in general: consider any sequence $$(b_i)$$ such that the series of variances $$(s_n)$$ stays bounded.

The other answer by Stephen Kolassa gives you an excellent analysis of the Lyapunov condition in this case. However, I think it is also fruitful to look at this problem using moment generating functions. In your problem you have independent values $$X_i \sim \text{Laplace}(0, \sigma_i/\sqrt{2})$$, so these random variables have scaled moment generating functions given by:

\begin{align} \varphi_{i}(t/n) \equiv \mathbb{E}(\exp(tX_i/n)) = \frac{1}{1 - \sigma_i^2 t^2/2n^2} &= 1 + \frac{\sigma_i^{2}}{2} \cdot \frac{t^2}{n^2} + \mathcal{O}(n^{-4}). \\[6pt] \end{align}

Letting $$\bar{X}_n \equiv \sum_{i=1}^n X_i/n$$ denote the sample mean of interest, this latter random variable has moment generating function, we have the characteristic function:

\begin{align} \varphi_{\bar{X}_n}(t) = \prod_{i=1}^n \varphi_{i}(t/n) &= \prod_{i=1}^n \frac{1}{1 - \sigma_i^2 t^2/2n^2}. \\[6pt] \end{align}

Taking $$n \rightarrow \infty$$ gives the asymptotic form:

\begin{align} \varphi_{\bar{X}_n}(t) &\rightarrow \prod_{i=1}^n \Bigg( 1 + \frac{\sigma_i^{2}}{2} \cdot \frac{t^{2}}{n^{2}} \Bigg). \\[6pt] \end{align}

In the special case where $$\sigma_1 = \sigma_2 = \sigma_3 = \cdots$$ this function converges to an exponential function in $$t^2$$, which is the moment generating function for the normal distribution. In the more general case, the moment generating function will not converge to the exponential function in $$t^2$$, and so the distribution of the sample mean does not converge to the normal distribution.

If you would like to go further than this, I suggest you look into conditions on the $$\sigma_i$$ values that will allow you to get a useful convergence result for the above asymptotic form. It may be possible to simplify this asymptotic form under some conditions on these values, but I will leave this to you to investigate.

• (+1). Ben, when I calculate the scaled MFG, I got $1/(1 \color{red}{-} \frac{s^{2}t^{2}}{2n^{2}})$ (note the minus sign in the denominator) but I could have made an error. Jul 29, 2020 at 6:32
• You're right --- corrected.
– Ben
Jul 29, 2020 at 9:54