Sum of two exponential series with equal means and variances

Question

Assuming $A$ and $B$ are two non-negative real-valued random variables such that

1. $\mathrm{E}(A)=\mathrm{E}(B)$ (equal means)
2. $\mathrm{Var}(A)=\mathrm{Var}(B)<\epsilon$ (equal small variances)

is there a way to prove that $\frac{1}{N}\sum_{j=1}^Ne^{-a_{j}}$ and $\frac{1}{N}\sum_{j=1}^Ne^{-b_{j}}$ are arbitrarily close to each other where $a_j$ and $b_j$ are realizations taken from $A$ and $B$, respectively. ($N$ can be assumed to be large as well)

Answer 1

Although it is not explicitly specified, I will assume that you intend for all the realisations of these random variables to be independent (i.e., I will assume joint independence of all the random variables in both series). The difference between the two series is the random variable defined by the function:

$$D(N) = \frac{1}{N} \sum_{i=1}^N (e^{-A_i} - e^{-B_i}).$$

Since $e^{-a} \leqslant 1$ for all $a \geqslant 0$, it follows that $\mathbb{V}(e^{-A}) \leqslant \mathbb{E}(e^{-2A}) \leqslant 1$ for any non-negative random variable $A$. Thus, we have:

$$\begin{equation} \begin{aligned} \mathbb{V}(D(N)) &= \frac{1}{N^2} \sum_{i=1}^N \Big( \mathbb{V}(e^{-A_i}) + \mathbb{V}(e^{-B_i}) \Big) \\[6pt] &\leqslant \frac{1}{N^2} \sum_{i=1}^N \Big( 1 + 1 \Big) \\[6pt] &= \frac{1}{N^2} \cdot 2N \\[6pt] &= \frac{2}{N}. \\[6pt] \end{aligned} \end{equation}$$

We therefore have $\lim_{N \rightarrow \infty} \mathbb{V}(D(N)) = 0$, so the variance of the difference converges to zero. We also have the constant mean difference:

$$\begin{equation} \begin{aligned} \mathbb{E}(D(N)) &= \frac{1}{N} \sum_{i=1}^N \Big( \mathbb{E}(e^{-A_i}) - \mathbb{E}(e^{-B_i}) \Big) \\[6pt] &= \mathbb{E}(e^{-A}) - \mathbb{E}(e^{-B}). \\[6pt] \end{aligned} \end{equation}$$

Combining these results we see that $D(N)$ converges in mean square to $\mathbb{E}(e^{-A}) - \mathbb{E}(e^{-B})$, which is a constant value. Since the variances of both random variables are small, this limiting value should be close to (but not necessarily equal to) zero. Thus, for large values of $N$ you would indeed expect the difference to converge towards a value near zero.

Answer 2

Just apply the law of large numbers, since $e^{-a_j}$ has finite variance because $a$ is positive and has finite variance it self. The result is that the variance of $e^{-a_j}$ is smaller than the variance of $a$. This is elaborated in this question: https://mathoverflow.net/questions/330348/proof-of-variance-bounds-for-transformed-random-variables/330357#330357