Kdjykuj

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)

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.

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