Question at the end of this long introduction!
When treating sums related to independent events or independent random variables, it seems to me that it is very common to use the exponential function. For two reasons:
The exponential transforms sums into products, and independence plays very well with sums.
Exponential has nice inequalities such as $(1 - x) \leq e^{-x}$ that are useful to eliminate the exponential itself.
The classical example is probably the hard part of Borel-Cantelli lema.
If $A_1, A_2, \dotsc$ are independent and $\sum P(A_n) = \infty$, then $P(\limsup A_n) = 1$.
Proof (using exponential):
Let $A = \limsup A_n$. Since $(A_n)^c = \Omega \setminus A_n$ are all independent, $$ \begin{align*} 0 &= \mathrm{e}^{-\sum_{n \geq N} P(A_n)} \\ &= \prod_{n \geq N} \mathrm{e}^{-P(A_n)} \\ &\geq \prod_{n \geq N} (1 - P(A_n)) \\ &= \prod_{n \geq N} P(A_n^c) \\ &= P(A_N^c \cap A_{N+1}^c \cap \dotsb) \\ &\rightarrow P(A^c). \end{align*} $$ Therefore, $P(A) = 1$.
However, this demonstration does not emphasize one aspect I consider very important: in a sense, the theorem is true because whenever you add a new set to $A_1 \cup \dotsb \cup A_n$, since this set is independent of $A_{n+1}$, a fraction of $A_{n+1}$ is already inside $A_1 \cup \dotsb \cup A_n$, and another fraction is "new". If this new part is sufficiently "big", $P(A_1 \cup \dotsb \cup A_n)$ has to converge to $1$...
Formally... Let $$ B_n = A_{n+1} \cup A_{n+2} \cup \dotsb $$ and $$ D_n = A_n \setminus B_n = A_n \cap B_n^c. $$ Because those sets are all disjoint, $$ \begin{align*} 1 &\geq \sum P(D_n) \\ &= \sum P(B_n^c) P(A_n). \end{align*} $$ Since $\sum P(A_n) = \infty$, we must have $P(B_n^c) \downarrow 0$. That is, $$ P(A) = \lim P(B_n) = 1. $$
Now, there is another theorem that says that
If $X_1, X_2, \dotsc$ are independent non-negative random variables, then $$ \sum X_n \text{ converges a.s.} \Leftrightarrow \sum E(X_n) < \infty. $$
The demonstration I know for the ($\Rightarrow$) part uses exponentiation and the inequality $\mathrm{e}^{-x} \leq 1 - ax \leq \mathrm{e}^{-ax}$, for a certain $a$. Exponentiation transforms sums into products, and the inequalities enable us to push "expectation" to inside the argument for the exponential.
Question starts here!
Now, I would like to be able to make a demonstration similar to what I did above, without using the exponential. My guess is to use sets similar to $$ A_n = \{X_n \geq E(X_n)\}. $$ They are all independent sets. Unfortunately, I was not able to provide a demonstration. Do you know how to do that? Any ideas?
Please, do not criticize me for wanting to do it without the exponentiation. Also, the answer to the question "why would you want to do that?" is: because I do! I know I should not have this aversion to the exponential... but I cannot help it! :-)
