Hello all, For information, I found the answer of my question in Cherubini's book. You can see the answer here below. Thank you all for your help.

Hello Anthony, For information, I found the answer of my question in Cherubini's book. You can see the answer here below. Anyway, thank you very much for your help.

Thank you for your answer. Does your definition of $Z_1, Z_2$ mean $f_{Z_1} = f_{N_1}(t)$ for $t \ge \frac{x}{2} $ and $f_{Z_1} = f_{N_2}(x-t)$ for $t \le \frac{x}{2} $ ? And besides, the statement $\{X_1+X_2\ge x\}\subset\{X_1\ge \frac x2\}\cup\{X_2\ge\frac x2\}$ doesn't seem correct. We have rather $\{X_1\ge \frac x2\}\cup\{X_2\ge\frac x2\} \subset \{X_1+X_2\ge x\} $ because for all $x \in \{X_1\ge \frac x2\}\cup\{X_2\ge\frac x2\}$, $x$ will also belong to $\{X_1+X_2\ge x\}$ but not in the other direction.

I don't know why the question is downvoted. About the upper bound, in the case where $n$ is an even number and $w_i = 1 \forall i=1,...,n$, if we take $X_1 = -X_2, X_3=-X_4,...$, the sum $S$ becomes $0$. So, the probability $P(S \leq x) = 1 \forall x \geq 0$. The upper bound is so equal to $1$ for $x \geq 0$. Perhaps there are some conditions on $n$ and $\{ w_i\} _{i=1,..n}$ such that we could contruct $X_i$ in order to obtain $S= 0$ and prove the upper bound is equal to $1$.

Thank you Brendan McKay, so $P(S \leq x)$ reaches the upper bound when $X_1,...,X_n$ are all identical ($X_1 =X_2 =...=X_n$). Do you know how to prove this? This seems to me correspond to the upper Fréchet–Hoeffding bound ($C_{+}=\min \{u_i \}_{i=1,...,n} $)? So, can we expect that the lower bound correspond to the lower Fréchet–Hoeffding bound (where the dependence structure is the copula $C_{-} = \max \{1-\sum_i^n (1-u_i),0 \} $ ) ?

Ah I see. The problem with this view become easy and evident, I don't know why I made it complicated. Thank you very much for your answer Robert!

Thanks Matt F, I suppose $Y$ and $Z$ follow normal distribution and can confirm that this formula is correct. In fact, $F^{-1}_{Y+\lambda Z}(\alpha) = q_{\alpha}(\sigma_X^2 +\lambda^2\sigma_Y^2)^\frac{1}{2}$. But how to prove this formula in general case, when we don't have explicite formula of quantile function? I have no idea.

The approximation of a usual Taylor formula is usually at the value of $\alpha$ and not at the functions. So, this formula is not evident for me and I don't know how to prove it (or find the coefficients). Yes, we can suppose that $Y$ and $Z$ have a joint distribution with a density.