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Let $X_1,...,X_n$ be $n$ gaussian random variables $N(0,1)$ not necessarily independent or jointly correlated, $S=\sum_{i=1}^n w_i X_i$ be the weighted sum of these gaussian variables (because $(X_i)_{i=1,..,n}$ are not jointly correlated, $S$ can be non normally distributed)

1/ What are the upper bound and/or lower bound of $P(S \leq x)$ $\forall x\in \mathbb{R}$?

2/ And what are the upper bound and/or lower bound of $P(S \leq x)$ if we know the covariance matrix $\Omega$ of these $n$ gaussian random variables?

Could you please recommend me some references on this topic?

Thank you in advance.

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    $\begingroup$ For your first question, the extremes are that they are all the same and that they cancel out identically. $\endgroup$ – Brendan McKay Dec 27 '20 at 3:33
  • $\begingroup$ 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 \} $ ) ? $\endgroup$ – NN2 Dec 27 '20 at 11:43
  • $\begingroup$ Maybe I am misinterpreting the question, but when all the $X_i$ are identical, while the sum $S = X_1+ \cdots + X_n$ has maximum probably of being large, still for each fixed $x$, $\mathbb{P}[S \le x]$ is not maximized. E.g. to simplify, take two Bernoulli $0$, $1$ random variables. If independent, then $\mathbb{P}[S \le 1] = \frac{3}{4}$; if equal then $\mathbb{P}[S \le 1] = \frac{1}{2}$. And in either case $\mathbb{P}[S \le 2] = 1$. $\endgroup$ – Mark Wildon Dec 27 '20 at 11:57
  • $\begingroup$ @MarkWildon I think $P[S\le x]$ should be minimised if the variables are identical, for $x\ge 0$. Not maximized. Is it wrong? $\endgroup$ – Brendan McKay Dec 27 '20 at 12:24
  • $\begingroup$ @Brendan McKay I agree with you, and therefore not with NN2's comment, since in his or her first line 'upper bound' should therefore be 'lower bound'. $\endgroup$ – Mark Wildon Dec 27 '20 at 12:30
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Here is a full answer to Question 1 in the special case $S=X_1+X_2$. I give an exact upper and lower bounds for $\mathbb P(S\ge x)$.

As mentioned in the comments, for $x\ge 0$ and $\mathbb P(S\le x)$ may be 1 (with a similar result for $\mathbb P(S\ge x)$ when $x\le 0$ by symmetry): if $X_2$ is taken to be $-X_1$, then $\mathbb P(S\le x)$ is 1 for all $x>0$.

The interesting remaining case is then $\mathbb P(S\ge x)$ when $x>0$ (or its symmetric version $\mathbb P(S\le x)$ when $x<0$). In this case, we show $\mathbb P(S\ge x)$ is $2\mathbb P(N\ge \frac x2)$.

To see this, fix $x>0$ and define a pair of random variables as follows:

Let $(Z_1,Z_2)$ be $(t,x-t)$ with one-dimensional probability density $f_{N_1}(t)$ for $t\in [\frac x2,\infty)$.

Let $(Z_1,Z_2)$ be $(x-t,t)$ with one-dimensional probability density $f_{N_2}(t)$ for $t\in [\frac x2,\infty)$.

Let $(Z_1,Z_2)$ be $(-\infty,-\infty)$ with the remaining probability.

Now for $t\ge \frac x2$, we can check that $\mathbb P(Z_1\ge t)=\mathbb P(N\ge t)$ and similarly with $Z_2$. Also $\mathbb P(Z_1\ge t)\le \mathbb P(N\ge t)$ for each $t<\frac x2$. In particular, we have $\mathbb P(Z_1\ge t),\mathbb P(Z_2\ge t)\le \mathbb P(N\ge t)$ for each $t$.

We can now define $(X_1,X_2)$ to be $(Z_1,Z_2)$ when the pair is finite, and to "fill in" the remaining probability (only on pairs with both coordinates less than $\frac x2$) to have the correct marginals. We see that $\mathbb P(X_1+X_2\ge x)=\mathbb P(X_1+X_2=x)=2\mathbb P(N\ge \frac x2)$.

Hence it is possible for $\mathbb P(X_1+X_2\ge x)$ to be as large as $2\mathbb P(N\ge \frac x2)$. On the other hand, $\{X_1+X_2\ge x\}\subset\{X_1\ge \frac x2\}\cup\{X_2\ge\frac x2\}$ so that $\mathbb P(X_1+X_2\ge x)\le \mathbb P(X_1\ge \frac x2)+\mathbb P(X_2\ge \frac x2)\le 2\mathbb P(N\ge \frac x2)$, giving a matching upper bound.

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  • $\begingroup$ 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. $\endgroup$ – NN2 Jan 4 at 21:30
  • $\begingroup$ If $a+b>x$, then at least one of $a$ and $b$ exceeds $x/2$. The converse is false. Hence $\{X_1+X_2\ge x\}\subset \{X_1\ge \frac x2\}\cup \{X_2\ge \frac x2\}$. I think the definition gives $f_{Z_1}(t)=f_N(t)$ for $t\ge \frac x2$ and $f_{Z_2}(t)=f_N(t)$ also for $t\ge \frac x2$. $\endgroup$ – Anthony Quas Jan 4 at 21:56
  • $\begingroup$ 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. $\endgroup$ – NN2 Jan 11 at 2:03
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I found the answer in the theorem 4.9, Bounds for Distribution Functions of Sums of n Random Variables, chapter 4 of Cherubini, Copula Methods in Finance (there is no proof).

Given $S = \sum_{i=1}^n w_i X_i $, the term $P(S \leq s) $ has the lower bound and upper bound as follows

Lower bound: $$F_L (s) = \sup_{\sum_{i=1}^n t_i =s} \max \{ \sum_{i=1}^n F_{w_i X_i}(t_i)-(n-1) ,0 \}$$

Upper bound: $$F_U (s) = \inf_{\sum_{i=1}^n t_i =s} \min \{ \sum_{i=1}^n F_{w_i X_i}(t_i) ,1 \}$$

In our particular case, where $X_i$ follows $N(0,1)$ with $i=1,...,n$, we have

$$F_L (s) = \sup_{\sum_{i=1}^n t_i =s} \max \{ \sum_{i=1}^n \Phi(\frac{t_i}{w_i})-(n-1) ,0 \}$$ $$F_U (s) = \inf_{\sum_{i=1}^n t_i =s} \min \{ \sum_{i=1}^n \Phi(\frac{t_i}{w_i}) ,1 \}$$

I don't think we can have close-form expression for these bounds. $$$$

For the second question, if we know the covariance matrix $\mathbf{\Omega}$. We can make a change of variables as $$S = \sum_{i=1}^n w_i X_i = \mathbf{w}^T \mathbf{X} = \mathbf{w}^T \sqrt{\mathbf{\Omega} }\mathbf{Z} = \sum_{i=1}^n w'_i Z_i$$ where $Z_i$ are $n$ gaussian random variables $N(0,1)$ with $i=1,...,n$, $w'_i$ are the element $i$ of the vector $ \sqrt{\mathbf{\Omega}} \mathbf{w}$, $\sqrt{\mathbf{\Omega}}$ is the square root of the matrix $\mathbf{\Omega}$.

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