# Bounds for the sum of dependent gaussian random variables

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?

• For your first question, the extremes are that they are all the same and that they cancel out identically. – Brendan McKay Dec 27 '20 at 3:33
• 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 $C_{-} = \max \{1-\sum_i^n (1-u_i),0 \}$ ) ? – NN2 Dec 27 '20 at 11:43
• 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 $\mathbb{P}[S \le 2] = 1$. – Mark Wildon Dec 27 '20 at 11:57
• @MarkWildon I think $P[S\le x]$ should be minimised if the variables are identical, for $x\ge 0$. Not maximized. Is it wrong? – Brendan McKay Dec 27 '20 at 12:24
• @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'. – Mark Wildon Dec 27 '20 at 12:30

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.

• 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. – NN2 Jan 4 at 21:30
• 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 $t\ge \frac x2$. – Anthony Quas Jan 4 at 21:56
• 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. – NN2 Jan 11 at 2:03

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 $$\mathbf{\Omega}$$.