0
$\begingroup$

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.

$\endgroup$
10
  • 1
    $\begingroup$ For your first question, the extremes are that they are all the same and that they cancel out identically. $\endgroup$ Dec 27, 2020 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, 2020 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 $\mathbb{P}[S \le 2] = 1$. $\endgroup$ Dec 27, 2020 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? $\endgroup$ Dec 27, 2020 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'. $\endgroup$ Dec 27, 2020 at 12:30

2 Answers 2

2
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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, 2021 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 $t\ge \frac x2$. $\endgroup$ Jan 4, 2021 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, 2021 at 2:03
0
$\begingroup$

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 $\mathbf{\Omega}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.