Timeline for Bounds for the sum of dependent gaussian random variables
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| Jan 11, 2021 at 8:48 | vote | accept | NN2 | ||
| Jan 11, 2021 at 2:04 | comment | added | NN2 | 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. | |
| Jan 11, 2021 at 2:01 | answer | added | NN2 | timeline score: 0 | |
| Jan 4, 2021 at 18:29 | comment | added | Anthony Quas | @BrendanMcKay: I gave a similar construction for Gaussians. Quite interestingly this doubles the probability of the sum being $x$ or greater compared to making the two variables equal; and this bound is sharp. It doesn't specifically rely on the Gaussian property. All that's used is $\rho(t)\le \rho(x-t)$ for $t\ge \frac x2$. | |
| Jan 4, 2021 at 18:26 | answer | added | Anthony Quas | timeline score: 2 | |
| Dec 28, 2020 at 13:23 | comment | added | NN2 | 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 $1$. | |
| Dec 28, 2020 at 12:04 | comment | added | Brendan McKay | @AnthonyQuas Nice example, but can that happen for gaussian variables? I didn't manage to prove it either way. | |
| Dec 28, 2020 at 11:50 | comment | added | Anthony Quas | @BrendanMcKay : I don’t think this is right. If $x$ is 2, say, then by making the rvs equal, you’re “wasting” some bigness if $X_1>1$. For a discrete example, if you are rolling 2 dice and $x=11$, it’s better if you couple 5 and 6 (if $X_1=5$, then $X_2=6$ and v.v. (Probability of exceeding 11 is 1/18) vs $X_1=X_2$ (probability of exceeding 11 is 1/36). | |
| Dec 27, 2020 at 12:30 | comment | added | Mark Wildon | @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'. | |
| Dec 27, 2020 at 12:24 | comment | added | Brendan McKay | @MarkWildon I think $P[S\le x]$ should be minimised if the variables are identical, for $x\ge 0$. Not maximized. Is it wrong? | |
| Dec 27, 2020 at 11:57 | comment | added | Mark Wildon | 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 $\mathbb{P}[S \le 2] = 1$. | |
| Dec 27, 2020 at 11:43 | comment | added | NN2 | 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 \} $ ) ? | |
| Dec 27, 2020 at 11:29 | history | edited | NN2 | CC BY-SA 4.0 |
added 14 characters in body
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| Dec 27, 2020 at 3:33 | comment | added | Brendan McKay | For your first question, the extremes are that they are all the same and that they cancel out identically. | |
| Dec 27, 2020 at 2:12 | review | Close votes | |||
| Jan 13, 2021 at 14:48 | |||||
| Dec 27, 2020 at 1:49 | history | edited | NN2 |
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| Dec 27, 2020 at 1:42 | history | asked | NN2 | CC BY-SA 4.0 |