Timeline for Sum of Products of Rademacher random variables
Current License: CC BY-SA 3.0
24 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Jul 14, 2017 at 19:09 | history | suggested | Glorfindel | CC BY-SA 3.0 |
typo corrected, thanks removed as per https://meta.stackexchange.com/q/2950/295232
|
| Jul 14, 2017 at 18:18 | review | Suggested edits | |||
| S Jul 14, 2017 at 19:09 | |||||
| Apr 17, 2014 at 22:31 | history | edited | Alecos Papadopoulos | CC BY-SA 3.0 |
Now we got the title right!
|
| Apr 17, 2014 at 20:26 | answer | added | whuber♦ | timeline score: 7 | |
| Apr 17, 2014 at 15:37 | answer | added | wolfies | timeline score: 3 | |
| Apr 16, 2014 at 14:31 | history | edited | Alecos Papadopoulos | CC BY-SA 3.0 |
Changed title to refelct the content more accurately
|
| Apr 16, 2014 at 14:30 | answer | added | Alecos Papadopoulos | timeline score: 3 | |
| Jul 10, 2013 at 1:30 | comment | added | cardinal | Unless my eyes are deceiving me, you are considering a sum of products, not a product of sums. :-) | |
| Jul 9, 2013 at 20:33 | history | edited | user1189053 | CC BY-SA 3.0 |
added 26 characters in body
|
| Jul 9, 2013 at 20:32 | comment | added | user1189053 | I am really sorry about a mistake of mine. I thought i had mentioned uniformally above. So p = 1/2 and we can take a and b larger than any constant (if needed) for the inequality to hold | |
| Jul 9, 2013 at 19:48 | comment | added | wolfies |
@user1189053 says: I have clearly stated in my question that x1…xa,y1…yb are independent random variables ..... Sooooooo, are these independent random variables $x_1$, $x_2$, ..., $x_a$, $y_1$, $y_2$, ..., $y_b$ IDENTICAL ... do they all share the same underlying probability of success parameter $p$ ... or does the parameter $p$ vary with the random variables. And why do you denote them with lower case notation (which is usually used for realisations rather than random variables)? The question leaves much unstated.
|
|
| Jul 9, 2013 at 18:10 | comment | added | whuber♦ | I'm sorry, the question has not been clarified by your comments. I have offered counterexamples. Perhaps the case of constant random variables is confusing, so consider $2a$ iid RVs $x_i$ and $y_j$ with $\Pr(x_i=1)=\Pr(y_j=1)=1-p$ and $\Pr(x_i=-1)=\Pr(y_j=-1)=p$. Then the chance that all these variables equal $1$ equals $(1-p)^{2a}$, whence the chance that your sum of products equals $a^2$ is at least this much. We can make it as close to $1$ as we wish by choosing sufficiently small $p$. You claim that $2e^{-c(a^2-1)/a}$ is an upper bound: for sufficiently large $a$, that's clearly false. | |
| Jul 9, 2013 at 17:32 | comment | added | user1189053 | I agree its not Bernoulli so yes we can call it $\{1,-1\}$ variables and I have clearly stated in my question that $x_1 \ldots x_a,y_1 \ldots y_b$ are independent random variables, they are not samples of a single random variable and I call them X and Y because specifying them in that manner makes specifying the product easier. I hope this makes the question clear | |
| Jul 9, 2013 at 16:48 | comment | added | wolfies | Moreover, the question is wrong ... it is not the sum of Bernoullis. This is because a Bernoulli rv is either 0 or 1 ... not -1 or 1 (as you specify). If $Z$ ~ Bernoulli($p$), then your random variable $X = 2Z - 1$. | |
| Jul 9, 2013 at 16:38 | comment | added | wolfies | @user1189053 As to notation ... do you have 2 random variables $X$ and $Y$, and a sample from each ... $x_1$ ... $x_a$, and $y_1$ ... $y_b$ ... OR ... do you have $a+b$ random variables ... if the latter, why do you call them $X$ and $Y$? | |
| Jul 9, 2013 at 15:56 | comment | added | user1189053 | The probability of all the variables being 1 goes down exponentially. I dont think I understand your comment. For $a=b$ and $t=a^2 -1$ the bound I stated is quite trivially true as the probability of the sum being greater than $t^2 - 1$ is $2^{-(a-1)} \leq e^{-ln(2)c(a-1/a)}$ | |
| Jul 9, 2013 at 13:48 | comment | added | whuber♦ | There is an obvious improvement in your bound for $t \gt ab$, for then the probability must be zero. It seems to me that's a "sub-Gaussian" tail :-). It also seems your bound is incorrect: variables that are constantly $1$ satisfy the conditions of this question. For $a=b$ and $t=a^2-1$ the probability is $1$ but your bound is asymptotically $2\exp(-ca) \to 0$ as $a$ grows large. | |
| Jul 9, 2013 at 13:47 | comment | added | Dilip Sarwate | Have you considered applying the Chernoff bound directly to $S$? You might be able to do something with $$E[\exp(\lambda S]=E\left[\lambda \sum_i\sum_j X_iY_j\right]=E\left[\lambda\left(\sum_i X_i\right)\left(\sum_j Y_j\right)\right]$$ | |
| Jul 9, 2013 at 11:01 | history | tweeted | twitter.com/#!/StackStats/status/354555948397887490 | ||
| S Jul 9, 2013 at 10:05 | history | suggested | Davide Giraudo | CC BY-SA 3.0 |
improved formatting.
|
| Jul 9, 2013 at 9:54 | review | Suggested edits | |||
| S Jul 9, 2013 at 10:05 | |||||
| Jul 9, 2013 at 7:00 | history | edited | COOLSerdash | CC BY-SA 3.0 |
added 5 characters in body
|
| Jul 9, 2013 at 6:37 | review | First posts | |||
| Jul 9, 2013 at 6:54 | |||||
| Jul 9, 2013 at 6:21 | history | asked | user1189053 | CC BY-SA 3.0 |