Obtain $ sumlimits_{substack{i}}^{} mathbf{N}(x_{i}) $ from $ sumlimits_{substack{i}}^{}...
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Is it possible to obtain $ sumlimits_{substack{i}}^{} mathbf{N}(x_{i}) $ from $ sumlimits_{substack{i}}^{} dfrac{1}{mathbf{N}(x_{i})},$
where, $mathbf{N}(x_{i})=Y+1$ is a random variable with $Y sim Bin(n-1, r_{n}).$
The variables $mathbf{N}(x_{i})$ for $i=1..n$ are not independent.
calculus probability sequences-and-series probability-theory
asked Dec 14 '18 at 0:17
Mounia HamidoucheMounia Hamidouche
32419
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add a comment |
0
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Is it possible to obtain $ sumlimits_{substack{i}}^{} mathbf{N}(x_{i}) $ from $ sumlimits_{substack{i}}^{} dfrac{1}{mathbf{N}(x_{i})},$
where, $mathbf{N}(x_{i})=Y+1$ is a random variable with $Y sim Bin(n-1, r_{n}).$
The variables $mathbf{N}(x_{i})$ for $i=1..n$ are not independent.
calculus probability sequences-and-series probability-theory
asked Dec 14 '18 at 0:17
Mounia HamidoucheMounia Hamidouche
32419
$endgroup$
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What is $mathbf{N}$?
$endgroup$
– Sean Roberson
Dec 14 '18 at 0:18
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Sorry, I did a mistake. I edited the question.
$endgroup$
– Mounia Hamidouche
Dec 14 '18 at 0:19
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Is the distribution of $mathbf{N}$ known? Otherwise we may only be able to obtain an estimate via the HM-AM-GM inequalities.
$endgroup$
– Sean Roberson
Dec 14 '18 at 0:21
$begingroup$
Yes, it is a binomially distributed (a modified distribution because I know that the random variable will never take 0 as a value). But the variables are not independent.
$endgroup$
– Mounia Hamidouche
Dec 14 '18 at 0:22
1
$begingroup$
You should write all details in the question itself and not in comments.
$endgroup$
– Martund
Dec 14 '18 at 0:25
add a comment |
0
0
0
$begingroup$
Is it possible to obtain $ sumlimits_{substack{i}}^{} mathbf{N}(x_{i}) $ from $ sumlimits_{substack{i}}^{} dfrac{1}{mathbf{N}(x_{i})},$
where, $mathbf{N}(x_{i})=Y+1$ is a random variable with $Y sim Bin(n-1, r_{n}).$
The variables $mathbf{N}(x_{i})$ for $i=1..n$ are not independent.
calculus probability sequences-and-series probability-theory
asked Dec 14 '18 at 0:17
Mounia HamidoucheMounia Hamidouche
32419
$endgroup$
Is it possible to obtain $ sumlimits_{substack{i}}^{} mathbf{N}(x_{i}) $ from $ sumlimits_{substack{i}}^{} dfrac{1}{mathbf{N}(x_{i})},$
where, $mathbf{N}(x_{i})=Y+1$ is a random variable with $Y sim Bin(n-1, r_{n}).$
The variables $mathbf{N}(x_{i})$ for $i=1..n$ are not independent.
calculus probability sequences-and-series probability-theory
calculus probability sequences-and-series probability-theory
asked Dec 14 '18 at 0:17
Mounia HamidoucheMounia Hamidouche
32419
asked Dec 14 '18 at 0:17
Mounia HamidoucheMounia Hamidouche
32419
edited Dec 14 '18 at 0:28
Mounia Hamidouche
asked Dec 14 '18 at 0:17
Mounia HamidoucheMounia Hamidouche
32419
asked Dec 14 '18 at 0:17
Mounia HamidoucheMounia Hamidouche
32419
asked Dec 14 '18 at 0:17
Mounia HamidoucheMounia Hamidouche
32419
32419
$begingroup$
What is $mathbf{N}$?
$endgroup$
– Sean Roberson
Dec 14 '18 at 0:18
$begingroup$
Sorry, I did a mistake. I edited the question.
$endgroup$
– Mounia Hamidouche
Dec 14 '18 at 0:19
$begingroup$
Is the distribution of $mathbf{N}$ known? Otherwise we may only be able to obtain an estimate via the HM-AM-GM inequalities.
$endgroup$
– Sean Roberson
Dec 14 '18 at 0:21
$begingroup$
Yes, it is a binomially distributed (a modified distribution because I know that the random variable will never take 0 as a value). But the variables are not independent.
$endgroup$
– Mounia Hamidouche
Dec 14 '18 at 0:22
1
$begingroup$
You should write all details in the question itself and not in comments.
$endgroup$
– Martund
Dec 14 '18 at 0:25
add a comment |
$begingroup$
What is $mathbf{N}$?
$endgroup$
– Sean Roberson
Dec 14 '18 at 0:18
$begingroup$
Sorry, I did a mistake. I edited the question.
$endgroup$
– Mounia Hamidouche
Dec 14 '18 at 0:19
$begingroup$
Is the distribution of $mathbf{N}$ known? Otherwise we may only be able to obtain an estimate via the HM-AM-GM inequalities.
$endgroup$
– Sean Roberson
Dec 14 '18 at 0:21
$begingroup$
Yes, it is a binomially distributed (a modified distribution because I know that the random variable will never take 0 as a value). But the variables are not independent.
$endgroup$
– Mounia Hamidouche
Dec 14 '18 at 0:22
1
$begingroup$
You should write all details in the question itself and not in comments.
$endgroup$
– Martund
Dec 14 '18 at 0:25
$begingroup$
What is $mathbf{N}$?
$endgroup$
– Sean Roberson
Dec 14 '18 at 0:18
$begingroup$
What is $mathbf{N}$?
$endgroup$
– Sean Roberson
Dec 14 '18 at 0:18
$begingroup$
Sorry, I did a mistake. I edited the question.
$endgroup$
– Mounia Hamidouche
Dec 14 '18 at 0:19
$begingroup$
Sorry, I did a mistake. I edited the question.
$endgroup$
– Mounia Hamidouche
Dec 14 '18 at 0:19
$begingroup$
Is the distribution of $mathbf{N}$ known? Otherwise we may only be able to obtain an estimate via the HM-AM-GM inequalities.
$endgroup$
– Sean Roberson
Dec 14 '18 at 0:21
$begingroup$
Is the distribution of $mathbf{N}$ known? Otherwise we may only be able to obtain an estimate via the HM-AM-GM inequalities.
$endgroup$
– Sean Roberson
Dec 14 '18 at 0:21
$begingroup$
Yes, it is a binomially distributed (a modified distribution because I know that the random variable will never take 0 as a value). But the variables are not independent.
$endgroup$
– Mounia Hamidouche
Dec 14 '18 at 0:22
$begingroup$
Yes, it is a binomially distributed (a modified distribution because I know that the random variable will never take 0 as a value). But the variables are not independent.
$endgroup$
– Mounia Hamidouche
Dec 14 '18 at 0:22
1
1
$begingroup$
You should write all details in the question itself and not in comments.
$endgroup$
– Martund
Dec 14 '18 at 0:25
$begingroup$
You should write all details in the question itself and not in comments.
$endgroup$
– Martund
Dec 14 '18 at 0:25
add a comment |
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$begingroup$
What is $mathbf{N}$?
$endgroup$
– Sean Roberson
Dec 14 '18 at 0:18
$begingroup$
Sorry, I did a mistake. I edited the question.
$endgroup$
– Mounia Hamidouche
Dec 14 '18 at 0:19
$begingroup$
Is the distribution of $mathbf{N}$ known? Otherwise we may only be able to obtain an estimate via the HM-AM-GM inequalities.
$endgroup$
– Sean Roberson
Dec 14 '18 at 0:21
$begingroup$
Yes, it is a binomially distributed (a modified distribution because I know that the random variable will never take 0 as a value). But the variables are not independent.
$endgroup$
– Mounia Hamidouche
Dec 14 '18 at 0:22
1
$begingroup$
You should write all details in the question itself and not in comments.
$endgroup$
– Martund
Dec 14 '18 at 0:25