Mixing
Is there a way to turn $mathbfE[(A/B)^2]$ into $mathbfE[A]-mathbfE[B]$ with inequality?
Clash Royale CLAN TAG#URR8PPP
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I try to upper-bound an error term $X$ with two random variables $A$ and $B$, but I only have closed form for $mathbfE[A] - mathbfE[B] = c$, and I derived $X$ into the following term:
$$
mathbfE[X] leq n - dfrac11+mathbfEleft[ left(dfracBA right)^2 right]
$$
where $n$ is a constant.
I wonder if there are any tricks or inequalities to use to push this further?
So far I know that $A>0$, $B>0$, and $mathbfE[A]>mathbfE[B]$
Edit:
Thanks for Robert pointing out that the information is not enough, so let me add in all the information I have:
Given $s$ probabilities $pi_1, pi_2, dots pi_s$, which are sampled from the distribution $Beta(alpha_s, beta_s)$ respectively. And both $alpha_s$ and $beta_s$ are sampled from the same distribution $mathcalP$ for all $s$.
For each probability $pi_t$, we sample $n$ random variables with Binomial:
$$
Y_i,t = Bin(2, pi_t)
$$
Therefore, there are total $ns$ Binomial random variables.
We define a set $mathcalT$, whose element is a pair of indices. A pair of indices is in $mathcalT$ if and only if the corresponding Binomial random variables are sampled from the same probability. In other words, for any two $Y_i, t$ and $Y_j, t'$, $(i, j) in mathcalT$ if and only if $t=t'$.
Now, back into the original question:
$$
A = dfrac1msum_(i,j)inmathcalTY_i,cdotY_j,cdot
$$
$$
B = dfrac1m'sum_(i,j)notinmathcalTY_i,cdotY_j,cdot
$$
where $m$ is the number of elements in $mathcalT$ and $m'$ is the total number of the rest pairs.
inequality
asked Aug 2 at 2:09
Haohan Wang
208110
add a comment |Â
up vote
0
down vote
favorite
I try to upper-bound an error term $X$ with two random variables $A$ and $B$, but I only have closed form for $mathbfE[A] - mathbfE[B] = c$, and I derived $X$ into the following term:
$$
mathbfE[X] leq n - dfrac11+mathbfEleft[ left(dfracBA right)^2 right]
$$
where $n$ is a constant.
I wonder if there are any tricks or inequalities to use to push this further?
So far I know that $A>0$, $B>0$, and $mathbfE[A]>mathbfE[B]$
Edit:
Thanks for Robert pointing out that the information is not enough, so let me add in all the information I have:
Given $s$ probabilities $pi_1, pi_2, dots pi_s$, which are sampled from the distribution $Beta(alpha_s, beta_s)$ respectively. And both $alpha_s$ and $beta_s$ are sampled from the same distribution $mathcalP$ for all $s$.
For each probability $pi_t$, we sample $n$ random variables with Binomial:
$$
Y_i,t = Bin(2, pi_t)
$$
Therefore, there are total $ns$ Binomial random variables.
We define a set $mathcalT$, whose element is a pair of indices. A pair of indices is in $mathcalT$ if and only if the corresponding Binomial random variables are sampled from the same probability. In other words, for any two $Y_i, t$ and $Y_j, t'$, $(i, j) in mathcalT$ if and only if $t=t'$.
Now, back into the original question:
$$
A = dfrac1msum_(i,j)inmathcalTY_i,cdotY_j,cdot
$$
$$
B = dfrac1m'sum_(i,j)notinmathcalTY_i,cdotY_j,cdot
$$
where $m$ is the number of elements in $mathcalT$ and $m'$ is the total number of the rest pairs.
inequality
asked Aug 2 at 2:09
Haohan Wang
208110
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I try to upper-bound an error term $X$ with two random variables $A$ and $B$, but I only have closed form for $mathbfE[A] - mathbfE[B] = c$, and I derived $X$ into the following term:
$$
mathbfE[X] leq n - dfrac11+mathbfEleft[ left(dfracBA right)^2 right]
$$
where $n$ is a constant.
I wonder if there are any tricks or inequalities to use to push this further?
So far I know that $A>0$, $B>0$, and $mathbfE[A]>mathbfE[B]$
Edit:
Thanks for Robert pointing out that the information is not enough, so let me add in all the information I have:
Given $s$ probabilities $pi_1, pi_2, dots pi_s$, which are sampled from the distribution $Beta(alpha_s, beta_s)$ respectively. And both $alpha_s$ and $beta_s$ are sampled from the same distribution $mathcalP$ for all $s$.
For each probability $pi_t$, we sample $n$ random variables with Binomial:
$$
Y_i,t = Bin(2, pi_t)
$$
Therefore, there are total $ns$ Binomial random variables.
We define a set $mathcalT$, whose element is a pair of indices. A pair of indices is in $mathcalT$ if and only if the corresponding Binomial random variables are sampled from the same probability. In other words, for any two $Y_i, t$ and $Y_j, t'$, $(i, j) in mathcalT$ if and only if $t=t'$.
Now, back into the original question:
$$
A = dfrac1msum_(i,j)inmathcalTY_i,cdotY_j,cdot
$$
$$
B = dfrac1m'sum_(i,j)notinmathcalTY_i,cdotY_j,cdot
$$
where $m$ is the number of elements in $mathcalT$ and $m'$ is the total number of the rest pairs.
inequality
asked Aug 2 at 2:09
Haohan Wang
208110
I try to upper-bound an error term $X$ with two random variables $A$ and $B$, but I only have closed form for $mathbfE[A] - mathbfE[B] = c$, and I derived $X$ into the following term:
$$
mathbfE[X] leq n - dfrac11+mathbfEleft[ left(dfracBA right)^2 right]
$$
where $n$ is a constant.
I wonder if there are any tricks or inequalities to use to push this further?
So far I know that $A>0$, $B>0$, and $mathbfE[A]>mathbfE[B]$
Edit:
Thanks for Robert pointing out that the information is not enough, so let me add in all the information I have:
Given $s$ probabilities $pi_1, pi_2, dots pi_s$, which are sampled from the distribution $Beta(alpha_s, beta_s)$ respectively. And both $alpha_s$ and $beta_s$ are sampled from the same distribution $mathcalP$ for all $s$.
For each probability $pi_t$, we sample $n$ random variables with Binomial:
$$
Y_i,t = Bin(2, pi_t)
$$
Therefore, there are total $ns$ Binomial random variables.
We define a set $mathcalT$, whose element is a pair of indices. A pair of indices is in $mathcalT$ if and only if the corresponding Binomial random variables are sampled from the same probability. In other words, for any two $Y_i, t$ and $Y_j, t'$, $(i, j) in mathcalT$ if and only if $t=t'$.
Now, back into the original question:
$$
A = dfrac1msum_(i,j)inmathcalTY_i,cdotY_j,cdot
$$
$$
B = dfrac1m'sum_(i,j)notinmathcalTY_i,cdotY_j,cdot
$$
where $m$ is the number of elements in $mathcalT$ and $m'$ is the total number of the rest pairs.
inequality
asked Aug 2 at 2:09
Haohan Wang
208110
edited Aug 3 at 14:39
asked Aug 2 at 2:09
Haohan Wang
208110
asked Aug 2 at 2:09
Haohan Wang
208110
asked Aug 2 at 2:09
Haohan Wang
208110
208110
add a comment |Â
add a comment |Â
1 Answer
1
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oldest
votes
up vote
1
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The given conditions on $A$ and $B$ don't let you say much about $E[(B/A)^2]$, only that it is positive.
Scenario 1: with probability $1$, $B = varepsilon$ and $A = c + varepsilon$, where $varepsilon > 0$ is small.
Then $E[(B/A)^2] = (varepsilon/(c+varepsilon))^2$ is small.
Scenario 2: $B=1$; with probability $1-varepsilon$, $A = varepsilon$ and with probability $varepsilon$, $A = (1+c)/varepsilon - 1 + varepsilon$.
The $E[(B/A)^2] approx (1+2c)/varepsilon^2$ is large.
answered Aug 2 at 2:41
Robert Israel
303k22200440
Thanks. I added all the information I have back to the original question. Could you please have a look?
– Haohan Wang
Aug 3 at 14:40
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
The given conditions on $A$ and $B$ don't let you say much about $E[(B/A)^2]$, only that it is positive.
Scenario 1: with probability $1$, $B = varepsilon$ and $A = c + varepsilon$, where $varepsilon > 0$ is small.
Then $E[(B/A)^2] = (varepsilon/(c+varepsilon))^2$ is small.
Scenario 2: $B=1$; with probability $1-varepsilon$, $A = varepsilon$ and with probability $varepsilon$, $A = (1+c)/varepsilon - 1 + varepsilon$.
The $E[(B/A)^2] approx (1+2c)/varepsilon^2$ is large.
answered Aug 2 at 2:41
Robert Israel
303k22200440
Thanks. I added all the information I have back to the original question. Could you please have a look?
– Haohan Wang
Aug 3 at 14:40
add a comment |Â
up vote
1
down vote
The given conditions on $A$ and $B$ don't let you say much about $E[(B/A)^2]$, only that it is positive.
Scenario 1: with probability $1$, $B = varepsilon$ and $A = c + varepsilon$, where $varepsilon > 0$ is small.
Then $E[(B/A)^2] = (varepsilon/(c+varepsilon))^2$ is small.
Scenario 2: $B=1$; with probability $1-varepsilon$, $A = varepsilon$ and with probability $varepsilon$, $A = (1+c)/varepsilon - 1 + varepsilon$.
The $E[(B/A)^2] approx (1+2c)/varepsilon^2$ is large.
answered Aug 2 at 2:41
Robert Israel
303k22200440
Thanks. I added all the information I have back to the original question. Could you please have a look?
– Haohan Wang
Aug 3 at 14:40
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The given conditions on $A$ and $B$ don't let you say much about $E[(B/A)^2]$, only that it is positive.
Scenario 1: with probability $1$, $B = varepsilon$ and $A = c + varepsilon$, where $varepsilon > 0$ is small.
Then $E[(B/A)^2] = (varepsilon/(c+varepsilon))^2$ is small.
Scenario 2: $B=1$; with probability $1-varepsilon$, $A = varepsilon$ and with probability $varepsilon$, $A = (1+c)/varepsilon - 1 + varepsilon$.
The $E[(B/A)^2] approx (1+2c)/varepsilon^2$ is large.
answered Aug 2 at 2:41
Robert Israel
303k22200440
The given conditions on $A$ and $B$ don't let you say much about $E[(B/A)^2]$, only that it is positive.
Scenario 1: with probability $1$, $B = varepsilon$ and $A = c + varepsilon$, where $varepsilon > 0$ is small.
Then $E[(B/A)^2] = (varepsilon/(c+varepsilon))^2$ is small.
Scenario 2: $B=1$; with probability $1-varepsilon$, $A = varepsilon$ and with probability $varepsilon$, $A = (1+c)/varepsilon - 1 + varepsilon$.
The $E[(B/A)^2] approx (1+2c)/varepsilon^2$ is large.
answered Aug 2 at 2:41
Robert Israel
303k22200440
answered Aug 2 at 2:41
Robert Israel
303k22200440
answered Aug 2 at 2:41
Robert Israel
303k22200440
answered Aug 2 at 2:41
Robert Israel
303k22200440
303k22200440
Thanks. I added all the information I have back to the original question. Could you please have a look?
– Haohan Wang
Aug 3 at 14:40
add a comment |Â
Thanks. I added all the information I have back to the original question. Could you please have a look?
– Haohan Wang
Aug 3 at 14:40
Thanks. I added all the information I have back to the original question. Could you please have a look?
– Haohan Wang
Aug 3 at 14:40
Thanks. I added all the information I have back to the original question. Could you please have a look?
– Haohan Wang
Aug 3 at 14:40
add a comment |Â
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