Calculating the expectation of a sum of dependent random variables
$begingroup$
Let $(X_i)_{i=1}^m$ be a sequence of i.i.d. Bernoulli random variables such that $Pr(X_i=1)=p<0.5$ and $Pr(X_i=0)=1-p$. Let $(Y_i)_{i=1}^m$ be defined as follows: $Y_1=X_1$, and for $2leq ileq m$
$$
Y_i = begin{cases}
1, &mathrm{if }pleft(1-frac{1}{i-1}sum_{j=1}^{i-1}Y_jright)+(1-p)frac{1}{i-1}sum_{j=1}^{i-1}Y_j<frac{1}{i}sum_{j=1}^iX_j\
0, & mathrm{otherwise}
end{cases}
$$
Finally, define
$$
Z_i=begin{cases}
Y_i, &mathrm{w.p. } 1-p\
1-Y_i, & mathrm{w.p. } p
end{cases}
$$
for $i=1,2,ldots,m$. I want to calculate or get an upper bound on the expectation $mathbb{E}left(sum_{i=1}^mZ_iright)$ and the variance $mathrm{Var}left(sum_{i=1}^mZ_iright)$.
For any $i$, we have $mathbb{E}Z_i = p+(1-2p)mathbb{E}Y_i$. So a trivial upper bound is
$$
mathbb{E}left(sum_{i=1}^mZ_iright) =mp+(1-2p)sum_{i=1}^mmathbb{E}Y_ileq mp+m(1-2p) = m(1-p)
$$
becasue $mathbb{E}Y_ileq1$. Unfortunately, when trying to calculate explicitly $mathbb{E}Y_i$ things quickly become quite involved. Is there any way to get better bounds? Numerical calculation (as well as analytical calculation of the first few) of the means of the random variables $Y_i$ suggest that $mathbb{E}Y_ileq mathbb{E}Y_1 = p$ for any $i$, which would give
$$
mathbb{E}left(sum_{i=1}^mZ_iright)leq 2mp(1-p)
$$
However, it is not clear if $mathbb{E}Y_ileq mathbb{E}Y_1$ is indeed true.
probability probability-theory
asked Dec 10 '18 at 14:42
Mike_DMike_D
162
$endgroup$
add a comment |
$begingroup$
Let $(X_i)_{i=1}^m$ be a sequence of i.i.d. Bernoulli random variables such that $Pr(X_i=1)=p<0.5$ and $Pr(X_i=0)=1-p$. Let $(Y_i)_{i=1}^m$ be defined as follows: $Y_1=X_1$, and for $2leq ileq m$
$$
Y_i = begin{cases}
1, &mathrm{if }pleft(1-frac{1}{i-1}sum_{j=1}^{i-1}Y_jright)+(1-p)frac{1}{i-1}sum_{j=1}^{i-1}Y_j<frac{1}{i}sum_{j=1}^iX_j\
0, & mathrm{otherwise}
end{cases}
$$
Finally, define
$$
Z_i=begin{cases}
Y_i, &mathrm{w.p. } 1-p\
1-Y_i, & mathrm{w.p. } p
end{cases}
$$
for $i=1,2,ldots,m$. I want to calculate or get an upper bound on the expectation $mathbb{E}left(sum_{i=1}^mZ_iright)$ and the variance $mathrm{Var}left(sum_{i=1}^mZ_iright)$.
For any $i$, we have $mathbb{E}Z_i = p+(1-2p)mathbb{E}Y_i$. So a trivial upper bound is
$$
mathbb{E}left(sum_{i=1}^mZ_iright) =mp+(1-2p)sum_{i=1}^mmathbb{E}Y_ileq mp+m(1-2p) = m(1-p)
$$
becasue $mathbb{E}Y_ileq1$. Unfortunately, when trying to calculate explicitly $mathbb{E}Y_i$ things quickly become quite involved. Is there any way to get better bounds? Numerical calculation (as well as analytical calculation of the first few) of the means of the random variables $Y_i$ suggest that $mathbb{E}Y_ileq mathbb{E}Y_1 = p$ for any $i$, which would give
$$
mathbb{E}left(sum_{i=1}^mZ_iright)leq 2mp(1-p)
$$
However, it is not clear if $mathbb{E}Y_ileq mathbb{E}Y_1$ is indeed true.
probability probability-theory
asked Dec 10 '18 at 14:42
Mike_DMike_D
162
$endgroup$
$begingroup$
So actually the actual question here is to compute (or bound) $E[Y_i]$, what follows is pretty irrelevant, right?
$endgroup$
– leonbloy
Dec 10 '18 at 16:30
$begingroup$
@leonbloy Yes, for the calculation of the mean of $sum_iZ_i$ (I'm also curious about the variance) it is suffice to compute or bound $E[Y_i]$ for any $i$.
$endgroup$
– Mike_D
Dec 10 '18 at 16:33
$begingroup$
This looks interesting. Could you elaborate on the interpretation of the success criteria for $Y_i~$, the $p(1-frac{Y_{i-1}}{i-1})+(1-p)frac{Y_{i-1}}{i-1}<frac{1}{i}sum_{j=1}^iX_j$? Does it come from a geometric or combinatorial scheme?
$endgroup$
– Lee David Chung Lin
Dec 10 '18 at 16:56
$begingroup$
@LeeDavidChungLin Yes. First note that I've fixed a typo in this condition. The meaning condition is if the relative number of 1's up to time $i$ in the sequence $(X_1,ldots,X_i)$ is greater than the binary convolution of the parameter $p$ with the relative number of 1's up to time $i-1$ in the sequence $(Y_1,ldots,Y_{i-1})$, then $Y_i=1$. Note that $sum_iZ_i$ is in fact just $mathrm{Binomial}(sum_{i=1}^mY_i,1-p)+mathrm{Binomial}(m-sum_{i=1}^mY_i,p)$ whose expectation, given the $sum_{i=1}^mY_i$ is exactly $m$ times the left hand side of the success criteria for $Y_m$.
$endgroup$
– Mike_D
Dec 10 '18 at 17:11
$begingroup$
oh, binary convolution, I see. Thanks for clarifying.
$endgroup$
– Lee David Chung Lin
Dec 10 '18 at 17:13
add a comment |
$begingroup$
Let $(X_i)_{i=1}^m$ be a sequence of i.i.d. Bernoulli random variables such that $Pr(X_i=1)=p<0.5$ and $Pr(X_i=0)=1-p$. Let $(Y_i)_{i=1}^m$ be defined as follows: $Y_1=X_1$, and for $2leq ileq m$
$$
Y_i = begin{cases}
1, &mathrm{if }pleft(1-frac{1}{i-1}sum_{j=1}^{i-1}Y_jright)+(1-p)frac{1}{i-1}sum_{j=1}^{i-1}Y_j<frac{1}{i}sum_{j=1}^iX_j\
0, & mathrm{otherwise}
end{cases}
$$
Finally, define
$$
Z_i=begin{cases}
Y_i, &mathrm{w.p. } 1-p\
1-Y_i, & mathrm{w.p. } p
end{cases}
$$
for $i=1,2,ldots,m$. I want to calculate or get an upper bound on the expectation $mathbb{E}left(sum_{i=1}^mZ_iright)$ and the variance $mathrm{Var}left(sum_{i=1}^mZ_iright)$.
For any $i$, we have $mathbb{E}Z_i = p+(1-2p)mathbb{E}Y_i$. So a trivial upper bound is
$$
mathbb{E}left(sum_{i=1}^mZ_iright) =mp+(1-2p)sum_{i=1}^mmathbb{E}Y_ileq mp+m(1-2p) = m(1-p)
$$
becasue $mathbb{E}Y_ileq1$. Unfortunately, when trying to calculate explicitly $mathbb{E}Y_i$ things quickly become quite involved. Is there any way to get better bounds? Numerical calculation (as well as analytical calculation of the first few) of the means of the random variables $Y_i$ suggest that $mathbb{E}Y_ileq mathbb{E}Y_1 = p$ for any $i$, which would give
$$
mathbb{E}left(sum_{i=1}^mZ_iright)leq 2mp(1-p)
$$
However, it is not clear if $mathbb{E}Y_ileq mathbb{E}Y_1$ is indeed true.
probability probability-theory
asked Dec 10 '18 at 14:42
Mike_DMike_D
162
$endgroup$
Let $(X_i)_{i=1}^m$ be a sequence of i.i.d. Bernoulli random variables such that $Pr(X_i=1)=p<0.5$ and $Pr(X_i=0)=1-p$. Let $(Y_i)_{i=1}^m$ be defined as follows: $Y_1=X_1$, and for $2leq ileq m$
$$
Y_i = begin{cases}
1, &mathrm{if }pleft(1-frac{1}{i-1}sum_{j=1}^{i-1}Y_jright)+(1-p)frac{1}{i-1}sum_{j=1}^{i-1}Y_j<frac{1}{i}sum_{j=1}^iX_j\
0, & mathrm{otherwise}
end{cases}
$$
Finally, define
$$
Z_i=begin{cases}
Y_i, &mathrm{w.p. } 1-p\
1-Y_i, & mathrm{w.p. } p
end{cases}
$$
for $i=1,2,ldots,m$. I want to calculate or get an upper bound on the expectation $mathbb{E}left(sum_{i=1}^mZ_iright)$ and the variance $mathrm{Var}left(sum_{i=1}^mZ_iright)$.
For any $i$, we have $mathbb{E}Z_i = p+(1-2p)mathbb{E}Y_i$. So a trivial upper bound is
$$
mathbb{E}left(sum_{i=1}^mZ_iright) =mp+(1-2p)sum_{i=1}^mmathbb{E}Y_ileq mp+m(1-2p) = m(1-p)
$$
becasue $mathbb{E}Y_ileq1$. Unfortunately, when trying to calculate explicitly $mathbb{E}Y_i$ things quickly become quite involved. Is there any way to get better bounds? Numerical calculation (as well as analytical calculation of the first few) of the means of the random variables $Y_i$ suggest that $mathbb{E}Y_ileq mathbb{E}Y_1 = p$ for any $i$, which would give
$$
mathbb{E}left(sum_{i=1}^mZ_iright)leq 2mp(1-p)
$$
However, it is not clear if $mathbb{E}Y_ileq mathbb{E}Y_1$ is indeed true.
probability probability-theory
probability probability-theory
asked Dec 10 '18 at 14:42
Mike_DMike_D
162
asked Dec 10 '18 at 14:42
Mike_DMike_D
162
edited Dec 11 '18 at 12:21
Mike_D
asked Dec 10 '18 at 14:42
Mike_DMike_D
162
asked Dec 10 '18 at 14:42
Mike_DMike_D
162
asked Dec 10 '18 at 14:42
Mike_DMike_D
162
162
$begingroup$
So actually the actual question here is to compute (or bound) $E[Y_i]$, what follows is pretty irrelevant, right?
$endgroup$
– leonbloy
Dec 10 '18 at 16:30
$begingroup$
@leonbloy Yes, for the calculation of the mean of $sum_iZ_i$ (I'm also curious about the variance) it is suffice to compute or bound $E[Y_i]$ for any $i$.
$endgroup$
– Mike_D
Dec 10 '18 at 16:33
$begingroup$
This looks interesting. Could you elaborate on the interpretation of the success criteria for $Y_i~$, the $p(1-frac{Y_{i-1}}{i-1})+(1-p)frac{Y_{i-1}}{i-1}<frac{1}{i}sum_{j=1}^iX_j$? Does it come from a geometric or combinatorial scheme?
$endgroup$
– Lee David Chung Lin
Dec 10 '18 at 16:56
$begingroup$
@LeeDavidChungLin Yes. First note that I've fixed a typo in this condition. The meaning condition is if the relative number of 1's up to time $i$ in the sequence $(X_1,ldots,X_i)$ is greater than the binary convolution of the parameter $p$ with the relative number of 1's up to time $i-1$ in the sequence $(Y_1,ldots,Y_{i-1})$, then $Y_i=1$. Note that $sum_iZ_i$ is in fact just $mathrm{Binomial}(sum_{i=1}^mY_i,1-p)+mathrm{Binomial}(m-sum_{i=1}^mY_i,p)$ whose expectation, given the $sum_{i=1}^mY_i$ is exactly $m$ times the left hand side of the success criteria for $Y_m$.
$endgroup$
– Mike_D
Dec 10 '18 at 17:11
$begingroup$
oh, binary convolution, I see. Thanks for clarifying.
$endgroup$
– Lee David Chung Lin
Dec 10 '18 at 17:13
add a comment |
$begingroup$
So actually the actual question here is to compute (or bound) $E[Y_i]$, what follows is pretty irrelevant, right?
$endgroup$
– leonbloy
Dec 10 '18 at 16:30
$begingroup$
@leonbloy Yes, for the calculation of the mean of $sum_iZ_i$ (I'm also curious about the variance) it is suffice to compute or bound $E[Y_i]$ for any $i$.
$endgroup$
– Mike_D
Dec 10 '18 at 16:33
$begingroup$
This looks interesting. Could you elaborate on the interpretation of the success criteria for $Y_i~$, the $p(1-frac{Y_{i-1}}{i-1})+(1-p)frac{Y_{i-1}}{i-1}<frac{1}{i}sum_{j=1}^iX_j$? Does it come from a geometric or combinatorial scheme?
$endgroup$
– Lee David Chung Lin
Dec 10 '18 at 16:56
$begingroup$
@LeeDavidChungLin Yes. First note that I've fixed a typo in this condition. The meaning condition is if the relative number of 1's up to time $i$ in the sequence $(X_1,ldots,X_i)$ is greater than the binary convolution of the parameter $p$ with the relative number of 1's up to time $i-1$ in the sequence $(Y_1,ldots,Y_{i-1})$, then $Y_i=1$. Note that $sum_iZ_i$ is in fact just $mathrm{Binomial}(sum_{i=1}^mY_i,1-p)+mathrm{Binomial}(m-sum_{i=1}^mY_i,p)$ whose expectation, given the $sum_{i=1}^mY_i$ is exactly $m$ times the left hand side of the success criteria for $Y_m$.
$endgroup$
– Mike_D
Dec 10 '18 at 17:11
$begingroup$
oh, binary convolution, I see. Thanks for clarifying.
$endgroup$
– Lee David Chung Lin
Dec 10 '18 at 17:13
$begingroup$
So actually the actual question here is to compute (or bound) $E[Y_i]$, what follows is pretty irrelevant, right?
$endgroup$
– leonbloy
Dec 10 '18 at 16:30
$begingroup$
So actually the actual question here is to compute (or bound) $E[Y_i]$, what follows is pretty irrelevant, right?
$endgroup$
– leonbloy
Dec 10 '18 at 16:30
$begingroup$
@leonbloy Yes, for the calculation of the mean of $sum_iZ_i$ (I'm also curious about the variance) it is suffice to compute or bound $E[Y_i]$ for any $i$.
$endgroup$
– Mike_D
Dec 10 '18 at 16:33
$begingroup$
@leonbloy Yes, for the calculation of the mean of $sum_iZ_i$ (I'm also curious about the variance) it is suffice to compute or bound $E[Y_i]$ for any $i$.
$endgroup$
– Mike_D
Dec 10 '18 at 16:33
$begingroup$
This looks interesting. Could you elaborate on the interpretation of the success criteria for $Y_i~$, the $p(1-frac{Y_{i-1}}{i-1})+(1-p)frac{Y_{i-1}}{i-1}<frac{1}{i}sum_{j=1}^iX_j$? Does it come from a geometric or combinatorial scheme?
$endgroup$
– Lee David Chung Lin
Dec 10 '18 at 16:56
$begingroup$
This looks interesting. Could you elaborate on the interpretation of the success criteria for $Y_i~$, the $p(1-frac{Y_{i-1}}{i-1})+(1-p)frac{Y_{i-1}}{i-1}<frac{1}{i}sum_{j=1}^iX_j$? Does it come from a geometric or combinatorial scheme?
$endgroup$
– Lee David Chung Lin
Dec 10 '18 at 16:56
$begingroup$
@LeeDavidChungLin Yes. First note that I've fixed a typo in this condition. The meaning condition is if the relative number of 1's up to time $i$ in the sequence $(X_1,ldots,X_i)$ is greater than the binary convolution of the parameter $p$ with the relative number of 1's up to time $i-1$ in the sequence $(Y_1,ldots,Y_{i-1})$, then $Y_i=1$. Note that $sum_iZ_i$ is in fact just $mathrm{Binomial}(sum_{i=1}^mY_i,1-p)+mathrm{Binomial}(m-sum_{i=1}^mY_i,p)$ whose expectation, given the $sum_{i=1}^mY_i$ is exactly $m$ times the left hand side of the success criteria for $Y_m$.
$endgroup$
– Mike_D
Dec 10 '18 at 17:11
$begingroup$
@LeeDavidChungLin Yes. First note that I've fixed a typo in this condition. The meaning condition is if the relative number of 1's up to time $i$ in the sequence $(X_1,ldots,X_i)$ is greater than the binary convolution of the parameter $p$ with the relative number of 1's up to time $i-1$ in the sequence $(Y_1,ldots,Y_{i-1})$, then $Y_i=1$. Note that $sum_iZ_i$ is in fact just $mathrm{Binomial}(sum_{i=1}^mY_i,1-p)+mathrm{Binomial}(m-sum_{i=1}^mY_i,p)$ whose expectation, given the $sum_{i=1}^mY_i$ is exactly $m$ times the left hand side of the success criteria for $Y_m$.
$endgroup$
– Mike_D
Dec 10 '18 at 17:11
$begingroup$
oh, binary convolution, I see. Thanks for clarifying.
$endgroup$
– Lee David Chung Lin
Dec 10 '18 at 17:13
$begingroup$
oh, binary convolution, I see. Thanks for clarifying.
$endgroup$
– Lee David Chung Lin
Dec 10 '18 at 17:13
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Not an answer. From simulations, it seems that $E[Y_i]$ has a suprisingly wild (chaotic?) behaviour, we should not expect to get a closed formula for it - nor even attempt to prove it's monotonous. Here's a graph for $p=0.15$ and $p=0.19$.
Now, in case you're thinking: "statistical noise"... it doesn't seem to be. If you look closely, I have drawn two graphs (orange and blue lines) for $p=0.15$ , for different runs ($10^6$ tries each), they are almost undistinguishable (look around $i=85$).
What seems to be true (already conjectured) is that $E[Y_i]le p$, for $p < frac12$ and $E[Y_i]ge p$, for $p > frac12$
answered Dec 11 '18 at 16:36
leonbloyleonbloy
40.9k645107
$endgroup$
$begingroup$
Thanks! Indeed this is exactly the behavior that I notices from my simulations too. The question is then whether it can be shown that $E(Y_i)leq p$ for $p<0.5$.
$endgroup$
– Mike_D
Dec 11 '18 at 19:04
add a comment |
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1 Answer
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$begingroup$
Not an answer. From simulations, it seems that $E[Y_i]$ has a suprisingly wild (chaotic?) behaviour, we should not expect to get a closed formula for it - nor even attempt to prove it's monotonous. Here's a graph for $p=0.15$ and $p=0.19$.
Now, in case you're thinking: "statistical noise"... it doesn't seem to be. If you look closely, I have drawn two graphs (orange and blue lines) for $p=0.15$ , for different runs ($10^6$ tries each), they are almost undistinguishable (look around $i=85$).
What seems to be true (already conjectured) is that $E[Y_i]le p$, for $p < frac12$ and $E[Y_i]ge p$, for $p > frac12$
answered Dec 11 '18 at 16:36
leonbloyleonbloy
40.9k645107
$endgroup$
$begingroup$
Thanks! Indeed this is exactly the behavior that I notices from my simulations too. The question is then whether it can be shown that $E(Y_i)leq p$ for $p<0.5$.
$endgroup$
– Mike_D
Dec 11 '18 at 19:04
add a comment |
$begingroup$
Not an answer. From simulations, it seems that $E[Y_i]$ has a suprisingly wild (chaotic?) behaviour, we should not expect to get a closed formula for it - nor even attempt to prove it's monotonous. Here's a graph for $p=0.15$ and $p=0.19$.
Now, in case you're thinking: "statistical noise"... it doesn't seem to be. If you look closely, I have drawn two graphs (orange and blue lines) for $p=0.15$ , for different runs ($10^6$ tries each), they are almost undistinguishable (look around $i=85$).
What seems to be true (already conjectured) is that $E[Y_i]le p$, for $p < frac12$ and $E[Y_i]ge p$, for $p > frac12$
answered Dec 11 '18 at 16:36
leonbloyleonbloy
40.9k645107
$endgroup$
$begingroup$
Thanks! Indeed this is exactly the behavior that I notices from my simulations too. The question is then whether it can be shown that $E(Y_i)leq p$ for $p<0.5$.
$endgroup$
– Mike_D
Dec 11 '18 at 19:04
add a comment |
$begingroup$
Not an answer. From simulations, it seems that $E[Y_i]$ has a suprisingly wild (chaotic?) behaviour, we should not expect to get a closed formula for it - nor even attempt to prove it's monotonous. Here's a graph for $p=0.15$ and $p=0.19$.
Now, in case you're thinking: "statistical noise"... it doesn't seem to be. If you look closely, I have drawn two graphs (orange and blue lines) for $p=0.15$ , for different runs ($10^6$ tries each), they are almost undistinguishable (look around $i=85$).
What seems to be true (already conjectured) is that $E[Y_i]le p$, for $p < frac12$ and $E[Y_i]ge p$, for $p > frac12$
answered Dec 11 '18 at 16:36
leonbloyleonbloy
40.9k645107
$endgroup$
Not an answer. From simulations, it seems that $E[Y_i]$ has a suprisingly wild (chaotic?) behaviour, we should not expect to get a closed formula for it - nor even attempt to prove it's monotonous. Here's a graph for $p=0.15$ and $p=0.19$.
Now, in case you're thinking: "statistical noise"... it doesn't seem to be. If you look closely, I have drawn two graphs (orange and blue lines) for $p=0.15$ , for different runs ($10^6$ tries each), they are almost undistinguishable (look around $i=85$).
What seems to be true (already conjectured) is that $E[Y_i]le p$, for $p < frac12$ and $E[Y_i]ge p$, for $p > frac12$
answered Dec 11 '18 at 16:36
leonbloyleonbloy
40.9k645107
edited Dec 11 '18 at 20:47
answered Dec 11 '18 at 16:36
leonbloyleonbloy
40.9k645107
answered Dec 11 '18 at 16:36
leonbloyleonbloy
40.9k645107
answered Dec 11 '18 at 16:36
leonbloyleonbloy
40.9k645107
40.9k645107
$begingroup$
Thanks! Indeed this is exactly the behavior that I notices from my simulations too. The question is then whether it can be shown that $E(Y_i)leq p$ for $p<0.5$.
$endgroup$
– Mike_D
Dec 11 '18 at 19:04
add a comment |
$begingroup$
Thanks! Indeed this is exactly the behavior that I notices from my simulations too. The question is then whether it can be shown that $E(Y_i)leq p$ for $p<0.5$.
$endgroup$
– Mike_D
Dec 11 '18 at 19:04
$begingroup$
Thanks! Indeed this is exactly the behavior that I notices from my simulations too. The question is then whether it can be shown that $E(Y_i)leq p$ for $p<0.5$.
$endgroup$
– Mike_D
Dec 11 '18 at 19:04
$begingroup$
Thanks! Indeed this is exactly the behavior that I notices from my simulations too. The question is then whether it can be shown that $E(Y_i)leq p$ for $p<0.5$.
$endgroup$
– Mike_D
Dec 11 '18 at 19:04
add a comment |
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So actually the actual question here is to compute (or bound) $E[Y_i]$, what follows is pretty irrelevant, right?
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– leonbloy
Dec 10 '18 at 16:30
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@leonbloy Yes, for the calculation of the mean of $sum_iZ_i$ (I'm also curious about the variance) it is suffice to compute or bound $E[Y_i]$ for any $i$.
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– Mike_D
Dec 10 '18 at 16:33
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This looks interesting. Could you elaborate on the interpretation of the success criteria for $Y_i~$, the $p(1-frac{Y_{i-1}}{i-1})+(1-p)frac{Y_{i-1}}{i-1}<frac{1}{i}sum_{j=1}^iX_j$? Does it come from a geometric or combinatorial scheme?
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– Lee David Chung Lin
Dec 10 '18 at 16:56
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@LeeDavidChungLin Yes. First note that I've fixed a typo in this condition. The meaning condition is if the relative number of 1's up to time $i$ in the sequence $(X_1,ldots,X_i)$ is greater than the binary convolution of the parameter $p$ with the relative number of 1's up to time $i-1$ in the sequence $(Y_1,ldots,Y_{i-1})$, then $Y_i=1$. Note that $sum_iZ_i$ is in fact just $mathrm{Binomial}(sum_{i=1}^mY_i,1-p)+mathrm{Binomial}(m-sum_{i=1}^mY_i,p)$ whose expectation, given the $sum_{i=1}^mY_i$ is exactly $m$ times the left hand side of the success criteria for $Y_m$.
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– Mike_D
Dec 10 '18 at 17:11
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oh, binary convolution, I see. Thanks for clarifying.
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– Lee David Chung Lin
Dec 10 '18 at 17:13