What is the first absolute central moment for number of succeses in Bernoulli scheme? The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Negative binomial distribution - deriving of the p.m.f. combinatoriallyBernoulli trials required for k successesProbability distribution for the number of successes for $N$ distinct trials with distinct probabilities of success and failureProbability distribution of k consecutive successes with n maximum trialsIntuitively: binomial probability as summation?Q: Two independent sequences of Bernoulli trialsHow to calculate the expected value on the difference between failed and success coin tossing trials?Expected Value where probability changes after successBernoulli trials (n,p) - probability for even/odd number of successesA marksman scores a bull's eye on 90% of his shots
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What is the first absolute central moment for number of succeses in Bernoulli scheme?
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Negative binomial distribution - deriving of the p.m.f. combinatoriallyBernoulli trials required for k successesProbability distribution for the number of successes for $N$ distinct trials with distinct probabilities of success and failureProbability distribution of k consecutive successes with n maximum trialsIntuitively: binomial probability as summation?Q: Two independent sequences of Bernoulli trialsHow to calculate the expected value on the difference between failed and success coin tossing trials?Expected Value where probability changes after successBernoulli trials (n,p) - probability for even/odd number of successesA marksman scores a bull's eye on 90% of his shots
$begingroup$
Given Bernoulli scheme with probability of single success = $p$. $mu=$ number of successes in $n$ trials. I need to calculate the first central absolute moment for the number of successes, that is
$$M(lvertmu-nprvert)$$
Formally, $M(lvertmu-nprvert)=sum_k=0^nlvert k-nprvertbinomnkp^kq^n-k$
Let's define $mequivlfloor nprfloor$, hence
$M(lvertmu-nprvert)=sum_k=0^m(k-np)binomnkp^kq^n-k+sum_k=m+1^n(np-k)binomnkp^kq^n-k$
... and I have no idea how to proceed further. Is it possible to get a closed form?
probability expected-value
asked Mar 31 at 14:49
NickNick
1514
$endgroup$
add a comment |
$begingroup$
Given Bernoulli scheme with probability of single success = $p$. $mu=$ number of successes in $n$ trials. I need to calculate the first central absolute moment for the number of successes, that is
$$M(lvertmu-nprvert)$$
Formally, $M(lvertmu-nprvert)=sum_k=0^nlvert k-nprvertbinomnkp^kq^n-k$
Let's define $mequivlfloor nprfloor$, hence
$M(lvertmu-nprvert)=sum_k=0^m(k-np)binomnkp^kq^n-k+sum_k=m+1^n(np-k)binomnkp^kq^n-k$
... and I have no idea how to proceed further. Is it possible to get a closed form?
probability expected-value
asked Mar 31 at 14:49
NickNick
1514
$endgroup$
$begingroup$
"closed form" is many times in the eye of the beholder. For $p=1/2$ there is a "nice" result: $2^-n leftlceil fracn2rightrceil binomnleftlceil fracn2rightrceil $. For other values of $p$ the $textHypergeometric2F1$ function shows up.
$endgroup$
– JimB
Mar 31 at 15:06
add a comment |
$begingroup$
Given Bernoulli scheme with probability of single success = $p$. $mu=$ number of successes in $n$ trials. I need to calculate the first central absolute moment for the number of successes, that is
$$M(lvertmu-nprvert)$$
Formally, $M(lvertmu-nprvert)=sum_k=0^nlvert k-nprvertbinomnkp^kq^n-k$
Let's define $mequivlfloor nprfloor$, hence
$M(lvertmu-nprvert)=sum_k=0^m(k-np)binomnkp^kq^n-k+sum_k=m+1^n(np-k)binomnkp^kq^n-k$
... and I have no idea how to proceed further. Is it possible to get a closed form?
probability expected-value
asked Mar 31 at 14:49
NickNick
1514
$endgroup$
Given Bernoulli scheme with probability of single success = $p$. $mu=$ number of successes in $n$ trials. I need to calculate the first central absolute moment for the number of successes, that is
$$M(lvertmu-nprvert)$$
Formally, $M(lvertmu-nprvert)=sum_k=0^nlvert k-nprvertbinomnkp^kq^n-k$
Let's define $mequivlfloor nprfloor$, hence
$M(lvertmu-nprvert)=sum_k=0^m(k-np)binomnkp^kq^n-k+sum_k=m+1^n(np-k)binomnkp^kq^n-k$
... and I have no idea how to proceed further. Is it possible to get a closed form?
probability expected-value
probability expected-value
asked Mar 31 at 14:49
NickNick
1514
asked Mar 31 at 14:49
NickNick
1514
asked Mar 31 at 14:49
NickNick
1514
asked Mar 31 at 14:49
NickNick
1514
asked Mar 31 at 14:49
NickNick
1514
1514
$begingroup$
"closed form" is many times in the eye of the beholder. For $p=1/2$ there is a "nice" result: $2^-n leftlceil fracn2rightrceil binomnleftlceil fracn2rightrceil $. For other values of $p$ the $textHypergeometric2F1$ function shows up.
$endgroup$
– JimB
Mar 31 at 15:06
add a comment |
$begingroup$
"closed form" is many times in the eye of the beholder. For $p=1/2$ there is a "nice" result: $2^-n leftlceil fracn2rightrceil binomnleftlceil fracn2rightrceil $. For other values of $p$ the $textHypergeometric2F1$ function shows up.
$endgroup$
– JimB
Mar 31 at 15:06
$begingroup$
"closed form" is many times in the eye of the beholder. For $p=1/2$ there is a "nice" result: $2^-n leftlceil fracn2rightrceil binomnleftlceil fracn2rightrceil $. For other values of $p$ the $textHypergeometric2F1$ function shows up.
$endgroup$
– JimB
Mar 31 at 15:06
$begingroup$
"closed form" is many times in the eye of the beholder. For $p=1/2$ there is a "nice" result: $2^-n leftlceil fracn2rightrceil binomnleftlceil fracn2rightrceil $. For other values of $p$ the $textHypergeometric2F1$ function shows up.
$endgroup$
– JimB
Mar 31 at 15:06
add a comment |
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$begingroup$
"closed form" is many times in the eye of the beholder. For $p=1/2$ there is a "nice" result: $2^-n leftlceil fracn2rightrceil binomnleftlceil fracn2rightrceil $. For other values of $p$ the $textHypergeometric2F1$ function shows up.
$endgroup$
– JimB
Mar 31 at 15:06