Equation with expected value of inverse of a binomial variableReciprocal of a BinomialExpected Value with Variable ProbabilityConvergence of binomial to normalIntegrating a special skew normal — the CDF of a convolution of a normal with a truncated normalBinomial probability expected valueExpected value of a particular sum of binomial random variableFinding expected value with binomial distributionexpected value and variance of a binomial variableExpected value of squared binomial variableConditional expected value of binomial random variable above a threshold?What is probability $P(X<x |X <Y)$
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Equation with expected value of inverse of a binomial variable
Reciprocal of a BinomialExpected Value with Variable ProbabilityConvergence of binomial to normalIntegrating a special skew normal — the CDF of a convolution of a normal with a truncated normalBinomial probability expected valueExpected value of a particular sum of binomial random variableFinding expected value with binomial distributionexpected value and variance of a binomial variableExpected value of squared binomial variableConditional expected value of binomial random variable above a threshold?What is probability $P(X<x |X <Y)$
$begingroup$
I need to solve an equation $Bbb Efrac1a + X = b$ where $Xsim mathrmBin(p; n)$ in terms of $p$, where $a,b> 0$ and integer $n$ are given. As far as I understood from this question, the LHS does not admit a particularly elegant closed form, so no big luck inverting it. On the other hand, I rather know an approximate value of $p$ for $n$ large enough. Perhaps, in that case it may make sense to assume that $X$ is almost normal: I have not found any results on how hard is to compute the LHS in that case. Any ideas?
probability
asked Aug 2 '17 at 13:32
IlyaIlya
25.1k453115
$endgroup$
add a comment |
$begingroup$
I need to solve an equation $Bbb Efrac1a + X = b$ where $Xsim mathrmBin(p; n)$ in terms of $p$, where $a,b> 0$ and integer $n$ are given. As far as I understood from this question, the LHS does not admit a particularly elegant closed form, so no big luck inverting it. On the other hand, I rather know an approximate value of $p$ for $n$ large enough. Perhaps, in that case it may make sense to assume that $X$ is almost normal: I have not found any results on how hard is to compute the LHS in that case. Any ideas?
probability
asked Aug 2 '17 at 13:32
IlyaIlya
25.1k453115
$endgroup$
$begingroup$
When $n$ increase the LHS will converge to $0$ except for $p=0$
$endgroup$
– lasen H
Aug 2 '17 at 14:16
add a comment |
$begingroup$
I need to solve an equation $Bbb Efrac1a + X = b$ where $Xsim mathrmBin(p; n)$ in terms of $p$, where $a,b> 0$ and integer $n$ are given. As far as I understood from this question, the LHS does not admit a particularly elegant closed form, so no big luck inverting it. On the other hand, I rather know an approximate value of $p$ for $n$ large enough. Perhaps, in that case it may make sense to assume that $X$ is almost normal: I have not found any results on how hard is to compute the LHS in that case. Any ideas?
probability
asked Aug 2 '17 at 13:32
IlyaIlya
25.1k453115
$endgroup$
I need to solve an equation $Bbb Efrac1a + X = b$ where $Xsim mathrmBin(p; n)$ in terms of $p$, where $a,b> 0$ and integer $n$ are given. As far as I understood from this question, the LHS does not admit a particularly elegant closed form, so no big luck inverting it. On the other hand, I rather know an approximate value of $p$ for $n$ large enough. Perhaps, in that case it may make sense to assume that $X$ is almost normal: I have not found any results on how hard is to compute the LHS in that case. Any ideas?
probability
probability
asked Aug 2 '17 at 13:32
IlyaIlya
25.1k453115
asked Aug 2 '17 at 13:32
IlyaIlya
25.1k453115
asked Aug 2 '17 at 13:32
IlyaIlya
25.1k453115
asked Aug 2 '17 at 13:32
IlyaIlya
25.1k453115
asked Aug 2 '17 at 13:32
IlyaIlya
25.1k453115
25.1k453115
$begingroup$
When $n$ increase the LHS will converge to $0$ except for $p=0$
$endgroup$
– lasen H
Aug 2 '17 at 14:16
add a comment |
$begingroup$
When $n$ increase the LHS will converge to $0$ except for $p=0$
$endgroup$
– lasen H
Aug 2 '17 at 14:16
$begingroup$
When $n$ increase the LHS will converge to $0$ except for $p=0$
$endgroup$
– lasen H
Aug 2 '17 at 14:16
$begingroup$
When $n$ increase the LHS will converge to $0$ except for $p=0$
$endgroup$
– lasen H
Aug 2 '17 at 14:16
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If you don't want to use the LHS closed-form from the answer in the your link, you can in that closed form substitute $Q=(1-p)^n , _2F_1left(a,-n;a+1;p/(p-1)right)$ by some approximation. To make one note that
$$Q^(m)(0)=frac(m!)^2 binomnmunderseti=0oversetmprod (a+i)quad,quad Q^(m)(1)=frac(m!)^2 binomnmunderseti=n-moversetnprod(a+i)$$
It doesn't make sense to consider just $n$ large, because the LHS converge to 0 pointwise except for $p=0$. (I suppose you want $0leq pleq1$)
answered Aug 2 '17 at 14:36
lasen Hlasen H
690112
$endgroup$
$begingroup$
I meant the asymptotics of $p$ as $n$ goes to infinity
$endgroup$
– Ilya
Aug 2 '17 at 17:56
add a comment |
$begingroup$
When $p$ is constant, we have that $Eleft[frac11+Xright]=Oleft(frac1ncdot pright)$
as $nrightarrowinfty$ as shown by Cribari-Neto et al. (2000). See also the paper by Chao, M. T., and W. E. Strawderman. “Negative Moments of Positive Random Variables.” Journal of the American Statistical Association, vol. 67, no. 338, 1972, pp. 429–431. JSTOR, www.jstor.org/stable/2284399.
answered Mar 15 at 0:33
Konstantinos I. StourasKonstantinos I. Stouras
212
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If you don't want to use the LHS closed-form from the answer in the your link, you can in that closed form substitute $Q=(1-p)^n , _2F_1left(a,-n;a+1;p/(p-1)right)$ by some approximation. To make one note that
$$Q^(m)(0)=frac(m!)^2 binomnmunderseti=0oversetmprod (a+i)quad,quad Q^(m)(1)=frac(m!)^2 binomnmunderseti=n-moversetnprod(a+i)$$
It doesn't make sense to consider just $n$ large, because the LHS converge to 0 pointwise except for $p=0$. (I suppose you want $0leq pleq1$)
answered Aug 2 '17 at 14:36
lasen Hlasen H
690112
$endgroup$
$begingroup$
I meant the asymptotics of $p$ as $n$ goes to infinity
$endgroup$
– Ilya
Aug 2 '17 at 17:56
add a comment |
$begingroup$
If you don't want to use the LHS closed-form from the answer in the your link, you can in that closed form substitute $Q=(1-p)^n , _2F_1left(a,-n;a+1;p/(p-1)right)$ by some approximation. To make one note that
$$Q^(m)(0)=frac(m!)^2 binomnmunderseti=0oversetmprod (a+i)quad,quad Q^(m)(1)=frac(m!)^2 binomnmunderseti=n-moversetnprod(a+i)$$
It doesn't make sense to consider just $n$ large, because the LHS converge to 0 pointwise except for $p=0$. (I suppose you want $0leq pleq1$)
answered Aug 2 '17 at 14:36
lasen Hlasen H
690112
$endgroup$
$begingroup$
I meant the asymptotics of $p$ as $n$ goes to infinity
$endgroup$
– Ilya
Aug 2 '17 at 17:56
add a comment |
$begingroup$
If you don't want to use the LHS closed-form from the answer in the your link, you can in that closed form substitute $Q=(1-p)^n , _2F_1left(a,-n;a+1;p/(p-1)right)$ by some approximation. To make one note that
$$Q^(m)(0)=frac(m!)^2 binomnmunderseti=0oversetmprod (a+i)quad,quad Q^(m)(1)=frac(m!)^2 binomnmunderseti=n-moversetnprod(a+i)$$
It doesn't make sense to consider just $n$ large, because the LHS converge to 0 pointwise except for $p=0$. (I suppose you want $0leq pleq1$)
answered Aug 2 '17 at 14:36
lasen Hlasen H
690112
$endgroup$
If you don't want to use the LHS closed-form from the answer in the your link, you can in that closed form substitute $Q=(1-p)^n , _2F_1left(a,-n;a+1;p/(p-1)right)$ by some approximation. To make one note that
$$Q^(m)(0)=frac(m!)^2 binomnmunderseti=0oversetmprod (a+i)quad,quad Q^(m)(1)=frac(m!)^2 binomnmunderseti=n-moversetnprod(a+i)$$
It doesn't make sense to consider just $n$ large, because the LHS converge to 0 pointwise except for $p=0$. (I suppose you want $0leq pleq1$)
answered Aug 2 '17 at 14:36
lasen Hlasen H
690112
answered Aug 2 '17 at 14:36
lasen Hlasen H
690112
answered Aug 2 '17 at 14:36
lasen Hlasen H
690112
answered Aug 2 '17 at 14:36
lasen Hlasen H
690112
690112
$begingroup$
I meant the asymptotics of $p$ as $n$ goes to infinity
$endgroup$
– Ilya
Aug 2 '17 at 17:56
add a comment |
$begingroup$
I meant the asymptotics of $p$ as $n$ goes to infinity
$endgroup$
– Ilya
Aug 2 '17 at 17:56
$begingroup$
I meant the asymptotics of $p$ as $n$ goes to infinity
$endgroup$
– Ilya
Aug 2 '17 at 17:56
$begingroup$
I meant the asymptotics of $p$ as $n$ goes to infinity
$endgroup$
– Ilya
Aug 2 '17 at 17:56
add a comment |
$begingroup$
When $p$ is constant, we have that $Eleft[frac11+Xright]=Oleft(frac1ncdot pright)$
as $nrightarrowinfty$ as shown by Cribari-Neto et al. (2000). See also the paper by Chao, M. T., and W. E. Strawderman. “Negative Moments of Positive Random Variables.” Journal of the American Statistical Association, vol. 67, no. 338, 1972, pp. 429–431. JSTOR, www.jstor.org/stable/2284399.
answered Mar 15 at 0:33
Konstantinos I. StourasKonstantinos I. Stouras
212
$endgroup$
add a comment |
$begingroup$
When $p$ is constant, we have that $Eleft[frac11+Xright]=Oleft(frac1ncdot pright)$
as $nrightarrowinfty$ as shown by Cribari-Neto et al. (2000). See also the paper by Chao, M. T., and W. E. Strawderman. “Negative Moments of Positive Random Variables.” Journal of the American Statistical Association, vol. 67, no. 338, 1972, pp. 429–431. JSTOR, www.jstor.org/stable/2284399.
answered Mar 15 at 0:33
Konstantinos I. StourasKonstantinos I. Stouras
212
$endgroup$
add a comment |
$begingroup$
When $p$ is constant, we have that $Eleft[frac11+Xright]=Oleft(frac1ncdot pright)$
as $nrightarrowinfty$ as shown by Cribari-Neto et al. (2000). See also the paper by Chao, M. T., and W. E. Strawderman. “Negative Moments of Positive Random Variables.” Journal of the American Statistical Association, vol. 67, no. 338, 1972, pp. 429–431. JSTOR, www.jstor.org/stable/2284399.
answered Mar 15 at 0:33
Konstantinos I. StourasKonstantinos I. Stouras
212
$endgroup$
When $p$ is constant, we have that $Eleft[frac11+Xright]=Oleft(frac1ncdot pright)$
as $nrightarrowinfty$ as shown by Cribari-Neto et al. (2000). See also the paper by Chao, M. T., and W. E. Strawderman. “Negative Moments of Positive Random Variables.” Journal of the American Statistical Association, vol. 67, no. 338, 1972, pp. 429–431. JSTOR, www.jstor.org/stable/2284399.
answered Mar 15 at 0:33
Konstantinos I. StourasKonstantinos I. Stouras
212
edited Mar 15 at 0:40
answered Mar 15 at 0:33
Konstantinos I. StourasKonstantinos I. Stouras
212
answered Mar 15 at 0:33
Konstantinos I. StourasKonstantinos I. Stouras
212
answered Mar 15 at 0:33
Konstantinos I. StourasKonstantinos I. Stouras
212
212
add a comment |
add a comment |
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$begingroup$
When $n$ increase the LHS will converge to $0$ except for $p=0$
$endgroup$
– lasen H
Aug 2 '17 at 14:16