Truncation of infinite summation involving a probability mass function
$begingroup$
Suppose we want to compute an infinite sum of the form $S := sum_{x=0}^infty f(x)p(x)$ within a tolerance error $epsilon$, where $f : mathcal{X} to mathbb{R}^+$ is some functional of interest and $p(x)$ is a probability mass function. For convenience, write $S = S_b + S_a$.
If $g(x) := f(x)p(x)$ achieves a single maximum, $g(n^star)$, I claim one can use some standard results on infinite series (e.g. the ones presented here) to approximate $S_a = sum_{k = n^star +1}^infty g(k)$ by exploiting the fact that for $n > n^star$ we have a sequence which is positive, decreasing with $r_n:= g(n+1)/g(n)$ such that $lim_{ntoinfty} r_n = 0$. If we define $z_n = frac{g(n + 1)}{1 - r_n}$ $-$ see equation (2) in the linked notes $-$, I think we can finally claim that there exists $n^prime(epsilon)$ such that
$$ left|S - left(S_b + S_{n'} + frac{z_{n'}}{2}right)right| < epsilon, $$
where $S_b := sum_{k= 0}^{n^star} g(k)$ and $S_{n^prime}:= sum_{j= n^star}^{n^prime} g(j)$.
Questions:
- I couldn't find these calculations anywhere on the web, suggesting it's not in the toolkit of statisticians. Is this technique useful in general or is the restriction of a single maximum a bit much?
- Is there a more general solution along the same lines that relaxes these assumptions (positivity of $f$ and the concavity of $pf$)?
probability summation numerical-methods truncation-error
asked Oct 29 '18 at 21:53
Luiz Max CarvalhoLuiz Max Carvalho
1019
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add a comment |
$begingroup$
Suppose we want to compute an infinite sum of the form $S := sum_{x=0}^infty f(x)p(x)$ within a tolerance error $epsilon$, where $f : mathcal{X} to mathbb{R}^+$ is some functional of interest and $p(x)$ is a probability mass function. For convenience, write $S = S_b + S_a$.
If $g(x) := f(x)p(x)$ achieves a single maximum, $g(n^star)$, I claim one can use some standard results on infinite series (e.g. the ones presented here) to approximate $S_a = sum_{k = n^star +1}^infty g(k)$ by exploiting the fact that for $n > n^star$ we have a sequence which is positive, decreasing with $r_n:= g(n+1)/g(n)$ such that $lim_{ntoinfty} r_n = 0$. If we define $z_n = frac{g(n + 1)}{1 - r_n}$ $-$ see equation (2) in the linked notes $-$, I think we can finally claim that there exists $n^prime(epsilon)$ such that
$$ left|S - left(S_b + S_{n'} + frac{z_{n'}}{2}right)right| < epsilon, $$
where $S_b := sum_{k= 0}^{n^star} g(k)$ and $S_{n^prime}:= sum_{j= n^star}^{n^prime} g(j)$.
Questions:
- I couldn't find these calculations anywhere on the web, suggesting it's not in the toolkit of statisticians. Is this technique useful in general or is the restriction of a single maximum a bit much?
- Is there a more general solution along the same lines that relaxes these assumptions (positivity of $f$ and the concavity of $pf$)?
probability summation numerical-methods truncation-error
asked Oct 29 '18 at 21:53
Luiz Max CarvalhoLuiz Max Carvalho
1019
$endgroup$
add a comment |
$begingroup$
Suppose we want to compute an infinite sum of the form $S := sum_{x=0}^infty f(x)p(x)$ within a tolerance error $epsilon$, where $f : mathcal{X} to mathbb{R}^+$ is some functional of interest and $p(x)$ is a probability mass function. For convenience, write $S = S_b + S_a$.
If $g(x) := f(x)p(x)$ achieves a single maximum, $g(n^star)$, I claim one can use some standard results on infinite series (e.g. the ones presented here) to approximate $S_a = sum_{k = n^star +1}^infty g(k)$ by exploiting the fact that for $n > n^star$ we have a sequence which is positive, decreasing with $r_n:= g(n+1)/g(n)$ such that $lim_{ntoinfty} r_n = 0$. If we define $z_n = frac{g(n + 1)}{1 - r_n}$ $-$ see equation (2) in the linked notes $-$, I think we can finally claim that there exists $n^prime(epsilon)$ such that
$$ left|S - left(S_b + S_{n'} + frac{z_{n'}}{2}right)right| < epsilon, $$
where $S_b := sum_{k= 0}^{n^star} g(k)$ and $S_{n^prime}:= sum_{j= n^star}^{n^prime} g(j)$.
Questions:
- I couldn't find these calculations anywhere on the web, suggesting it's not in the toolkit of statisticians. Is this technique useful in general or is the restriction of a single maximum a bit much?
- Is there a more general solution along the same lines that relaxes these assumptions (positivity of $f$ and the concavity of $pf$)?
probability summation numerical-methods truncation-error
asked Oct 29 '18 at 21:53
Luiz Max CarvalhoLuiz Max Carvalho
1019
$endgroup$
Suppose we want to compute an infinite sum of the form $S := sum_{x=0}^infty f(x)p(x)$ within a tolerance error $epsilon$, where $f : mathcal{X} to mathbb{R}^+$ is some functional of interest and $p(x)$ is a probability mass function. For convenience, write $S = S_b + S_a$.
If $g(x) := f(x)p(x)$ achieves a single maximum, $g(n^star)$, I claim one can use some standard results on infinite series (e.g. the ones presented here) to approximate $S_a = sum_{k = n^star +1}^infty g(k)$ by exploiting the fact that for $n > n^star$ we have a sequence which is positive, decreasing with $r_n:= g(n+1)/g(n)$ such that $lim_{ntoinfty} r_n = 0$. If we define $z_n = frac{g(n + 1)}{1 - r_n}$ $-$ see equation (2) in the linked notes $-$, I think we can finally claim that there exists $n^prime(epsilon)$ such that
$$ left|S - left(S_b + S_{n'} + frac{z_{n'}}{2}right)right| < epsilon, $$
where $S_b := sum_{k= 0}^{n^star} g(k)$ and $S_{n^prime}:= sum_{j= n^star}^{n^prime} g(j)$.
Questions:
- I couldn't find these calculations anywhere on the web, suggesting it's not in the toolkit of statisticians. Is this technique useful in general or is the restriction of a single maximum a bit much?
- Is there a more general solution along the same lines that relaxes these assumptions (positivity of $f$ and the concavity of $pf$)?
probability summation numerical-methods truncation-error
probability summation numerical-methods truncation-error
asked Oct 29 '18 at 21:53
Luiz Max CarvalhoLuiz Max Carvalho
1019
asked Oct 29 '18 at 21:53
Luiz Max CarvalhoLuiz Max Carvalho
1019
edited Dec 11 '18 at 11:08
Luiz Max Carvalho
asked Oct 29 '18 at 21:53
Luiz Max CarvalhoLuiz Max Carvalho
1019
asked Oct 29 '18 at 21:53
Luiz Max CarvalhoLuiz Max Carvalho
1019
asked Oct 29 '18 at 21:53
Luiz Max CarvalhoLuiz Max Carvalho
1019
1019
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