Can we compute the infinite series of the cdf of a standard normal distribution $sum^{infty}_{n=0}...
I've tried plugging the series in Wolfram to first of all check if it converges but it doesn't return anything useful. Moreover, just computing $Phi(-100)$ returns something useless. Looking at the distribution function of the standard normal distribution it is obvious that $Phi(-100) approx 0$, but I guess the same can be said about $1/100$, while the infinite series of $1/n$ is infinity.
Is this series even defined, i.e. can it be computed, or are some problems that don't allow computation?
probability sequences-and-series statistics convergence normal-distribution
asked Nov 21 '18 at 20:30
S. Crim
12710
add a comment |
I've tried plugging the series in Wolfram to first of all check if it converges but it doesn't return anything useful. Moreover, just computing $Phi(-100)$ returns something useless. Looking at the distribution function of the standard normal distribution it is obvious that $Phi(-100) approx 0$, but I guess the same can be said about $1/100$, while the infinite series of $1/n$ is infinity.
Is this series even defined, i.e. can it be computed, or are some problems that don't allow computation?
probability sequences-and-series statistics convergence normal-distribution
asked Nov 21 '18 at 20:30
S. Crim
12710
What is the "useless" thing that it returns? If it's some obviously incorrect value there might be some errors arising from the actual calculation, i.e. unstable algorithm for very large negative numbers.
– TrostAft
Nov 21 '18 at 20:31
I guess it might not be completely useless. It returns: $frac{1}{2}$erfc$(50sqrt{2})$. Which is very close to zero, but the erfc term throws me off.
– S. Crim
Nov 21 '18 at 20:34
This makes sense though right? $Phi(-100) = P[Z leq -100]$ where $Z$ is a standard normal random variable, which should be very very small. As for whether the series converges, unsure at the moment.
– TrostAft
Nov 21 '18 at 20:38
add a comment |
I've tried plugging the series in Wolfram to first of all check if it converges but it doesn't return anything useful. Moreover, just computing $Phi(-100)$ returns something useless. Looking at the distribution function of the standard normal distribution it is obvious that $Phi(-100) approx 0$, but I guess the same can be said about $1/100$, while the infinite series of $1/n$ is infinity.
Is this series even defined, i.e. can it be computed, or are some problems that don't allow computation?
probability sequences-and-series statistics convergence normal-distribution
asked Nov 21 '18 at 20:30
S. Crim
12710
I've tried plugging the series in Wolfram to first of all check if it converges but it doesn't return anything useful. Moreover, just computing $Phi(-100)$ returns something useless. Looking at the distribution function of the standard normal distribution it is obvious that $Phi(-100) approx 0$, but I guess the same can be said about $1/100$, while the infinite series of $1/n$ is infinity.
Is this series even defined, i.e. can it be computed, or are some problems that don't allow computation?
probability sequences-and-series statistics convergence normal-distribution
probability sequences-and-series statistics convergence normal-distribution
asked Nov 21 '18 at 20:30
S. Crim
12710
asked Nov 21 '18 at 20:30
S. Crim
12710
asked Nov 21 '18 at 20:30
S. Crim
12710
asked Nov 21 '18 at 20:30
S. Crim
12710
asked Nov 21 '18 at 20:30
S. Crim
12710
12710
What is the "useless" thing that it returns? If it's some obviously incorrect value there might be some errors arising from the actual calculation, i.e. unstable algorithm for very large negative numbers.
– TrostAft
Nov 21 '18 at 20:31
I guess it might not be completely useless. It returns: $frac{1}{2}$erfc$(50sqrt{2})$. Which is very close to zero, but the erfc term throws me off.
– S. Crim
Nov 21 '18 at 20:34
This makes sense though right? $Phi(-100) = P[Z leq -100]$ where $Z$ is a standard normal random variable, which should be very very small. As for whether the series converges, unsure at the moment.
– TrostAft
Nov 21 '18 at 20:38
add a comment |
What is the "useless" thing that it returns? If it's some obviously incorrect value there might be some errors arising from the actual calculation, i.e. unstable algorithm for very large negative numbers.
– TrostAft
Nov 21 '18 at 20:31
I guess it might not be completely useless. It returns: $frac{1}{2}$erfc$(50sqrt{2})$. Which is very close to zero, but the erfc term throws me off.
– S. Crim
Nov 21 '18 at 20:34
This makes sense though right? $Phi(-100) = P[Z leq -100]$ where $Z$ is a standard normal random variable, which should be very very small. As for whether the series converges, unsure at the moment.
– TrostAft
Nov 21 '18 at 20:38
What is the "useless" thing that it returns? If it's some obviously incorrect value there might be some errors arising from the actual calculation, i.e. unstable algorithm for very large negative numbers.
– TrostAft
Nov 21 '18 at 20:31
What is the "useless" thing that it returns? If it's some obviously incorrect value there might be some errors arising from the actual calculation, i.e. unstable algorithm for very large negative numbers.
– TrostAft
Nov 21 '18 at 20:31
I guess it might not be completely useless. It returns: $frac{1}{2}$erfc$(50sqrt{2})$. Which is very close to zero, but the erfc term throws me off.
– S. Crim
Nov 21 '18 at 20:34
I guess it might not be completely useless. It returns: $frac{1}{2}$erfc$(50sqrt{2})$. Which is very close to zero, but the erfc term throws me off.
– S. Crim
Nov 21 '18 at 20:34
This makes sense though right? $Phi(-100) = P[Z leq -100]$ where $Z$ is a standard normal random variable, which should be very very small. As for whether the series converges, unsure at the moment.
– TrostAft
Nov 21 '18 at 20:38
This makes sense though right? $Phi(-100) = P[Z leq -100]$ where $Z$ is a standard normal random variable, which should be very very small. As for whether the series converges, unsure at the moment.
– TrostAft
Nov 21 '18 at 20:38
add a comment |
1 Answer
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$Phi(-t) sim frac{e^{-t^2/2}}{sqrt{2pi} t}$ as $t to infty$, so your series does converge. According to Maple, the sum is approximately
$$ .83116074272973488211268081565524436110357388769594$$
I doubt that there is a closed form expression for this.
answered Nov 21 '18 at 20:48
Robert Israel
318k23208457
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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votes
$Phi(-t) sim frac{e^{-t^2/2}}{sqrt{2pi} t}$ as $t to infty$, so your series does converge. According to Maple, the sum is approximately
$$ .83116074272973488211268081565524436110357388769594$$
I doubt that there is a closed form expression for this.
answered Nov 21 '18 at 20:48
Robert Israel
318k23208457
add a comment |
$Phi(-t) sim frac{e^{-t^2/2}}{sqrt{2pi} t}$ as $t to infty$, so your series does converge. According to Maple, the sum is approximately
$$ .83116074272973488211268081565524436110357388769594$$
I doubt that there is a closed form expression for this.
answered Nov 21 '18 at 20:48
Robert Israel
318k23208457
add a comment |
$Phi(-t) sim frac{e^{-t^2/2}}{sqrt{2pi} t}$ as $t to infty$, so your series does converge. According to Maple, the sum is approximately
$$ .83116074272973488211268081565524436110357388769594$$
I doubt that there is a closed form expression for this.
answered Nov 21 '18 at 20:48
Robert Israel
318k23208457
$Phi(-t) sim frac{e^{-t^2/2}}{sqrt{2pi} t}$ as $t to infty$, so your series does converge. According to Maple, the sum is approximately
$$ .83116074272973488211268081565524436110357388769594$$
I doubt that there is a closed form expression for this.
answered Nov 21 '18 at 20:48
Robert Israel
318k23208457
answered Nov 21 '18 at 20:48
Robert Israel
318k23208457
answered Nov 21 '18 at 20:48
Robert Israel
318k23208457
answered Nov 21 '18 at 20:48
Robert Israel
318k23208457
318k23208457
add a comment |
add a comment |
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What is the "useless" thing that it returns? If it's some obviously incorrect value there might be some errors arising from the actual calculation, i.e. unstable algorithm for very large negative numbers.
– TrostAft
Nov 21 '18 at 20:31
I guess it might not be completely useless. It returns: $frac{1}{2}$erfc$(50sqrt{2})$. Which is very close to zero, but the erfc term throws me off.
– S. Crim
Nov 21 '18 at 20:34
This makes sense though right? $Phi(-100) = P[Z leq -100]$ where $Z$ is a standard normal random variable, which should be very very small. As for whether the series converges, unsure at the moment.
– TrostAft
Nov 21 '18 at 20:38