Integral over an interval of the normal distribution
$begingroup$
Assume that $f$ is the normal distribution function with mean equal to $0$ and variance equal to $sigma^2$.
Set some interval $I=[a,b]$ in $mathbb{R}^+$.
Denote the integral of $f$ over $I$ by $S_I(sigma^2)$.
My question: is $S_I(sigma^2)$ monotonous (decreasing?) as $sigma^2$ increases continuously? Or is it dependent on $I$ or the initial value of $sigma^2$?
Thank you!
probability
asked Jan 16 at 12:07
DaugmentedDaugmented
326212
$endgroup$
add a comment |
$begingroup$
Assume that $f$ is the normal distribution function with mean equal to $0$ and variance equal to $sigma^2$.
Set some interval $I=[a,b]$ in $mathbb{R}^+$.
Denote the integral of $f$ over $I$ by $S_I(sigma^2)$.
My question: is $S_I(sigma^2)$ monotonous (decreasing?) as $sigma^2$ increases continuously? Or is it dependent on $I$ or the initial value of $sigma^2$?
Thank you!
probability
asked Jan 16 at 12:07
DaugmentedDaugmented
326212
$endgroup$
add a comment |
$begingroup$
Assume that $f$ is the normal distribution function with mean equal to $0$ and variance equal to $sigma^2$.
Set some interval $I=[a,b]$ in $mathbb{R}^+$.
Denote the integral of $f$ over $I$ by $S_I(sigma^2)$.
My question: is $S_I(sigma^2)$ monotonous (decreasing?) as $sigma^2$ increases continuously? Or is it dependent on $I$ or the initial value of $sigma^2$?
Thank you!
probability
asked Jan 16 at 12:07
DaugmentedDaugmented
326212
$endgroup$
Assume that $f$ is the normal distribution function with mean equal to $0$ and variance equal to $sigma^2$.
Set some interval $I=[a,b]$ in $mathbb{R}^+$.
Denote the integral of $f$ over $I$ by $S_I(sigma^2)$.
My question: is $S_I(sigma^2)$ monotonous (decreasing?) as $sigma^2$ increases continuously? Or is it dependent on $I$ or the initial value of $sigma^2$?
Thank you!
probability
probability
asked Jan 16 at 12:07
DaugmentedDaugmented
326212
asked Jan 16 at 12:07
DaugmentedDaugmented
326212
asked Jan 16 at 12:07
DaugmentedDaugmented
326212
asked Jan 16 at 12:07
DaugmentedDaugmented
326212
asked Jan 16 at 12:07
DaugmentedDaugmented
326212
326212
add a comment |
add a comment |
1 Answer
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$begingroup$
Consider this plot and the interval around $2$. It's clear that the integral is not monotonous with $sigma$. So in general case, it depends both on interval (for symmetric or wide enough intervals, the integral is monotonous) and on initial value of $sigma^2$ (for each interval there is such $sigma_0$, so for $sigma>sigma_0$ the integral is monotonous).
answered Jan 16 at 12:37
Vasily MitchVasily Mitch
2,6791312
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Consider this plot and the interval around $2$. It's clear that the integral is not monotonous with $sigma$. So in general case, it depends both on interval (for symmetric or wide enough intervals, the integral is monotonous) and on initial value of $sigma^2$ (for each interval there is such $sigma_0$, so for $sigma>sigma_0$ the integral is monotonous).
answered Jan 16 at 12:37
Vasily MitchVasily Mitch
2,6791312
$endgroup$
add a comment |
$begingroup$
Consider this plot and the interval around $2$. It's clear that the integral is not monotonous with $sigma$. So in general case, it depends both on interval (for symmetric or wide enough intervals, the integral is monotonous) and on initial value of $sigma^2$ (for each interval there is such $sigma_0$, so for $sigma>sigma_0$ the integral is monotonous).
answered Jan 16 at 12:37
Vasily MitchVasily Mitch
2,6791312
$endgroup$
add a comment |
$begingroup$
Consider this plot and the interval around $2$. It's clear that the integral is not monotonous with $sigma$. So in general case, it depends both on interval (for symmetric or wide enough intervals, the integral is monotonous) and on initial value of $sigma^2$ (for each interval there is such $sigma_0$, so for $sigma>sigma_0$ the integral is monotonous).
answered Jan 16 at 12:37
Vasily MitchVasily Mitch
2,6791312
$endgroup$
Consider this plot and the interval around $2$. It's clear that the integral is not monotonous with $sigma$. So in general case, it depends both on interval (for symmetric or wide enough intervals, the integral is monotonous) and on initial value of $sigma^2$ (for each interval there is such $sigma_0$, so for $sigma>sigma_0$ the integral is monotonous).
answered Jan 16 at 12:37
Vasily MitchVasily Mitch
2,6791312
answered Jan 16 at 12:37
Vasily MitchVasily Mitch
2,6791312
answered Jan 16 at 12:37
Vasily MitchVasily Mitch
2,6791312
answered Jan 16 at 12:37
Vasily MitchVasily Mitch
2,6791312
2,6791312
add a comment |
add a comment |
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