Distribution of $|sin(Theta)|$ if $Theta$ is uniform on $(0,2pi)$ and arcsine distribution












1












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Suppose $Theta sim Unif(0,2pi)$ and $s in (0,1)$. I am trying to find $P(mid sin(theta) mid < sqrt s)$.



My attempt:



We have that



$$P(mid sin(theta) mid < sqrt s) = P(-sqrt s < sin(theta) < sqrt s) = \
= P(sin(theta) < sqrt s) - P(sin(theta) < - sqrt s) = \ =P(theta < arcsin(sqrt s)) - P(theta < arcsin(-sqrt s)) = \
= frac{arcsin(sqrt s)}{2pi} - frac{arcsin(-sqrt s)}{2pi} = frac{arcsin(sqrt s)}{pi}$$



where I've used the oddity of the arcsine function.



However, taking the derivative with respect to $s$ of this laxt expression yields



$$frac{d}{ds}frac{arcsin(sqrt s)}{pi} = frac{1}{pisqrt{ s(1-s)}}ds$$



which integrates to $frac{1}{2}$. Where did I miss a $2$ along the way?






probability proof-verification probability-distributions







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  • $begingroup$
    Comments are not for extended discussion; this conversation has been moved to chat.
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    – Aloizio Macedo♦
    Jan 31 at 11:23










  • $begingroup$
    (Comment reposted) Why omit every reference to the question which prompted this one?
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    – Did
    Jan 31 at 12:39












  • $begingroup$
    (Comment reposted) You should review the definition of the arcsine...
    $endgroup$
    – Did
    Jan 31 at 12:39










  • $begingroup$
    You have all the answers in the chat.
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    – Easymode44
    Jan 31 at 12:46
















1












$begingroup$


Suppose $Theta sim Unif(0,2pi)$ and $s in (0,1)$. I am trying to find $P(mid sin(theta) mid < sqrt s)$.



My attempt:



We have that



$$P(mid sin(theta) mid < sqrt s) = P(-sqrt s < sin(theta) < sqrt s) = \
= P(sin(theta) < sqrt s) - P(sin(theta) < - sqrt s) = \ =P(theta < arcsin(sqrt s)) - P(theta < arcsin(-sqrt s)) = \
= frac{arcsin(sqrt s)}{2pi} - frac{arcsin(-sqrt s)}{2pi} = frac{arcsin(sqrt s)}{pi}$$



where I've used the oddity of the arcsine function.



However, taking the derivative with respect to $s$ of this laxt expression yields



$$frac{d}{ds}frac{arcsin(sqrt s)}{pi} = frac{1}{pisqrt{ s(1-s)}}ds$$



which integrates to $frac{1}{2}$. Where did I miss a $2$ along the way?






probability proof-verification probability-distributions







share|cite|improve this question











$endgroup$












  • $begingroup$
    Comments are not for extended discussion; this conversation has been moved to chat.
    $endgroup$
    – Aloizio Macedo♦
    Jan 31 at 11:23










  • $begingroup$
    (Comment reposted) Why omit every reference to the question which prompted this one?
    $endgroup$
    – Did
    Jan 31 at 12:39












  • $begingroup$
    (Comment reposted) You should review the definition of the arcsine...
    $endgroup$
    – Did
    Jan 31 at 12:39










  • $begingroup$
    You have all the answers in the chat.
    $endgroup$
    – Easymode44
    Jan 31 at 12:46














1












1








1


0



$begingroup$


Suppose $Theta sim Unif(0,2pi)$ and $s in (0,1)$. I am trying to find $P(mid sin(theta) mid < sqrt s)$.



My attempt:



We have that



$$P(mid sin(theta) mid < sqrt s) = P(-sqrt s < sin(theta) < sqrt s) = \
= P(sin(theta) < sqrt s) - P(sin(theta) < - sqrt s) = \ =P(theta < arcsin(sqrt s)) - P(theta < arcsin(-sqrt s)) = \
= frac{arcsin(sqrt s)}{2pi} - frac{arcsin(-sqrt s)}{2pi} = frac{arcsin(sqrt s)}{pi}$$



where I've used the oddity of the arcsine function.



However, taking the derivative with respect to $s$ of this laxt expression yields



$$frac{d}{ds}frac{arcsin(sqrt s)}{pi} = frac{1}{pisqrt{ s(1-s)}}ds$$



which integrates to $frac{1}{2}$. Where did I miss a $2$ along the way?






probability proof-verification probability-distributions







share|cite|improve this question











$endgroup$




Suppose $Theta sim Unif(0,2pi)$ and $s in (0,1)$. I am trying to find $P(mid sin(theta) mid < sqrt s)$.



My attempt:



We have that



$$P(mid sin(theta) mid < sqrt s) = P(-sqrt s < sin(theta) < sqrt s) = \
= P(sin(theta) < sqrt s) - P(sin(theta) < - sqrt s) = \ =P(theta < arcsin(sqrt s)) - P(theta < arcsin(-sqrt s)) = \
= frac{arcsin(sqrt s)}{2pi} - frac{arcsin(-sqrt s)}{2pi} = frac{arcsin(sqrt s)}{pi}$$



where I've used the oddity of the arcsine function.



However, taking the derivative with respect to $s$ of this laxt expression yields



$$frac{d}{ds}frac{arcsin(sqrt s)}{pi} = frac{1}{pisqrt{ s(1-s)}}ds$$



which integrates to $frac{1}{2}$. Where did I miss a $2$ along the way?






probability proof-verification probability-distributions




probability proof-verification probability-distributions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 30 at 10:41









Did

249k23227466




249k23227466










asked Jan 29 at 16:05









Easymode44Easymode44

409213




409213












  • $begingroup$
    Comments are not for extended discussion; this conversation has been moved to chat.
    $endgroup$
    – Aloizio Macedo♦
    Jan 31 at 11:23










  • $begingroup$
    (Comment reposted) Why omit every reference to the question which prompted this one?
    $endgroup$
    – Did
    Jan 31 at 12:39












  • $begingroup$
    (Comment reposted) You should review the definition of the arcsine...
    $endgroup$
    – Did
    Jan 31 at 12:39










  • $begingroup$
    You have all the answers in the chat.
    $endgroup$
    – Easymode44
    Jan 31 at 12:46


















  • $begingroup$
    Comments are not for extended discussion; this conversation has been moved to chat.
    $endgroup$
    – Aloizio Macedo♦
    Jan 31 at 11:23










  • $begingroup$
    (Comment reposted) Why omit every reference to the question which prompted this one?
    $endgroup$
    – Did
    Jan 31 at 12:39












  • $begingroup$
    (Comment reposted) You should review the definition of the arcsine...
    $endgroup$
    – Did
    Jan 31 at 12:39










  • $begingroup$
    You have all the answers in the chat.
    $endgroup$
    – Easymode44
    Jan 31 at 12:46
















$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Jan 31 at 11:23




$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Jan 31 at 11:23












$begingroup$
(Comment reposted) Why omit every reference to the question which prompted this one?
$endgroup$
– Did
Jan 31 at 12:39






$begingroup$
(Comment reposted) Why omit every reference to the question which prompted this one?
$endgroup$
– Did
Jan 31 at 12:39














$begingroup$
(Comment reposted) You should review the definition of the arcsine...
$endgroup$
– Did
Jan 31 at 12:39




$begingroup$
(Comment reposted) You should review the definition of the arcsine...
$endgroup$
– Did
Jan 31 at 12:39












$begingroup$
You have all the answers in the chat.
$endgroup$
– Easymode44
Jan 31 at 12:46




$begingroup$
You have all the answers in the chat.
$endgroup$
– Easymode44
Jan 31 at 12:46










1 Answer
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For $xin [0,1]$,
$$
mathsf{P}(sin(theta)le -x)=mathsf{P}(sin(theta)ge x)=mathsf{P}(arcsin(x)le thetale pi-arcsin(x)).
$$
Thus,
$$
mathsf{P}(|sin(theta)|le x)=1-2mathsf{P}(arcsin(x)le thetale pi-arcsin(x))=frac{2arcsin(x)}{pi}.
$$






share|cite|improve this answer











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  • $begingroup$
    thank you for your answer, I appreciate the input.
    $endgroup$
    – Easymode44
    Jan 31 at 9:50












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1 Answer
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1 Answer
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$begingroup$

For $xin [0,1]$,
$$
mathsf{P}(sin(theta)le -x)=mathsf{P}(sin(theta)ge x)=mathsf{P}(arcsin(x)le thetale pi-arcsin(x)).
$$
Thus,
$$
mathsf{P}(|sin(theta)|le x)=1-2mathsf{P}(arcsin(x)le thetale pi-arcsin(x))=frac{2arcsin(x)}{pi}.
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    thank you for your answer, I appreciate the input.
    $endgroup$
    – Easymode44
    Jan 31 at 9:50
















1












$begingroup$

For $xin [0,1]$,
$$
mathsf{P}(sin(theta)le -x)=mathsf{P}(sin(theta)ge x)=mathsf{P}(arcsin(x)le thetale pi-arcsin(x)).
$$
Thus,
$$
mathsf{P}(|sin(theta)|le x)=1-2mathsf{P}(arcsin(x)le thetale pi-arcsin(x))=frac{2arcsin(x)}{pi}.
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    thank you for your answer, I appreciate the input.
    $endgroup$
    – Easymode44
    Jan 31 at 9:50














1












1








1





$begingroup$

For $xin [0,1]$,
$$
mathsf{P}(sin(theta)le -x)=mathsf{P}(sin(theta)ge x)=mathsf{P}(arcsin(x)le thetale pi-arcsin(x)).
$$
Thus,
$$
mathsf{P}(|sin(theta)|le x)=1-2mathsf{P}(arcsin(x)le thetale pi-arcsin(x))=frac{2arcsin(x)}{pi}.
$$






share|cite|improve this answer











$endgroup$



For $xin [0,1]$,
$$
mathsf{P}(sin(theta)le -x)=mathsf{P}(sin(theta)ge x)=mathsf{P}(arcsin(x)le thetale pi-arcsin(x)).
$$
Thus,
$$
mathsf{P}(|sin(theta)|le x)=1-2mathsf{P}(arcsin(x)le thetale pi-arcsin(x))=frac{2arcsin(x)}{pi}.
$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 30 at 10:36

























answered Jan 30 at 10:30









d.k.o.d.k.o.

10.6k630




10.6k630












  • $begingroup$
    thank you for your answer, I appreciate the input.
    $endgroup$
    – Easymode44
    Jan 31 at 9:50


















  • $begingroup$
    thank you for your answer, I appreciate the input.
    $endgroup$
    – Easymode44
    Jan 31 at 9:50
















$begingroup$
thank you for your answer, I appreciate the input.
$endgroup$
– Easymode44
Jan 31 at 9:50




$begingroup$
thank you for your answer, I appreciate the input.
$endgroup$
– Easymode44
Jan 31 at 9:50


















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