Mixing
Any good approximation for $Phi^{-1}(1-Phi(a))$?
$begingroup$
Let $Phi$ be the standard Gaussian CDF and $a > 0$.
Question
Is there any good approximation for $Phi^{-1}(1-Phi(a))$ ?
normal-distribution approximation gaussian-integral
asked Feb 3 at 12:59
dohmatobdohmatob
3,737629
$endgroup$
add a comment |
$begingroup$
Let $Phi$ be the standard Gaussian CDF and $a > 0$.
Question
Is there any good approximation for $Phi^{-1}(1-Phi(a))$ ?
normal-distribution approximation gaussian-integral
asked Feb 3 at 12:59
dohmatobdohmatob
3,737629
$endgroup$
1
$begingroup$
I don't have any nonobvious suggestions but I am curious what would be considered 'good' here?
$endgroup$
– Tony S.F.
Feb 3 at 13:06
$begingroup$
Ya, mine was a really stupid question indeed. Thanks for the input
$endgroup$
– dohmatob
Feb 3 at 14:06
add a comment |
$begingroup$
Let $Phi$ be the standard Gaussian CDF and $a > 0$.
Question
Is there any good approximation for $Phi^{-1}(1-Phi(a))$ ?
normal-distribution approximation gaussian-integral
asked Feb 3 at 12:59
dohmatobdohmatob
3,737629
$endgroup$
Let $Phi$ be the standard Gaussian CDF and $a > 0$.
Question
Is there any good approximation for $Phi^{-1}(1-Phi(a))$ ?
normal-distribution approximation gaussian-integral
normal-distribution approximation gaussian-integral
asked Feb 3 at 12:59
dohmatobdohmatob
3,737629
asked Feb 3 at 12:59
dohmatobdohmatob
3,737629
asked Feb 3 at 12:59
dohmatobdohmatob
3,737629
asked Feb 3 at 12:59
dohmatobdohmatob
3,737629
asked Feb 3 at 12:59
dohmatobdohmatob
3,737629
3,737629
1
$begingroup$
I don't have any nonobvious suggestions but I am curious what would be considered 'good' here?
$endgroup$
– Tony S.F.
Feb 3 at 13:06
$begingroup$
Ya, mine was a really stupid question indeed. Thanks for the input
$endgroup$
– dohmatob
Feb 3 at 14:06
add a comment |
1
$begingroup$
I don't have any nonobvious suggestions but I am curious what would be considered 'good' here?
$endgroup$
– Tony S.F.
Feb 3 at 13:06
$begingroup$
Ya, mine was a really stupid question indeed. Thanks for the input
$endgroup$
– dohmatob
Feb 3 at 14:06
1
1
$begingroup$
I don't have any nonobvious suggestions but I am curious what would be considered 'good' here?
$endgroup$
– Tony S.F.
Feb 3 at 13:06
$begingroup$
I don't have any nonobvious suggestions but I am curious what would be considered 'good' here?
$endgroup$
– Tony S.F.
Feb 3 at 13:06
$begingroup$
Ya, mine was a really stupid question indeed. Thanks for the input
$endgroup$
– dohmatob
Feb 3 at 14:06
$begingroup$
Ya, mine was a really stupid question indeed. Thanks for the input
$endgroup$
– dohmatob
Feb 3 at 14:06
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For $a in mathbb{R}$ and not just $a > 0$, the exact solution is
$$Phi^{-1}(1-Phi(a)) = -a \
because 1 - Phi(a) = Phi(-a) $$
due to the symmetry with respect to zero. This applies to any symmetric density, or equivalently, any CDF that are rotationally symmetric with respect to $(0,frac12)$.
answered Feb 3 at 13:14
Lee David Chung LinLee David Chung Lin
4,50841342
$endgroup$
$begingroup$
Ya, mine was a really stupid question indeed. Thanks for the input. I was initially interested in approximating $Phi^{-1}(x)$, which I managed to show to be roughly $sqrt{2log(1 / (1-x))}$, BTW.
$endgroup$
– dohmatob
Feb 3 at 14:06
$begingroup$
I see. Approximating $Phi^{-1}$ is as respectable a task one can take on as any other. Cheers : )
$endgroup$
– Lee David Chung Lin
Feb 3 at 14:13
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For $a in mathbb{R}$ and not just $a > 0$, the exact solution is
$$Phi^{-1}(1-Phi(a)) = -a \
because 1 - Phi(a) = Phi(-a) $$
due to the symmetry with respect to zero. This applies to any symmetric density, or equivalently, any CDF that are rotationally symmetric with respect to $(0,frac12)$.
answered Feb 3 at 13:14
Lee David Chung LinLee David Chung Lin
4,50841342
$endgroup$
$begingroup$
Ya, mine was a really stupid question indeed. Thanks for the input. I was initially interested in approximating $Phi^{-1}(x)$, which I managed to show to be roughly $sqrt{2log(1 / (1-x))}$, BTW.
$endgroup$
– dohmatob
Feb 3 at 14:06
$begingroup$
I see. Approximating $Phi^{-1}$ is as respectable a task one can take on as any other. Cheers : )
$endgroup$
– Lee David Chung Lin
Feb 3 at 14:13
add a comment |
$begingroup$
For $a in mathbb{R}$ and not just $a > 0$, the exact solution is
$$Phi^{-1}(1-Phi(a)) = -a \
because 1 - Phi(a) = Phi(-a) $$
due to the symmetry with respect to zero. This applies to any symmetric density, or equivalently, any CDF that are rotationally symmetric with respect to $(0,frac12)$.
answered Feb 3 at 13:14
Lee David Chung LinLee David Chung Lin
4,50841342
$endgroup$
$begingroup$
Ya, mine was a really stupid question indeed. Thanks for the input. I was initially interested in approximating $Phi^{-1}(x)$, which I managed to show to be roughly $sqrt{2log(1 / (1-x))}$, BTW.
$endgroup$
– dohmatob
Feb 3 at 14:06
$begingroup$
I see. Approximating $Phi^{-1}$ is as respectable a task one can take on as any other. Cheers : )
$endgroup$
– Lee David Chung Lin
Feb 3 at 14:13
add a comment |
$begingroup$
For $a in mathbb{R}$ and not just $a > 0$, the exact solution is
$$Phi^{-1}(1-Phi(a)) = -a \
because 1 - Phi(a) = Phi(-a) $$
due to the symmetry with respect to zero. This applies to any symmetric density, or equivalently, any CDF that are rotationally symmetric with respect to $(0,frac12)$.
answered Feb 3 at 13:14
Lee David Chung LinLee David Chung Lin
4,50841342
$endgroup$
For $a in mathbb{R}$ and not just $a > 0$, the exact solution is
$$Phi^{-1}(1-Phi(a)) = -a \
because 1 - Phi(a) = Phi(-a) $$
due to the symmetry with respect to zero. This applies to any symmetric density, or equivalently, any CDF that are rotationally symmetric with respect to $(0,frac12)$.
answered Feb 3 at 13:14
Lee David Chung LinLee David Chung Lin
4,50841342
answered Feb 3 at 13:14
Lee David Chung LinLee David Chung Lin
4,50841342
answered Feb 3 at 13:14
Lee David Chung LinLee David Chung Lin
4,50841342
answered Feb 3 at 13:14
Lee David Chung LinLee David Chung Lin
4,50841342
4,50841342
$begingroup$
Ya, mine was a really stupid question indeed. Thanks for the input. I was initially interested in approximating $Phi^{-1}(x)$, which I managed to show to be roughly $sqrt{2log(1 / (1-x))}$, BTW.
$endgroup$
– dohmatob
Feb 3 at 14:06
$begingroup$
I see. Approximating $Phi^{-1}$ is as respectable a task one can take on as any other. Cheers : )
$endgroup$
– Lee David Chung Lin
Feb 3 at 14:13
add a comment |
$begingroup$
Ya, mine was a really stupid question indeed. Thanks for the input. I was initially interested in approximating $Phi^{-1}(x)$, which I managed to show to be roughly $sqrt{2log(1 / (1-x))}$, BTW.
$endgroup$
– dohmatob
Feb 3 at 14:06
$begingroup$
I see. Approximating $Phi^{-1}$ is as respectable a task one can take on as any other. Cheers : )
$endgroup$
– Lee David Chung Lin
Feb 3 at 14:13
$begingroup$
Ya, mine was a really stupid question indeed. Thanks for the input. I was initially interested in approximating $Phi^{-1}(x)$, which I managed to show to be roughly $sqrt{2log(1 / (1-x))}$, BTW.
$endgroup$
– dohmatob
Feb 3 at 14:06
$begingroup$
Ya, mine was a really stupid question indeed. Thanks for the input. I was initially interested in approximating $Phi^{-1}(x)$, which I managed to show to be roughly $sqrt{2log(1 / (1-x))}$, BTW.
$endgroup$
– dohmatob
Feb 3 at 14:06
$begingroup$
I see. Approximating $Phi^{-1}$ is as respectable a task one can take on as any other. Cheers : )
$endgroup$
– Lee David Chung Lin
Feb 3 at 14:13
$begingroup$
I see. Approximating $Phi^{-1}$ is as respectable a task one can take on as any other. Cheers : )
$endgroup$
– Lee David Chung Lin
Feb 3 at 14:13
add a comment |
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1
$begingroup$
I don't have any nonobvious suggestions but I am curious what would be considered 'good' here?
$endgroup$
– Tony S.F.
Feb 3 at 13:06
$begingroup$
Ya, mine was a really stupid question indeed. Thanks for the input
$endgroup$
– dohmatob
Feb 3 at 14:06