Probability:Normal distribution
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My question is a prove question as follows.
I have solved it in full length which i am uploading as images below.
Now my doubt is that in page 3, is the derivation of psi(-x) wrong? or is there some other mistakes too? Please help.
probability-theory discrete-mathematics normal-distribution
asked Nov 23 at 4:19
user524694
162
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up vote
1
down vote
favorite
My question is a prove question as follows.
I have solved it in full length which i am uploading as images below.
Now my doubt is that in page 3, is the derivation of psi(-x) wrong? or is there some other mistakes too? Please help.
probability-theory discrete-mathematics normal-distribution
asked Nov 23 at 4:19
user524694
162
Thank you so much
– user524694
Nov 23 at 13:44
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
My question is a prove question as follows.
I have solved it in full length which i am uploading as images below.
Now my doubt is that in page 3, is the derivation of psi(-x) wrong? or is there some other mistakes too? Please help.
probability-theory discrete-mathematics normal-distribution
asked Nov 23 at 4:19
user524694
162
My question is a prove question as follows.
I have solved it in full length which i am uploading as images below.
Now my doubt is that in page 3, is the derivation of psi(-x) wrong? or is there some other mistakes too? Please help.
probability-theory discrete-mathematics normal-distribution
probability-theory discrete-mathematics normal-distribution
asked Nov 23 at 4:19
user524694
162
asked Nov 23 at 4:19
user524694
162
asked Nov 23 at 4:19
user524694
162
asked Nov 23 at 4:19
user524694
162
asked Nov 23 at 4:19
user524694
162
162
Thank you so much
– user524694
Nov 23 at 13:44
add a comment |
Thank you so much
– user524694
Nov 23 at 13:44
Thank you so much
– user524694
Nov 23 at 13:44
Thank you so much
– user524694
Nov 23 at 13:44
add a comment |
2 Answers
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The problem looks wrong. $Phi(x)$ is an even function of $x$.
Therefore $int_{-x}^xPhi(u)du=2int_0^xPhi(u)du=2psi(x)$.
The correct statement would have $psi(x)=int_{-infty}^xPhi(u)du$ Using the fact that $Phi(x)$ is even, $2psi(x)=int_{-infty}^x Phi(u)du+int_{-x}^infty Phi(u)du=int_{-x}^0Phi(u)du +int_0^xPhi(u)du +int_{-infty}^0Phi(u)du +int_0^inftyPhi(u)du=int_{-x}^xPhi(u)du+1$.
answered Nov 23 at 4:50
herb steinberg
2,4332310
Thank yo so much
– user524694
Nov 23 at 13:45
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I think you just forgot to write an integral and an integration limit seems to be wrong.
In the context of the normal distribution $Phi$ stands for the cumulative distribution function:
$Phi (x) = frac{1}{sqrt{2pi}}int_{-infty}^x e^{-frac{t^2}{2}}, dt$.
Then, with $psi (x) = frac{1}{sqrt{2pi}}int_{color{blue}{-x}}^x e^{-frac{t^2}{2}}, dt$ you get
$$2Phi(x) - 1 = psi(x)$$
Note that because of symmetry you get:
$$Phi(x) = frac{1}{2}left(1- psi(x)right) + psi(x) Rightarrow 2Phi(x) = 1+psi(x)$$
answered Nov 23 at 8:07
trancelocation
8,8601521
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2 Answers
2
active
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2 Answers
2
active
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up vote
1
down vote
The problem looks wrong. $Phi(x)$ is an even function of $x$.
Therefore $int_{-x}^xPhi(u)du=2int_0^xPhi(u)du=2psi(x)$.
The correct statement would have $psi(x)=int_{-infty}^xPhi(u)du$ Using the fact that $Phi(x)$ is even, $2psi(x)=int_{-infty}^x Phi(u)du+int_{-x}^infty Phi(u)du=int_{-x}^0Phi(u)du +int_0^xPhi(u)du +int_{-infty}^0Phi(u)du +int_0^inftyPhi(u)du=int_{-x}^xPhi(u)du+1$.
answered Nov 23 at 4:50
herb steinberg
2,4332310
Thank yo so much
– user524694
Nov 23 at 13:45
add a comment |
up vote
1
down vote
The problem looks wrong. $Phi(x)$ is an even function of $x$.
Therefore $int_{-x}^xPhi(u)du=2int_0^xPhi(u)du=2psi(x)$.
The correct statement would have $psi(x)=int_{-infty}^xPhi(u)du$ Using the fact that $Phi(x)$ is even, $2psi(x)=int_{-infty}^x Phi(u)du+int_{-x}^infty Phi(u)du=int_{-x}^0Phi(u)du +int_0^xPhi(u)du +int_{-infty}^0Phi(u)du +int_0^inftyPhi(u)du=int_{-x}^xPhi(u)du+1$.
answered Nov 23 at 4:50
herb steinberg
2,4332310
Thank yo so much
– user524694
Nov 23 at 13:45
add a comment |
up vote
1
down vote
up vote
1
down vote
The problem looks wrong. $Phi(x)$ is an even function of $x$.
Therefore $int_{-x}^xPhi(u)du=2int_0^xPhi(u)du=2psi(x)$.
The correct statement would have $psi(x)=int_{-infty}^xPhi(u)du$ Using the fact that $Phi(x)$ is even, $2psi(x)=int_{-infty}^x Phi(u)du+int_{-x}^infty Phi(u)du=int_{-x}^0Phi(u)du +int_0^xPhi(u)du +int_{-infty}^0Phi(u)du +int_0^inftyPhi(u)du=int_{-x}^xPhi(u)du+1$.
answered Nov 23 at 4:50
herb steinberg
2,4332310
The problem looks wrong. $Phi(x)$ is an even function of $x$.
Therefore $int_{-x}^xPhi(u)du=2int_0^xPhi(u)du=2psi(x)$.
The correct statement would have $psi(x)=int_{-infty}^xPhi(u)du$ Using the fact that $Phi(x)$ is even, $2psi(x)=int_{-infty}^x Phi(u)du+int_{-x}^infty Phi(u)du=int_{-x}^0Phi(u)du +int_0^xPhi(u)du +int_{-infty}^0Phi(u)du +int_0^inftyPhi(u)du=int_{-x}^xPhi(u)du+1$.
answered Nov 23 at 4:50
herb steinberg
2,4332310
edited Nov 23 at 5:32
answered Nov 23 at 4:50
herb steinberg
2,4332310
answered Nov 23 at 4:50
herb steinberg
2,4332310
answered Nov 23 at 4:50
herb steinberg
2,4332310
2,4332310
Thank yo so much
– user524694
Nov 23 at 13:45
add a comment |
Thank yo so much
– user524694
Nov 23 at 13:45
Thank yo so much
– user524694
Nov 23 at 13:45
Thank yo so much
– user524694
Nov 23 at 13:45
add a comment |
up vote
0
down vote
I think you just forgot to write an integral and an integration limit seems to be wrong.
In the context of the normal distribution $Phi$ stands for the cumulative distribution function:
$Phi (x) = frac{1}{sqrt{2pi}}int_{-infty}^x e^{-frac{t^2}{2}}, dt$.
Then, with $psi (x) = frac{1}{sqrt{2pi}}int_{color{blue}{-x}}^x e^{-frac{t^2}{2}}, dt$ you get
$$2Phi(x) - 1 = psi(x)$$
Note that because of symmetry you get:
$$Phi(x) = frac{1}{2}left(1- psi(x)right) + psi(x) Rightarrow 2Phi(x) = 1+psi(x)$$
answered Nov 23 at 8:07
trancelocation
8,8601521
add a comment |
up vote
0
down vote
I think you just forgot to write an integral and an integration limit seems to be wrong.
In the context of the normal distribution $Phi$ stands for the cumulative distribution function:
$Phi (x) = frac{1}{sqrt{2pi}}int_{-infty}^x e^{-frac{t^2}{2}}, dt$.
Then, with $psi (x) = frac{1}{sqrt{2pi}}int_{color{blue}{-x}}^x e^{-frac{t^2}{2}}, dt$ you get
$$2Phi(x) - 1 = psi(x)$$
Note that because of symmetry you get:
$$Phi(x) = frac{1}{2}left(1- psi(x)right) + psi(x) Rightarrow 2Phi(x) = 1+psi(x)$$
answered Nov 23 at 8:07
trancelocation
8,8601521
add a comment |
up vote
0
down vote
up vote
0
down vote
I think you just forgot to write an integral and an integration limit seems to be wrong.
In the context of the normal distribution $Phi$ stands for the cumulative distribution function:
$Phi (x) = frac{1}{sqrt{2pi}}int_{-infty}^x e^{-frac{t^2}{2}}, dt$.
Then, with $psi (x) = frac{1}{sqrt{2pi}}int_{color{blue}{-x}}^x e^{-frac{t^2}{2}}, dt$ you get
$$2Phi(x) - 1 = psi(x)$$
Note that because of symmetry you get:
$$Phi(x) = frac{1}{2}left(1- psi(x)right) + psi(x) Rightarrow 2Phi(x) = 1+psi(x)$$
answered Nov 23 at 8:07
trancelocation
8,8601521
I think you just forgot to write an integral and an integration limit seems to be wrong.
In the context of the normal distribution $Phi$ stands for the cumulative distribution function:
$Phi (x) = frac{1}{sqrt{2pi}}int_{-infty}^x e^{-frac{t^2}{2}}, dt$.
Then, with $psi (x) = frac{1}{sqrt{2pi}}int_{color{blue}{-x}}^x e^{-frac{t^2}{2}}, dt$ you get
$$2Phi(x) - 1 = psi(x)$$
Note that because of symmetry you get:
$$Phi(x) = frac{1}{2}left(1- psi(x)right) + psi(x) Rightarrow 2Phi(x) = 1+psi(x)$$
answered Nov 23 at 8:07
trancelocation
8,8601521
edited Nov 23 at 8:20
answered Nov 23 at 8:07
trancelocation
8,8601521
answered Nov 23 at 8:07
trancelocation
8,8601521
answered Nov 23 at 8:07
trancelocation
8,8601521
8,8601521
add a comment |
add a comment |
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Thank you so much
– user524694
Nov 23 at 13:44