Integral of pdf and cdf normal standard distribution
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$$ int_{-infty}^{infty}Phi(a+bx)phi(x)dx=Phileft(frac{a}{sqrt{1+b^2}}right) $$
I have a problem with showing the above result, where $phi(x)$ and $Phi(x)$ respectively are the pdf and cdf of the standard normal distribution. I've tried to prove it by calculating the pdf $phi(a+bx)$ first and combine it with the pdf $phi(x)$ but the result is complicated and doesn't yield $phileft(dfrac{a}{sqrt{1+b^2}}right) $ . I really need your help. Thanks.
integration statistics normal-distribution density-function
New contributor
add a comment |
up vote
1
down vote
favorite
$$ int_{-infty}^{infty}Phi(a+bx)phi(x)dx=Phileft(frac{a}{sqrt{1+b^2}}right) $$
I have a problem with showing the above result, where $phi(x)$ and $Phi(x)$ respectively are the pdf and cdf of the standard normal distribution. I've tried to prove it by calculating the pdf $phi(a+bx)$ first and combine it with the pdf $phi(x)$ but the result is complicated and doesn't yield $phileft(dfrac{a}{sqrt{1+b^2}}right) $ . I really need your help. Thanks.
integration statistics normal-distribution density-function
New contributor
Is $Phi(a+bx)$ pdf or cdf?
– callculus
2 days ago
$Phi (a+bx) $ is cdf of normal standard and $phi(x) $ is pdf of normal standard
– Kildah Namariq
2 days ago
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
$$ int_{-infty}^{infty}Phi(a+bx)phi(x)dx=Phileft(frac{a}{sqrt{1+b^2}}right) $$
I have a problem with showing the above result, where $phi(x)$ and $Phi(x)$ respectively are the pdf and cdf of the standard normal distribution. I've tried to prove it by calculating the pdf $phi(a+bx)$ first and combine it with the pdf $phi(x)$ but the result is complicated and doesn't yield $phileft(dfrac{a}{sqrt{1+b^2}}right) $ . I really need your help. Thanks.
integration statistics normal-distribution density-function
New contributor
$$ int_{-infty}^{infty}Phi(a+bx)phi(x)dx=Phileft(frac{a}{sqrt{1+b^2}}right) $$
I have a problem with showing the above result, where $phi(x)$ and $Phi(x)$ respectively are the pdf and cdf of the standard normal distribution. I've tried to prove it by calculating the pdf $phi(a+bx)$ first and combine it with the pdf $phi(x)$ but the result is complicated and doesn't yield $phileft(dfrac{a}{sqrt{1+b^2}}right) $ . I really need your help. Thanks.
integration statistics normal-distribution density-function
integration statistics normal-distribution density-function
New contributor
New contributor
edited 2 days ago
Semiclassical
11k32464
11k32464
New contributor
asked 2 days ago
Kildah Namariq
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83
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New contributor
Is $Phi(a+bx)$ pdf or cdf?
– callculus
2 days ago
$Phi (a+bx) $ is cdf of normal standard and $phi(x) $ is pdf of normal standard
– Kildah Namariq
2 days ago
add a comment |
Is $Phi(a+bx)$ pdf or cdf?
– callculus
2 days ago
$Phi (a+bx) $ is cdf of normal standard and $phi(x) $ is pdf of normal standard
– Kildah Namariq
2 days ago
Is $Phi(a+bx)$ pdf or cdf?
– callculus
2 days ago
Is $Phi(a+bx)$ pdf or cdf?
– callculus
2 days ago
$Phi (a+bx) $ is cdf of normal standard and $phi(x) $ is pdf of normal standard
– Kildah Namariq
2 days ago
$Phi (a+bx) $ is cdf of normal standard and $phi(x) $ is pdf of normal standard
– Kildah Namariq
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Using $N(0,,1)$ iids $U,,V$ and defining $W:=U-bVsim N(0,,1+b^2)$, we can write the integral as $$int_{Bbb R}P(Ule a+bx)dP(Vle x)=P(Ule a+bV)=P(Wle a)=Phibigg(frac{a}{sqrt{1+b^2}}bigg).$$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Using $N(0,,1)$ iids $U,,V$ and defining $W:=U-bVsim N(0,,1+b^2)$, we can write the integral as $$int_{Bbb R}P(Ule a+bx)dP(Vle x)=P(Ule a+bV)=P(Wle a)=Phibigg(frac{a}{sqrt{1+b^2}}bigg).$$
add a comment |
up vote
2
down vote
accepted
Using $N(0,,1)$ iids $U,,V$ and defining $W:=U-bVsim N(0,,1+b^2)$, we can write the integral as $$int_{Bbb R}P(Ule a+bx)dP(Vle x)=P(Ule a+bV)=P(Wle a)=Phibigg(frac{a}{sqrt{1+b^2}}bigg).$$
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Using $N(0,,1)$ iids $U,,V$ and defining $W:=U-bVsim N(0,,1+b^2)$, we can write the integral as $$int_{Bbb R}P(Ule a+bx)dP(Vle x)=P(Ule a+bV)=P(Wle a)=Phibigg(frac{a}{sqrt{1+b^2}}bigg).$$
Using $N(0,,1)$ iids $U,,V$ and defining $W:=U-bVsim N(0,,1+b^2)$, we can write the integral as $$int_{Bbb R}P(Ule a+bx)dP(Vle x)=P(Ule a+bV)=P(Wle a)=Phibigg(frac{a}{sqrt{1+b^2}}bigg).$$
answered 2 days ago
J.G.
17.9k11830
17.9k11830
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add a comment |
Kildah Namariq is a new contributor. Be nice, and check out our Code of Conduct.
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Is $Phi(a+bx)$ pdf or cdf?
– callculus
2 days ago
$Phi (a+bx) $ is cdf of normal standard and $phi(x) $ is pdf of normal standard
– Kildah Namariq
2 days ago