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.










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  • Is $Phi(a+bx)$ pdf or cdf?
    – callculus
    Nov 15 at 17:11










  • $Phi (a+bx) $ is cdf of normal standard and $phi(x) $ is pdf of normal standard
    – Kildah Namariq
    Nov 15 at 17:39

















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.










share|cite|improve this question









New contributor




Kildah Namariq is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • Is $Phi(a+bx)$ pdf or cdf?
    – callculus
    Nov 15 at 17:11










  • $Phi (a+bx) $ is cdf of normal standard and $phi(x) $ is pdf of normal standard
    – Kildah Namariq
    Nov 15 at 17:39















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.










share|cite|improve this question









New contributor




Kildah Namariq is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











$$ int_{-infty}^{infty}Phi(a+bx)phi(x)dx=Phileft(frac{a}{sqrt{1+b^2}}right) $$



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






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edited Nov 15 at 18:50









Semiclassical

11k32464




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asked Nov 15 at 6:12









Kildah Namariq

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New contributor





Kildah Namariq is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Kildah Namariq is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • Is $Phi(a+bx)$ pdf or cdf?
    – callculus
    Nov 15 at 17:11










  • $Phi (a+bx) $ is cdf of normal standard and $phi(x) $ is pdf of normal standard
    – Kildah Namariq
    Nov 15 at 17:39




















  • Is $Phi(a+bx)$ pdf or cdf?
    – callculus
    Nov 15 at 17:11










  • $Phi (a+bx) $ is cdf of normal standard and $phi(x) $ is pdf of normal standard
    – Kildah Namariq
    Nov 15 at 17:39


















Is $Phi(a+bx)$ pdf or cdf?
– callculus
Nov 15 at 17:11




Is $Phi(a+bx)$ pdf or cdf?
– callculus
Nov 15 at 17:11












$Phi (a+bx) $ is cdf of normal standard and $phi(x) $ is pdf of normal standard
– Kildah Namariq
Nov 15 at 17:39






$Phi (a+bx) $ is cdf of normal standard and $phi(x) $ is pdf of normal standard
– Kildah Namariq
Nov 15 at 17:39












1 Answer
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Using $N(0,,1)$ iids $U,,V$ and defining $W:=U-bVsim N(0,,1+b^2)$, we can write the integral as $$int_{Bbb R}P(Ule a+bx)dP(Vle x)=P(Ule a+bV)=P(Wle a)=Phibigg(frac{a}{sqrt{1+b^2}}bigg).$$






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    1 Answer
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    active

    oldest

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    1 Answer
    1






    active

    oldest

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    active

    oldest

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    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 $$int_{Bbb R}P(Ule a+bx)dP(Vle x)=P(Ule a+bV)=P(Wle a)=Phibigg(frac{a}{sqrt{1+b^2}}bigg).$$






    share|cite|improve this answer

























      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 $$int_{Bbb R}P(Ule a+bx)dP(Vle x)=P(Ule a+bV)=P(Wle a)=Phibigg(frac{a}{sqrt{1+b^2}}bigg).$$






      share|cite|improve this answer























        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 $$int_{Bbb R}P(Ule a+bx)dP(Vle x)=P(Ule a+bV)=P(Wle a)=Phibigg(frac{a}{sqrt{1+b^2}}bigg).$$






        share|cite|improve this answer












        Using $N(0,,1)$ iids $U,,V$ and defining $W:=U-bVsim N(0,,1+b^2)$, we can write the integral as $$int_{Bbb R}P(Ule a+bx)dP(Vle x)=P(Ule a+bV)=P(Wle a)=Phibigg(frac{a}{sqrt{1+b^2}}bigg).$$







        share|cite|improve this answer












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        answered Nov 15 at 19:40









        J.G.

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        18k11830






















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