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Thread: Difficult Integral standard normal pdf/cdf

  1. #16
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    Quote Originally Posted by BGM View Post
    The first four terms vanish due to \phi(z)
    The last integral again can be transformed as a bivariate normal integrals

    Let X, Y, Z be independent normal random variables with mean 0
    and variance 1, 1, \frac {1} {3} respectively
    \int_{-\infty}^{+\infty} \Phi(z)^2 \phi(\sqrt{3}z)dz
    = \frac {1} {\sqrt{3}} \int_{-\infty}^{+\infty} 
\Phi(z)\Phi(z) \sqrt{3}\phi(\sqrt{3}z)dz
    = \frac {1} {\sqrt{3}} \int_{-\infty}^{+\infty} 
Pr\{X \leq z\}Pr\{Y \leq z\} f_Z(z) dz
    = \frac {1} {\sqrt{3}} \int_{-\infty}^{+\infty} 
Pr\{X \leq z, Y \leq z\} f_Z(z) dz
    = \frac {1} {\sqrt{3}} Pr\{X \leq Z, Y \leq Z\}
    = \frac {1} {\sqrt{3}} 
Pr\{\frac {X - Z} {\frac {2} {\sqrt{3}}} \leq 0, 
\frac {Y - Z} {\frac {2} {\sqrt{3}}}\leq 0\}
    The correlation (covariance)
    = (\frac {\sqrt{3}} {2})^2 (\frac {1} {3}) = \frac {1} {4}

    As a conclusion,
    Let
    \begin{bmatrix} Z_1 \\ Z_2 \end{bmatrix} \sim 
N(\begin{bmatrix} 0 \\ 0 \end{bmatrix},
\begin{bmatrix} 1 & \frac {1} {4} \\
\frac {1} {4} & 1 \end{bmatrix})

    Then
    p = Pr\{Z_1 \leq 0, Z_2 \leq 0\}
= \int_{-\infty}^{+\infty} \Phi(z)^2 \sqrt{3}\phi(\sqrt{3}z)dz
    and p approximately equal to 0.290215311627583

    The required integral
    I = \frac {1} {4\sqrt5\pi^2} + \frac {16} {\sqrt{3}\pi}p
    and approximately equal to 0.864683184411258
    Great - thanks

  2. #17
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    BGM: It occurred to me that analytical expressions are possible. That is, given your results and using the standard normal bivariate pdf as you suggested we have for the first integral

    I_{1}=\int_{-\infty }^{\infty }\left ( z^{2} -1\right )\left \{\Phi \left ( z \right )  \right \}^{4}\phi \left ( z \right )dz =\frac{\sqrt{3}\arctan\left ( \sqrt{\frac{5}{3}} \right )}{\pi ^{2}}=0.160004...


    and for the second integral

    I_{2}=\int_{-\infty }^{\infty }\left ( z^{4} +z^{2} -4\right )\left \{\Phi \left ( z \right )  \right \}^{4}\phi \left ( z \right )dz

    where

    I_{2}=\frac{3\sqrt{5}+160\sqrt{3}\left (\pi -\arctan\left ( \sqrt{15} \right )  \right )}{60\pi ^{2}}=0.86468318...

    and where \Phi \left ( z \right ) is the standard normal cdf and \phi \left ( z \right ) and is standard normal pdf.

    And, that should do it.
    Last edited by Dragan; 05-31-2010 at 06:00 PM. Reason: For clarity

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    Thanks. Learn many things about the normal integration.

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