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Thread: Difficult Integral standard normal pdf/cdf

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    Difficult Integral standard normal pdf/cdf




    I would like to get an explicit analytical form for the integral (below)...or something close to it with an associated (small) error term. (Note that I realize it is going to take some special techniques.)

    I\left ( z \right )=\int_{-\infty }^{\infty }\left ( z^{2} -1\right )\left \{\Phi \left ( z \right )  \right \}^{4}\phi \left ( z \right )dz

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

    Note, that I can get a numerical solution...but apparently that's not good enough for the reviewers of this journal I am working with.

    Note also that the standard normal cdf can be expressed as an infinite series as follows:

    \Phi \left ( z \right )=\frac{1}{2}+\phi \left ( z \right )\left \{ z+\frac{z^{3}}{3}+\frac{z^{5}}{3\cdot 5}+\frac{z^{7}}{3\cdot 5\cdot 7}+\frac{z^{9}}{3\cdot 5\cdot 7\cdot 9} +\cdots \right \}

    Any ideas: Martingale, BGM, Ecologist, or Ritchie?

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    I just truncated the series you provided for \Phi(z) up to z^9 and then integrate in quickmath, i got

    \frac {1773206471387788} {5046844482421875\sqrt{5}\pi^2} + \frac {761134} {1240029\sqrt{3}\pi}

    Is it close enough to your numerical solution?

    btw, the integral should be independent of the dummy variable z?

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    I have try another thing.

    Using the results
    d\Phi(z) = \phi(z)dz
    -d\phi(z) = z\phi(z)dz
    -d(z\phi(z)) =  (z^2 - 1)\phi(z)dz


    \int (z^2 - 1)\Phi(z)^4\phi(z)dz
    = -\int \Phi(z)^4d(z\phi(z))
    = -z\phi(z)\Phi(z)^4 + 4\int z\Phi(z)^3\phi(z)dz
    = -z\phi(z)\Phi(z)^4 - 4\int \Phi(z)^3d\phi(z)
    = -z\phi(z)\Phi(z)^4 - 4\Phi(z)^3\phi(z) 
+ 12\int \Phi(z)^2\phi(z)dz
    = -z\phi(z)\Phi(z)^4 - 4\Phi(z)^3\phi(z) 
+ 12\int \Phi(z)^2d\Phi(z)
    = [-z\Phi(z)^4 - 4\Phi(z)^3]\phi(z) + 4\Phi(z)^3 + C

    Interesting when evaluating this function at z = \pm\infty,
    the first term vanish due to \phi(z)
    and the later part give you 4. Anything wrong here?

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    Im not big math guy but i think there is something wrong with the integration by parts.


    = -z\phi(z)\Phi(z)^4 + 4\int z\Phi(z)^3\phi(z)dz

    but

    d\Phi(z)^{4} = \Phi(z)^{3} \phi(z) 4dz

    so should be squared pdf in the integral?

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    Thanks very much for spotting my careless mistake

    I try another one
    -\frac {1} {2} d\phi(z)^2 = -\frac {1} {2} 2\phi(z)(-z\phi(z))dz 
= z\phi(z)^2dz
    \phi(z)^2 = \frac {1} {\sqrt{2\pi}} \phi(\sqrt{2}z)

    \int (z^2 - 1)\Phi(z)^4\phi(z)dz
    = -\int \Phi(z)^4d(z\phi(z))
    = -z\phi(z)\Phi(z)^4 + 4\int z\Phi(z)^3\phi(z)^2dz
    = -z\phi(z)\Phi(z)^4 - 2\int \Phi(z)^3d\phi(z)^2
    = -z\phi(z)\Phi(z)^4 - 2\Phi(z)^3\phi(z)^2 
+ 6\int \Phi(z)^2\phi(z)^2 dz
    = -z\phi(z)\Phi(z)^4 - 2\Phi(z)^3\phi(z)^2 
+ \frac {6} {\sqrt{2\pi}} \int \Phi(z)^2\phi(\sqrt{2}z) dz
    = -z\phi(z)\Phi(z)^4 - 2\Phi(z)^3\phi(z)^2 
+ \frac {3} {\sqrt{\pi}} \int \Phi(z)^2d\Phi(\sqrt{2}z)
    = -z\phi(z)\Phi(z)^4 - 2\Phi(z)^3\phi(z)^2 
+ \frac {3} {\sqrt{\pi}} \Phi(z)^2\Phi(\sqrt{2}z) 
- \frac {6} {\sqrt{\pi}}\int\Phi(z)\Phi(\sqrt{2}z)dz

    The latter integral
    \int\Phi(z)\Phi(\sqrt{2}z)dz
    = z\Phi(z)\Phi(\sqrt{2}z) - \int z\Phi(\sqrt{2}z)\phi(z) dz
- \sqrt{2}\int z\Phi(z)\phi(\sqrt{2}z)dz
    = z\Phi(z)\Phi(\sqrt{2}z) + \int \Phi(\sqrt{2}z)d\phi(z)
+ \frac {1} {\sqrt{2}} \int \Phi(z)d\phi(\sqrt{2}z)
    = z\Phi(z)\Phi(\sqrt{2}z) + \Phi(\sqrt{2}z)\phi(z) 
- \sqrt{2} \int \phi(z) \phi(\sqrt{2}z) dz + \frac {1} {\sqrt{2}} \Phi(z)\phi(\sqrt{2}z)
- \frac {1} {\sqrt{2}} \int \phi(\sqrt{2}z)\phi(z) dz
    = z\Phi(z)\Phi(\sqrt{2}z) + \Phi(\sqrt{2}z)\phi(z) 
+ \frac {1} {\sqrt{2}} \Phi(z)\phi(\sqrt{2}z)
- \frac {3} {2\sqrt{\pi}} \int \phi(\sqrt{3}z) dz
    = z\Phi(z)\Phi(\sqrt{2}z) + \Phi(\sqrt{2}z)\phi(z) 
+ \frac {1} {\sqrt{2}} \Phi(z)\phi(\sqrt{2}z)
- \frac {1} {2} \sqrt{\frac {3} {\pi}} \Phi(\sqrt{3}z)

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    Thanks BGM, I am going to check the result (above) later today when I get to my office.

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    Quote Originally Posted by BGM View Post
    Thanks very much for spotting my careless mistake

    I try another one
    -\frac {1} {2} d\phi(z)^2 = -\frac {1} {2} 2\phi(z)(-z\phi(z))dz 
= z\phi(z)^2dz
    \phi(z)^2 = \frac {1} {\sqrt{2\pi}} \phi(\sqrt{2}z)

    \int (z^2 - 1)\Phi(z)^4\phi(z)dz
    = -\int \Phi(z)^4d(z\phi(z))
    = -z\phi(z)\Phi(z)^4 + 4\int z\Phi(z)^3\phi(z)^2dz
    = -z\phi(z)\Phi(z)^4 - 2\int \Phi(z)^3d\phi(z)^2
    = -z\phi(z)\Phi(z)^4 - 2\Phi(z)^3\phi(z)^2 
+ 6\int \Phi(z)^2\phi(z)^2 dz
    = -z\phi(z)\Phi(z)^4 - 2\Phi(z)^3\phi(z)^2 
+ \frac {6} {\sqrt{2\pi}} \int \Phi(z)^2\phi(\sqrt{2}z) dz
    = -z\phi(z)\Phi(z)^4 - 2\Phi(z)^3\phi(z)^2 
+ \frac {3} {\sqrt{\pi}} \int \Phi(z)^2d\Phi(\sqrt{2}z)
    = -z\phi(z)\Phi(z)^4 - 2\Phi(z)^3\phi(z)^2 
+ \frac {3} {\sqrt{\pi}} \Phi(z)^2\Phi(\sqrt{2}z) 
- \frac {6} {\sqrt{\pi}}\int\Phi(z)\Phi(\sqrt{2}z)dz

    The latter integral
    \int\Phi(z)\Phi(\sqrt{2}z)dz
    = z\Phi(z)\Phi(\sqrt{2}z) - \int z\Phi(\sqrt{2}z)\phi(z) dz
- \sqrt{2}\int z\Phi(z)\phi(\sqrt{2}z)dz
    = z\Phi(z)\Phi(\sqrt{2}z) + \int \Phi(\sqrt{2}z)d\phi(z)
+ \frac {1} {\sqrt{2}} \int \Phi(z)d\phi(\sqrt{2}z)
    = z\Phi(z)\Phi(\sqrt{2}z) + \Phi(\sqrt{2}z)\phi(z) 
- \sqrt{2} \int \phi(z) \phi(\sqrt{2}z) dz + \frac {1} {\sqrt{2}} \Phi(z)\phi(\sqrt{2}z)
- \frac {1} {\sqrt{2}} \int \phi(\sqrt{2}z)\phi(z) dz
    = z\Phi(z)\Phi(\sqrt{2}z) + \Phi(\sqrt{2}z)\phi(z) 
+ \frac {1} {\sqrt{2}} \Phi(z)\phi(\sqrt{2}z)
- \frac {3} {2\sqrt{\pi}} \int \phi(\sqrt{3}z) dz
    = z\Phi(z)\Phi(\sqrt{2}z) + \Phi(\sqrt{2}z)\phi(z) 
+ \frac {1} {\sqrt{2}} \Phi(z)\phi(\sqrt{2}z)
- \frac {1} {2} \sqrt{\frac {3} {\pi}} \Phi(\sqrt{3}z)




    Okay, now that I am at my office:

    First, the correct numerical solution is:



    I\left ( z \right )=\int_{-\infty }^{\infty }\left ( z^{2} -1\right )\left \{\Phi \left ( z \right )  \right \}^{4}\phi \left ( z \right )dz = 0.160004087194126565200736


    Thus, what I am looking for is an analytical expression that would give me the numerical result above...I could even live with an expression that has some small remainder attached to it.

    Note: I can integrate the function when I am raising the standard normal cdf to odd powers e.g. 1,3,and 5 but the even power of 4 gives me trouble.

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    Oops Sorry actually I repeat the same mistakes for a several time
    Always forgetting d\Phi(z) = \phi(z)dz -.-

    \int (z^2 - 1)\Phi(z)^4\phi(z)dz
    = -\int \Phi(z)^4d(z\phi(z))
    = -z\phi(z)\Phi(z)^4 + 4\int z\Phi(z)^3\phi(z)^2dz
    = -z\phi(z)\Phi(z)^4 - 2\int \Phi(z)^3d\phi(z)^2
    = -z\phi(z)\Phi(z)^4 - 2\Phi(z)^3\phi(z)^2 + 6\int \Phi(z)^2\phi(z)^3 dz
    = -z\phi(z)\Phi(z)^4 - 2\Phi(z)^3\phi(z)^2 + \frac {3} {\pi} \int \Phi(z)^2\phi(\sqrt{3}z) dz
    = -z\phi(z)\Phi(z)^4 - 2\Phi(z)^3\phi(z)^2 + \frac {\sqrt{3}} {\pi} \int \Phi(z)^2d\Phi(\sqrt{3}z)
    = -z\phi(z)\Phi(z)^4 - 2\Phi(z)^3\phi(z)^2 + \frac {\sqrt{3}} {\pi} \Phi(z)^2\Phi(\sqrt{3}z) - \frac {2\sqrt{3}} {\pi} \int \Phi(z)\Phi(\sqrt{3}z)\phi(z)dz

    Not sure if this is the right direction
    It is late here so probably need to work tmr.

    Sorry for giving a wrong answer

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    I have just find a method to evaluate the last integral through bivariate normal.
    (not indefinite though)

    Let X, Y, Z be three independent standard normal random variables.

    \int_{-\infty}^{\infty} \Phi(z)\Phi(\sqrt{3}z)\phi(z)dz
    = \int_{-\infty}^{\infty} Pr\{X \leq z\}Pr\{Y \leq \sqrt{3}z\}f_Z(z)dz
    = \int_{-\infty}^{\infty} Pr\{X \leq z, Y \leq \sqrt{3}z\}f_Z(z)dz
    = Pr\{X \leq Z, Y \leq \sqrt{3}Z\}
    = Pr\{\frac {X - Z} {\sqrt{2}} \leq 0, \frac {Y - \sqrt{3}Z} {2} \leq 0\}
    Note E[\frac {X - Z} {\sqrt{2}}] = E[\frac {Y - \sqrt{3}Z} {2}] = 0
    Var[\frac {X - Z} {\sqrt{2}}] = Var[\frac {Y - \sqrt{3}Z} {2}] = 1
    Cov[\frac {X - Z} {\sqrt{2}}, \frac {Y - \sqrt{3}Z} {2}]= (-\frac {1} {\sqrt{2}})(-\frac {\sqrt{3}} {2})Cov[Z, Z]= \frac {\sqrt{6}} {4}

    So the problem just converted to evaluating this bivariate normal probability
    I have just use the "pmvnorm" function inside R library "mvtnorm"
    to evaluate the probability. It approximately equal to 0.3548923

    Combining the previous result,
    \frac {\sqrt{3}} {\pi} (1 - 2 Pr\{\frac {X - Z} {\sqrt{2}} \leq 0, \frac {Y - \sqrt{3}Z} {2} \leq 0\} )

    I have check this value in R. It is approximately 0.1600041 which agree to
    your answer. Hope this help.

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    Quote Originally Posted by BGM View Post
    I have just find a method to evaluate the last integral through bivariate normal.
    (not indefinite though)

    Let X, Y, Z be three independent standard normal random variables.

    \int_{-\infty}^{\infty} \Phi(z)\Phi(\sqrt{3}z)\phi(z)dz
    = \int_{-\infty}^{\infty} Pr\{X \leq z\}Pr\{Y \leq \sqrt{3}z\}f_Z(z)dz
    = \int_{-\infty}^{\infty} Pr\{X \leq z, Y \leq \sqrt{3}z\}f_Z(z)dz
    = Pr\{X \leq Z, Y \leq \sqrt{3}Z\}
    = Pr\{\frac {X - Z} {\sqrt{2}} \leq 0, 
\frac {Y - \sqrt{3}Z} {2} \leq 0\}
    Note E[\frac {X - Z} {\sqrt{2}}] = E[\frac {Y - \sqrt{3}Z} {2}] = 0
    Var[\frac {X - Z} {\sqrt{2}}] = Var[\frac {Y - \sqrt{3}Z} {2}] = 1
    Cov[\frac {X - Z} {\sqrt{2}}, \frac {Y - \sqrt{3}Z} {2}]
= (-\frac {1} {\sqrt{2}})(-\frac {\sqrt{3}} {2})Cov[Z, Z]
= \frac {\sqrt{6}} {4}

    So the problem just converted to evaluating this bivariate normal probability
    I have just use the "pmvnorm" function inside R library "mvtnorm"
    to evaluate the probability. It approximately equal to 0.3548923

    Combining the previous result,
    \frac {\sqrt{3}} {\pi} (1 - 2 Pr\{\frac {X - Z} {\sqrt{2}} \leq 0, 
\frac {Y - \sqrt{3}Z} {2} \leq 0\} )

    I have check this value in R. It is approximately 0.1600041 which agree to
    your answer. Hope this help.

    Thanks BGM: I'm trying to evaluate the standard bivariate pdf to verify your result and I'm getting a slightly smaller result .353283. Note that I am using Mathematica.

    So, let me just ask to be clear, when you are evaluating the pdf what lower and upper limits are you using for X and Y? I am also setting the correlation to be Sqrt[6.]/4.

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    Suppose
    \begin{bmatrix} Z_1 \\ Z_2 \end{bmatrix} \sim 
N(\begin{bmatrix} 0 \\ 0 \end{bmatrix},
\begin{bmatrix} 1, \frac {\sqrt{6}} {4} \\
\frac {\sqrt{6}} {4}, 1 \end{bmatrix})

    Then
    Pr\{Z_1 \leq 0, Z_2 \leq 0\}
= \int_{-\infty}^{+\infty} \Phi(z)\Phi(\sqrt{3}z)\phi(z)dz

    And the required integral
    I = \frac {\sqrt{3}} {\pi} 
[1 - 2\int_{-\infty}^{+\infty} \Phi(z)\Phi(\sqrt{3}z)\phi(z)dz]

    as
    \phi(z) \rightarrow 0 when z \rightarrow \pm\infty, the first two terms vanish.

    I do not know the accuracy of the R program nor mathematica.
    But that result match with your post earlier on, so I just thought that
    should be a correct one.

    I check with that R function once more, it has 15 digits of accuracy:
    (error bounded by 10^(-15))
    It gives 0.3548923441862084
    and I = 0.1600040871941266

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    Smile

    Quote Originally Posted by BGM View Post
    Suppose
    \begin{bmatrix} Z_1 \\ Z_2 \end{bmatrix} \sim 
N(\begin{bmatrix} 0 \\ 0 \end{bmatrix},
\begin{bmatrix} 1, \frac {\sqrt{6}} {4} \\
\frac {\sqrt{6}} {4}, 1 \end{bmatrix})

    Then
    Pr\{Z_1 \leq 0, Z_2 \leq 0\}
= \int_{-\infty}^{+\infty} \Phi(z)\Phi(\sqrt{3}z)\phi(z)dz

    And the required integral
    I = \frac {\sqrt{3}} {\pi} 
[1 - 2\int_{-\infty}^{+\infty} \Phi(z)\Phi(\sqrt{3}z)\phi(z)dz]

    as
    \phi(z) \rightarrow 0 when z \rightarrow \pm\infty, the first two terms vanish.

    I do not know the accuracy of the R program nor mathematica.
    But that result match with your post earlier on, so I just thought that
    should be a correct one.

    I check with that R function once more, it has 15 digits of accuracy:
    (error bounded by 10^(-15))
    It gives 0.3548923441862084
    and I = 0.1600040871941266

    I've got it correct now. ...Your result is correct and I've got Mathematica to match what you did with the R function.

    Thanks again.


    By the way, in this particular article I have been working on, there were a total of 8 integrals of similar form. I managed to get solutions to 6 of the 8 integrals in closed form. The one above is the 7th.


    The remaining Integral is similar but a bit more complicated. I am wondering if you think this same technique can be applied to this (last) case --- but note Z is now raised to the power of 4. Here it is:

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

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    -d(z^3 + 4z)\phi(z) = (z^4 + z^2 - 4)\phi(z)dz
    -\frac {1} {2}d(z^2 + 5)\phi(z)^2 = (z^3 + 4z)\phi(z)^2 dz
    -\frac {1} {3}dz\phi(z)^3 = (z^2 - \frac {1} {3} )\phi(z)^3 dz
    -\frac {1} {4}d\phi(z)^4 = z\phi(z)^4dz

    \int (z^4 + z^2 - 4)\Phi(z)^4\phi(z)dz
    = -\int \Phi(z)^4d(z^3 + 4z)\phi(z)
    = -\Phi(z)^4(z^3 + 4z)\phi(z) + 4\int (z^3 + 4z)\Phi(z)^3\phi(z)^2dz
    = -\Phi(z)^4(z^3 + 4z)\phi(z) - 2\int \Phi(z)^3d(z^2+5)\phi(z)^2
    = -\Phi(z)^4(z^3 + 4z)\phi(z) - 2\Phi(z)^3(z^2+5)\phi(z)^2 + 6 \int (z^2+5)\Phi(z)^2\phi(z)^3 dz
    = -\Phi(z)^4(z^3 + 4z)\phi(z) - 2\Phi(z)^3(z^2+5)\phi(z)^2 - 2 \int \Phi(z)^2dz\phi(z)^3 + 32 \int \Phi(z)^2\phi(z)^3dz
    = -\Phi(z)^4(z^3 + 4z)\phi(z) - 2\Phi(z)^3(z^2+5)\phi(z)^2 - 2z\Phi(z)^2\phi(z)^3 + 4\int z\Phi(z)\phi(z)^4dz+ \frac {16} {\pi} \int \Phi(z)^2\phi(\sqrt{3}z)dz
    = -\Phi(z)^4(z^3 + 4z)\phi(z) - 2\Phi(z)^3(z^2+5)\phi(z)^2 - 2z\Phi(z)^2\phi(z)^3 - \int \Phi(z)d\phi(z)^4+ \frac {16} {\pi} \int \Phi(z)^2\phi(\sqrt{3}z)dz
    = -\Phi(z)^4(z^3 + 4z)\phi(z) - 2\Phi(z)^3(z^2+5)\phi(z)^2 - 2z\Phi(z)^2\phi(z)^3 - \Phi(z)\phi(z)^4 + \int \phi(z)^5 dz
    + \frac {16} {\pi} \int \Phi(z)^2\phi(\sqrt{3}z)dz
    = -\Phi(z)^4(z^3 + 4z)\phi(z) - 2\Phi(z)^3(z^2+5)\phi(z)^2 - 2z\Phi(z)^2\phi(z)^3 - \Phi(z)\phi(z)^4
    + \frac {1} {4\pi^2} \int \phi(\sqrt{5}z) dz+ \frac {16} {\pi} \int \Phi(z)^2\phi(\sqrt{3}z)dz
    = -\Phi(z)^4(z^3 + 4z)\phi(z) - 2\Phi(z)^3(z^2+5)\phi(z)^2 - 2z\Phi(z)^2\phi(z)^3 - \Phi(z)\phi(z)^4
    + \frac {1} {4\sqrt{5}\pi^2} \Phi(\sqrt{5}z)+ \frac {16} {\pi} \int \Phi(z)^2\phi(\sqrt{3}z)dz

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    Quote Originally Posted by BGM View Post
    -d(z^3 + 4z)\phi(z) = (z^4 + z^2 - 4)\phi(z)dz
    -\frac {1} {2}d(z^2 + 5)\phi(z)^2 = (z^3 + 4z)\phi(z)^2 dz
    -\frac {1} {3}dz\phi(z)^3 = (z^2 - \frac {1} {3} )\phi(z)^2 dz

    \int (z^4 + z^2 - 4)\Phi(z)^4\phi(z)dz
    = -\int \Phi(z)^4d(z^3 + 4z)\phi(z)
    = -\Phi(z)^4(z^3 + 4z)\phi(z) + 4\int (z^3 + 4z)\Phi(z)^3\phi(z)^2dz
    = -\Phi(z)^4(z^3 + 4z)\phi(z) - 2\int \Phi(z)^3d(z^2+5)\phi(z)^2
    = -\Phi(z)^4(z^3 + 4z)\phi(z) - 2\Phi(z)^3(z^2+5)\phi(z)^2 
+ 6 \int (z^2+5)\Phi(z)^2\phi(z)^3 dz

    Yes, this is correct. Nice Work!

    BGM: So you don't have to spend too much time, can you just provide me, like you did above in your last post (one above mine regarding the previous integral) where you gave z1, z2, with means of zero and unit variance the covariance between z1 and z2 and then the integral of concern.

    As a check ( I forgot to mention):

    \int (z^4 + z^2 - 4)\Phi(z)^4\phi(z)dz = 0.864683184411258697525039

    Again, thanks for your time.

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    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

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