# Difficult Integral standard normal pdf/cdf

#### Dragan

##### Super Moderator
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.)

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

where [Math] \Phi \left ( z \right ) [/Math] is the standard normal cdf and [Math] \phi \left ( z \right )[/Math] 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?

#### BGM

##### TS Contributor
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 $z$?

#### BGM

##### TS Contributor
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?

#### fed1

##### TS Contributor
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?

#### BGM

##### TS Contributor
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
$\int\Phi(z)\Phi(\sqrt{2}z)dz$
$= z\Phi(z)\Phi(\sqrt{2}z) - \int z\Phi(\sqrt{2}z)\phi(z) dz - \sqrt{2}\int z\Phi(z)\phi(\sqrt{2}z)dz$
$= z\Phi(z)\Phi(\sqrt{2}z) + \int \Phi(\sqrt{2}z)d\phi(z) + \frac {1} {\sqrt{2}} \int \Phi(z)d\phi(\sqrt{2}z)$
$= z\Phi(z)\Phi(\sqrt{2}z) + \Phi(\sqrt{2}z)\phi(z) - \sqrt{2} \int \phi(z) \phi(\sqrt{2}z) dz$ $+ \frac {1} {\sqrt{2}} \Phi(z)\phi(\sqrt{2}z) - \frac {1} {\sqrt{2}} \int \phi(\sqrt{2}z)\phi(z) dz$
$= z\Phi(z)\Phi(\sqrt{2}z) + \Phi(\sqrt{2}z)\phi(z) + \frac {1} {\sqrt{2}} \Phi(z)\phi(\sqrt{2}z) - \frac {3} {2\sqrt{\pi}} \int \phi(\sqrt{3}z) dz$
$= z\Phi(z)\Phi(\sqrt{2}z) + \Phi(\sqrt{2}z)\phi(z) + \frac {1} {\sqrt{2}} \Phi(z)\phi(\sqrt{2}z) - \frac {1} {2} \sqrt{\frac {3} {\pi}} \Phi(\sqrt{3}z)$

#### Dragan

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

#### Dragan

##### Super Moderator
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:

[Math]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[/Math]

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.

#### BGM

##### TS Contributor
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

#### BGM

##### TS Contributor
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

#### Dragan

##### Super Moderator
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

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.

#### BGM

##### TS Contributor
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

#### Dragan

##### Super Moderator
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:

[Math]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 [/Math]

#### BGM

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

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

#### Dragan

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

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

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):

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

#### BGM

##### TS Contributor
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

#### Dragan

##### Super Moderator
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

#### Dragan

##### Super Moderator
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 [Math] \Phi \left ( z \right ) [/Math] is the standard normal cdf and [Math] \phi \left ( z \right )[/Math] and is standard normal pdf.

And, that should do it.

Last edited:

#### BGM

##### TS Contributor
Thanks. Learn many things about the normal integration.