It is a property of the two-dimensional normal distribution that the joint density of
y
1
{\displaystyle y_{1}}
and
y
2
{\displaystyle y_{2}}
depends only on their covariance and is given explicitly by the expression
p
(
y
1
,
y
2
)
=
1
2
π
1
−
ρ
2
e
−
y
1
2
+
y
2
2
−
2
ρ
y
1
y
2
2
(
1
−
ρ
2
)
{\displaystyle p(y_{1},y_{2})={\frac {1}{2\pi {\sqrt {1-\rho ^{2}}}}}e^{-{\frac {y_{1}^{2}+y_{2}^{2}-2\rho y_{1}y_{2}}{2(1-\rho ^{2})}}}}
where
y
1
{\displaystyle y_{1}}
and
y
2
{\displaystyle y_{2}}
are standard Gaussian random variables with correlation
ϕ
y
1
y
2
=
ρ
{\displaystyle \phi _{y_{1}y_{2}}=\rho }
.
Assume that
r
2
=
Q
(
y
2
)
{\displaystyle r_{2}=Q(y_{2})}
, the correlation between
y
1
{\displaystyle y_{1}}
and
r
2
{\displaystyle r_{2}}
is,
ϕ
y
1
r
2
=
1
2
π
1
−
ρ
2
∫
−
∞
∞
∫
−
∞
∞
y
1
Q
(
y
2
)
e
−
y
1
2
+
y
2
2
−
2
ρ
y
1
y
2
2
(
1
−
ρ
2
)
d
y
1
d
y
2
{\displaystyle \phi _{y_{1}r_{2}}={\frac {1}{2\pi {\sqrt {1-\rho ^{2}}}}}\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }y_{1}Q(y_{2})e^{-{\frac {y_{1}^{2}+y_{2}^{2}-2\rho y_{1}y_{2}}{2(1-\rho ^{2})}}}\,dy_{1}dy_{2}}
.
Since
∫
−
∞
∞
y
1
e
−
1
2
(
1
−
ρ
2
)
y
1
2
+
ρ
y
2
1
−
ρ
2
y
1
d
y
1
=
ρ
2
π
(
1
−
ρ
2
)
y
2
e
ρ
2
y
2
2
2
(
1
−
ρ
2
)
{\displaystyle \int _{-\infty }^{\infty }y_{1}e^{-{\frac {1}{2(1-\rho ^{2})}}y_{1}^{2}+{\frac {\rho y_{2}}{1-\rho ^{2}}}y_{1}}\,dy_{1}=\rho {\sqrt {2\pi (1-\rho ^{2})}}y_{2}e^{\frac {\rho ^{2}y_{2}^{2}}{2(1-\rho ^{2})}}}
,
the correlation
ϕ
y
1
r
2
{\displaystyle \phi _{y_{1}r_{2}}}
may be simplified as
ϕ
y
1
r
2
=
ρ
2
π
∫
−
∞
∞
y
2
Q
(
y
2
)
e
−
y
2
2
2
d
y
2
{\displaystyle \phi _{y_{1}r_{2}}={\frac {\rho }{\sqrt {2\pi }}}\int _{-\infty }^{\infty }y_{2}Q(y_{2})e^{-{\frac {y_{2}^{2}}{2}}}\,dy_{2}}
.
The integral above is seen to depend only on the distortion characteristic
Q
(
)
{\displaystyle Q()}
and is independent of
ρ
{\displaystyle \rho }
.
Remembering that
ρ
=
ϕ
y
1
y
2
{\displaystyle \rho =\phi _{y_{1}y_{2}}}
, we observe that for a given distortion characteristic
Q
(
)
{\displaystyle Q()}
, the ratio
ϕ
y
1
r
2
ϕ
y
1
y
2
{\displaystyle {\frac {\phi _{y_{1}r_{2}}}{\phi _{y_{1}y_{2}}}}}
is
K
Q
=
1
2
π
∫
−
∞
∞
y
2
Q
(
y
2
)
e
−
y
2
2
2
d
y
2
{\displaystyle K_{Q}={\frac {1}{\sqrt {2\pi }}}\int _{-\infty }^{\infty }y_{2}Q(y_{2})e^{-{\frac {y_{2}^{2}}{2}}}\,dy_{2}}
.
Therefore, the correlation can be rewritten in the form
ϕ
y
1
r
2
=
K
Q
ϕ
y
1
y
2
{\displaystyle \phi _{y_{1}r_{2}}=K_{Q}\phi _{y_{1}y_{2}}}
.
The above equation is the mathematical expression of the stated "Bussgang‘s theorem".
If
Q
(
x
)
=
sign
(
x
)
{\displaystyle Q(x)={\text{sign}}(x)}
, or called one-bit quantization, then
K
Q
=
2
2
π
∫
0
∞
y
2
e
−
y
2
2
2
d
y
2
=
2
π
{\displaystyle K_{Q}={\frac {2}{\sqrt {2\pi }}}\int _{0}^{\infty }y_{2}e^{-{\frac {y_{2}^{2}}{2}}}\,dy_{2}={\sqrt {\frac {2}{\pi }}}}
.
[ 2] [ 3] [ 1] [ 4]
If the two random variables are both distorted, i.e.,
r
1
=
Q
(
y
1
)
,
r
2
=
Q
(
y
2
)
{\displaystyle r_{1}=Q(y_{1}),r_{2}=Q(y_{2})}
, the correlation of
r
1
{\displaystyle r_{1}}
and
r
2
{\displaystyle r_{2}}
is
ϕ
r
1
r
2
=
∫
−
∞
∞
∫
−
∞
∞
Q
(
y
1
)
Q
(
y
2
)
p
(
y
1
,
y
2
)
d
y
1
d
y
2
{\displaystyle \phi _{r_{1}r_{2}}=\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }Q(y_{1})Q(y_{2})p(y_{1},y_{2})\,dy_{1}dy_{2}}
.
When
Q
(
x
)
=
sign
(
x
)
{\displaystyle Q(x)={\text{sign}}(x)}
, the expression becomes,
ϕ
r
1
r
2
=
1
2
π
1
−
ρ
2
[
∫
0
∞
∫
0
∞
e
−
α
d
y
1
d
y
2
+
∫
−
∞
0
∫
−
∞
0
e
−
α
d
y
1
d
y
2
−
∫
0
∞
∫
−
∞
0
e
−
α
d
y
1
d
y
2
−
∫
−
∞
0
∫
0
∞
e
−
α
d
y
1
d
y
2
]
{\displaystyle \phi _{r_{1}r_{2}}={\frac {1}{2\pi {\sqrt {1-\rho ^{2}}}}}\left[\int _{0}^{\infty }\int _{0}^{\infty }e^{-\alpha }\,dy_{1}dy_{2}+\int _{-\infty }^{0}\int _{-\infty }^{0}e^{-\alpha }\,dy_{1}dy_{2}-\int _{0}^{\infty }\int _{-\infty }^{0}e^{-\alpha }\,dy_{1}dy_{2}-\int _{-\infty }^{0}\int _{0}^{\infty }e^{-\alpha }\,dy_{1}dy_{2}\right]}
where
α
=
y
1
2
+
y
2
2
−
2
ρ
y
1
y
2
2
(
1
−
ρ
2
)
{\displaystyle \alpha ={\frac {y_{1}^{2}+y_{2}^{2}-2\rho y_{1}y_{2}}{2(1-\rho ^{2})}}}
.
Noticing that
∫
−
∞
∞
∫
−
∞
∞
p
(
y
1
,
y
2
)
d
y
1
d
y
2
=
1
2
π
1
−
ρ
2
[
∫
0
∞
∫
0
∞
e
−
α
d
y
1
d
y
2
+
∫
−
∞
0
∫
−
∞
0
e
−
α
d
y
1
d
y
2
+
∫
0
∞
∫
−
∞
0
e
−
α
d
y
1
d
y
2
+
∫
−
∞
0
∫
0
∞
e
−
α
d
y
1
d
y
2
]
=
1
{\displaystyle \int _{-\infty }^{\infty }\int _{-\infty }^{\infty }p(y_{1},y_{2})\,dy_{1}dy_{2}={\frac {1}{2\pi {\sqrt {1-\rho ^{2}}}}}\left[\int _{0}^{\infty }\int _{0}^{\infty }e^{-\alpha }\,dy_{1}dy_{2}+\int _{-\infty }^{0}\int _{-\infty }^{0}e^{-\alpha }\,dy_{1}dy_{2}+\int _{0}^{\infty }\int _{-\infty }^{0}e^{-\alpha }\,dy_{1}dy_{2}+\int _{-\infty }^{0}\int _{0}^{\infty }e^{-\alpha }\,dy_{1}dy_{2}\right]=1}
,
and
∫
0
∞
∫
0
∞
e
−
α
d
y
1
d
y
2
=
∫
−
∞
0
∫
−
∞
0
e
−
α
d
y
1
d
y
2
{\displaystyle \int _{0}^{\infty }\int _{0}^{\infty }e^{-\alpha }\,dy_{1}dy_{2}=\int _{-\infty }^{0}\int _{-\infty }^{0}e^{-\alpha }\,dy_{1}dy_{2}}
,
∫
0
∞
∫
−
∞
0
e
−
α
d
y
1
d
y
2
=
∫
−
∞
0
∫
0
∞
e
−
α
d
y
1
d
y
2
{\displaystyle \int _{0}^{\infty }\int _{-\infty }^{0}e^{-\alpha }\,dy_{1}dy_{2}=\int _{-\infty }^{0}\int _{0}^{\infty }e^{-\alpha }\,dy_{1}dy_{2}}
,
we can simplify the expression of
ϕ
r
1
r
2
{\displaystyle \phi _{r_{1}r_{2}}}
as
ϕ
r
1
r
2
=
4
2
π
1
−
ρ
2
∫
0
∞
∫
0
∞
e
−
α
d
y
1
d
y
2
−
1
{\displaystyle \phi _{r_{1}r_{2}}={\frac {4}{2\pi {\sqrt {1-\rho ^{2}}}}}\int _{0}^{\infty }\int _{0}^{\infty }e^{-\alpha }\,dy_{1}dy_{2}-1}
Also, it is convenient to introduce the polar coordinate
y
1
=
R
cos
θ
,
y
2
=
R
sin
θ
{\displaystyle y_{1}=R\cos \theta ,y_{2}=R\sin \theta }
. It is thus found that
ϕ
r
1
r
2
=
4
2
π
1
−
ρ
2
∫
0
π
/
2
∫
0
∞
e
−
R
2
−
2
R
2
ρ
cos
θ
sin
θ
2
(
1
−
ρ
2
)
R
d
R
d
θ
−
1
=
4
2
π
1
−
ρ
2
∫
0
π
/
2
∫
0
∞
e
−
R
2
(
1
−
ρ
sin
2
θ
)
2
(
1
−
ρ
2
)
R
d
R
d
θ
−
1
{\displaystyle \phi _{r_{1}r_{2}}={\frac {4}{2\pi {\sqrt {1-\rho ^{2}}}}}\int _{0}^{\pi /2}\int _{0}^{\infty }e^{-{\frac {R^{2}-2R^{2}\rho \cos \theta \sin \theta \ }{2(1-\rho ^{2})}}}R\,dRd\theta -1={\frac {4}{2\pi {\sqrt {1-\rho ^{2}}}}}\int _{0}^{\pi /2}\int _{0}^{\infty }e^{-{\frac {R^{2}(1-\rho \sin 2\theta )}{2(1-\rho ^{2})}}}R\,dRd\theta -1}
.
Integration gives
ϕ
r
1
r
2
=
2
1
−
ρ
2
π
∫
0
π
/
2
d
θ
1
−
ρ
sin
2
θ
−
1
=
−
2
π
arctan
(
ρ
−
tan
θ
1
−
ρ
2
)
|
0
π
/
2
−
1
=
2
π
arcsin
(
ρ
)
{\displaystyle \phi _{r_{1}r_{2}}={\frac {2{\sqrt {1-\rho ^{2}}}}{\pi }}\int _{0}^{\pi /2}{\frac {d\theta }{1-\rho \sin 2\theta }}-1=-{\frac {2}{\pi }}\arctan \left({\frac {\rho -\tan \theta }{\sqrt {1-\rho ^{2}}}}\right){\Bigg |}_{0}^{\pi /2}-1={\frac {2}{\pi }}\arcsin(\rho )}
,
This is called "Arcsine law", which was first found by J. H. Van Vleck in 1943 and republished in 1966.[ 2] [ 3] The "Arcsine law" can also be proved in a simpler way by applying Price's Theorem.[ 4] [ 5]
The function
f
(
x
)
=
2
π
arcsin
x
{\displaystyle f(x)={\frac {2}{\pi }}\arcsin x}
can be approximated as
f
(
x
)
≈
2
π
x
{\displaystyle f(x)\approx {\frac {2}{\pi }}x}
when
x
{\displaystyle x}
is small.
Given two jointly normal random variables
y
1
{\displaystyle y_{1}}
and
y
2
{\displaystyle y_{2}}
with joint probability function
p
(
y
1
,
y
2
)
=
1
2
π
1
−
ρ
2
e
−
y
1
2
+
y
2
2
−
2
ρ
y
1
y
2
2
(
1
−
ρ
2
)
{\displaystyle {\displaystyle p(y_{1},y_{2})={\frac {1}{2\pi {\sqrt {1-\rho ^{2}}}}}e^{-{\frac {y_{1}^{2}+y_{2}^{2}-2\rho y_{1}y_{2}}{2(1-\rho ^{2})}}}}}
,
we form the mean
I
(
ρ
)
=
E
(
g
(
y
1
,
y
2
)
)
=
∫
−
∞
+
∞
∫
−
∞
+
∞
g
(
y
1
,
y
2
)
p
(
y
1
,
y
2
)
d
y
1
d
y
2
{\displaystyle I(\rho )=E(g(y_{1},y_{2}))=\int _{-\infty }^{+\infty }\int _{-\infty }^{+\infty }g(y_{1},y_{2})p(y_{1},y_{2})\,dy_{1}dy_{2}}
of some function
g
(
y
1
,
y
2
)
{\displaystyle g(y_{1},y_{2})}
of
(
y
1
,
y
2
)
{\displaystyle (y_{1},y_{2})}
. If
g
(
y
1
,
y
2
)
p
(
y
1
,
y
2
)
→
0
{\displaystyle g(y_{1},y_{2})p(y_{1},y_{2})\rightarrow 0}
as
(
y
1
,
y
2
)
→
0
{\displaystyle (y_{1},y_{2})\rightarrow 0}
, then
∂
n
I
(
ρ
)
∂
ρ
n
=
∫
−
∞
∞
∫
−
∞
∞
∂
2
n
g
(
y
1
,
y
2
)
∂
y
1
n
∂
y
2
n
p
(
y
1
,
y
2
)
d
y
1
d
y
2
=
E
(
∂
2
n
g
(
y
1
,
y
2
)
∂
y
1
n
∂
y
2
n
)
{\displaystyle {\frac {\partial ^{n}I(\rho )}{\partial \rho ^{n}}}=\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }{\frac {\partial ^{2n}g(y_{1},y_{2})}{\partial y_{1}^{n}\partial y_{2}^{n}}}p(y_{1},y_{2})\,dy_{1}dy_{2}=E\left({\frac {\partial ^{2n}g(y_{1},y_{2})}{\partial y_{1}^{n}\partial y_{2}^{n}}}\right)}
.
Proof. The joint characteristic function of the random variables
y
1
{\displaystyle y_{1}}
and
y
2
{\displaystyle y_{2}}
is by definition the integral
Φ
(
ω
1
,
ω
2
)
=
∫
−
∞
∞
∫
−
∞
∞
p
(
y
1
,
y
2
)
e
j
(
ω
1
y
1
+
ω
2
y
2
)
d
y
1
d
y
2
=
exp
{
−
ω
1
2
+
ω
2
2
+
2
ρ
ω
1
ω
2
2
}
{\displaystyle \Phi (\omega _{1},\omega _{2})=\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }p(y_{1},y_{2})e^{j(\omega _{1}y_{1}+\omega _{2}y_{2})}\,dy_{1}dy_{2}=\exp \left\{-{\frac {\omega _{1}^{2}+\omega _{2}^{2}+2\rho \omega _{1}\omega _{2}}{2}}\right\}}
.
From the two-dimensional inversion formula of Fourier transform, it follows that
p
(
y
1
,
y
2
)
=
1
4
π
2
∫
−
∞
∞
∫
−
∞
∞
Φ
(
ω
1
,
ω
2
)
e
−
j
(
ω
1
y
1
+
ω
2
y
2
)
d
ω
1
d
ω
2
=
1
4
π
2
∫
−
∞
∞
∫
−
∞
∞
exp
{
−
ω
1
2
+
ω
2
2
+
2
ρ
ω
1
ω
2
2
}
e
−
j
(
ω
1
y
1
+
ω
2
y
2
)
d
ω
1
d
ω
2
{\displaystyle p(y_{1},y_{2})={\frac {1}{4\pi ^{2}}}\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\Phi (\omega _{1},\omega _{2})e^{-j(\omega _{1}y_{1}+\omega _{2}y_{2})}\,d\omega _{1}d\omega _{2}={\frac {1}{4\pi ^{2}}}\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\exp \left\{-{\frac {\omega _{1}^{2}+\omega _{2}^{2}+2\rho \omega _{1}\omega _{2}}{2}}\right\}e^{-j(\omega _{1}y_{1}+\omega _{2}y_{2})}\,d\omega _{1}d\omega _{2}}
.
Therefore, plugging the expression of
p
(
y
1
,
y
2
)
{\displaystyle p(y_{1},y_{2})}
into
I
(
ρ
)
{\displaystyle I(\rho )}
, and differentiating with respect to
ρ
{\displaystyle \rho }
, we obtain
∂
n
I
(
ρ
)
∂
ρ
n
=
∫
−
∞
∞
∫
−
∞
∞
g
(
y
1
,
y
2
)
p
(
y
1
,
y
2
)
d
y
1
d
y
2
=
∫
−
∞
∞
∫
−
∞
∞
g
(
y
1
,
y
2
)
(
1
4
π
2
∫
−
∞
∞
∫
−
∞
∞
∂
n
Φ
(
ω
1
,
ω
2
)
∂
ρ
n
e
−
j
(
ω
1
y
1
+
ω
2
y
2
)
d
ω
1
d
ω
2
)
d
y
1
d
y
2
=
∫
−
∞
∞
∫
−
∞
∞
g
(
y
1
,
y
2
)
(
(
−
1
)
n
4
π
2
∫
−
∞
∞
∫
−
∞
∞
ω
1
n
ω
2
n
Φ
(
ω
1
,
ω
2
)
e
−
j
(
ω
1
y
1
+
ω
2
y
2
)
d
ω
1
d
ω
2
)
d
y
1
d
y
2
=
∫
−
∞
∞
∫
−
∞
∞
g
(
y
1
,
y
2
)
(
1
4
π
2
∫
−
∞
∞
∫
−
∞
∞
Φ
(
ω
1
,
ω
2
)
∂
2
n
e
−
j
(
ω
1
y
1
+
ω
2
y
2
)
∂
y
1
n
∂
y
2
n
d
ω
1
d
ω
2
)
d
y
1
d
y
2
=
∫
−
∞
∞
∫
−
∞
∞
g
(
y
1
,
y
2
)
∂
2
n
p
(
y
1
,
y
2
)
∂
y
1
n
∂
y
2
n
d
y
1
d
y
2
{\displaystyle {\begin{aligned}{\frac {\partial ^{n}I(\rho )}{\partial \rho ^{n}}}&=\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }g(y_{1},y_{2})p(y_{1},y_{2})\,dy_{1}dy_{2}\\&=\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }g(y_{1},y_{2})\left({\frac {1}{4\pi ^{2}}}\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }{\frac {\partial ^{n}\Phi (\omega _{1},\omega _{2})}{\partial \rho ^{n}}}e^{-j(\omega _{1}y_{1}+\omega _{2}y_{2})}\,d\omega _{1}d\omega _{2}\right)\,dy_{1}dy_{2}\\&=\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }g(y_{1},y_{2})\left({\frac {(-1)^{n}}{4\pi ^{2}}}\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\omega _{1}^{n}\omega _{2}^{n}\Phi (\omega _{1},\omega _{2})e^{-j(\omega _{1}y_{1}+\omega _{2}y_{2})}\,d\omega _{1}d\omega _{2}\right)\,dy_{1}dy_{2}\\&=\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }g(y_{1},y_{2})\left({\frac {1}{4\pi ^{2}}}\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\Phi (\omega _{1},\omega _{2}){\frac {\partial ^{2n}e^{-j(\omega _{1}y_{1}+\omega _{2}y_{2})}}{\partial y_{1}^{n}\partial y_{2}^{n}}}\,d\omega _{1}d\omega _{2}\right)\,dy_{1}dy_{2}\\&=\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }g(y_{1},y_{2}){\frac {\partial ^{2n}p(y_{1},y_{2})}{\partial y_{1}^{n}\partial y_{2}^{n}}}\,dy_{1}dy_{2}\\\end{aligned}}}
After repeated integration by parts and using the condition at
∞
{\displaystyle \infty }
, we obtain the Price's theorem.
∂
n
I
(
ρ
)
∂
ρ
n
=
∫
−
∞
∞
∫
−
∞
∞
g
(
y
1
,
y
2
)
∂
2
n
p
(
y
1
,
y
2
)
∂
y
1
n
∂
y
2
n
d
y
1
d
y
2
=
∫
−
∞
∞
∫
−
∞
∞
∂
2
g
(
y
1
,
y
2
)
∂
y
1
∂
y
2
∂
2
n
−
2
p
(
y
1
,
y
2
)
∂
y
1
n
−
1
∂
y
2
n
−
1
d
y
1
d
y
2
=
⋯
=
∫
−
∞
∞
∫
−
∞
∞
∂
2
n
g
(
y
1
,
y
2
)
∂
y
1
n
∂
y
2
n
p
(
y
1
,
y
2
)
d
y
1
d
y
2
{\displaystyle {\begin{aligned}{\frac {\partial ^{n}I(\rho )}{\partial \rho ^{n}}}&=\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }g(y_{1},y_{2}){\frac {\partial ^{2n}p(y_{1},y_{2})}{\partial y_{1}^{n}\partial y_{2}^{n}}}\,dy_{1}dy_{2}\\&=\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }{\frac {\partial ^{2}g(y_{1},y_{2})}{\partial y_{1}\partial y_{2}}}{\frac {\partial ^{2n-2}p(y_{1},y_{2})}{\partial y_{1}^{n-1}\partial y_{2}^{n-1}}}\,dy_{1}dy_{2}\\&=\cdots \\&=\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }{\frac {\partial ^{2n}g(y_{1},y_{2})}{\partial y_{1}^{n}\partial y_{2}^{n}}}p(y_{1},y_{2})\,dy_{1}dy_{2}\end{aligned}}}
[ 4] [ 5]
Proof of Arcsine law by Price's Theorem
edit
If
g
(
y
1
,
y
2
)
=
sign
(
y
1
)
sign
(
y
2
)
{\displaystyle g(y_{1},y_{2})={\text{sign}}(y_{1}){\text{sign}}(y_{2})}
, then
∂
2
g
(
y
1
,
y
2
)
∂
y
1
∂
y
2
=
4
δ
(
y
1
)
δ
(
y
2
)
{\displaystyle {\frac {\partial ^{2}g(y_{1},y_{2})}{\partial y_{1}\partial y_{2}}}=4\delta (y_{1})\delta (y_{2})}
where
δ
(
)
{\displaystyle \delta ()}
is the Dirac delta function.
Substituting into Price's Theorem, we obtain,
∂
E
(
sign
(
y
1
)
sign
(
y
2
)
)
∂
ρ
=
∂
I
(
ρ
)
∂
ρ
=
∫
−
∞
∞
∫
−
∞
∞
4
δ
(
y
1
)
δ
(
y
2
)
p
(
y
1
,
y
2
)
d
y
1
d
y
2
=
2
π
1
−
ρ
2
{\displaystyle {\frac {\partial E({\text{sign}}(y_{1}){\text{sign}}(y_{2}))}{\partial \rho }}={\frac {\partial I(\rho )}{\partial \rho }}=\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }4\delta (y_{1})\delta (y_{2})p(y_{1},y_{2})\,dy_{1}dy_{2}={\frac {2}{\pi {\sqrt {1-\rho ^{2}}}}}}
.
When
ρ
=
0
{\displaystyle \rho =0}
,
I
(
ρ
)
=
0
{\displaystyle I(\rho )=0}
. Thus
E
(
sign
(
y
1
)
sign
(
y
2
)
)
=
I
(
ρ
)
=
2
π
∫
0
ρ
1
1
−
ρ
2
d
ρ
=
2
π
arcsin
(
ρ
)
{\displaystyle E\left({\text{sign}}(y_{1}){\text{sign}}(y_{2})\right)=I(\rho )={\frac {2}{\pi }}\int _{0}^{\rho }{\frac {1}{\sqrt {1-\rho ^{2}}}}\,d\rho ={\frac {2}{\pi }}\arcsin(\rho )}
,
which is Van Vleck's well-known result of "Arcsine law".
[ 2] [ 3]