公式分析，首先假设q和k都是服从期望为0，方差为1的独立的随机变量。
Assume：
X
=
q
i
X=q_{i}
X=qi，
Y
=
k
i
Y=k_{i}
Y=ki，那么：
1、
E
(
X
Y
)
=
E
(
X
)
E
(
Y
)
=
0
∗
0
=
0
E(XY)=E(X)E(Y)=0*0=0
E(XY)=E(X)E(Y)=0∗0=0
2、
D
(
X
Y
)
=
E
(
X
2
Y
2
)
−
[
E
(
X
Y
)
]
2
D(XY)=E(X^{2}Y^{2})-[E(XY)]^{2}
D(XY)=E(X2Y2)−[E(XY)]2
=
E
(
X
2
)
E
(
Y
2
)
−
[
E
(
X
)
E
(
Y
)
]
2
=E(X^{2})E(Y^{2})-[E(X)E(Y)]^{2}
=E(X2)E(Y2)−[E(X)E(Y)]2
=
E
(
X
2
−
0
2
)
E
(
Y
2
−
0
2
)
−
[
E
(
X
)
E
(
Y
)
]
2
=E(X^{2}-0^{2})E(Y^{2}-0^{2})-[E(X)E(Y)]^{2}
=E(X2−02)E(Y2−02)−[E(X)E(Y)]2
=
E
(
X
2
−
[
E
(
X
)
]
2
)
E
(
Y
2
−
[
E
(
Y
)
]
2
)
−
[
E
(
X
)
E
(
Y
)
]
2
=E(X^{2}-[E(X)]^{2})E(Y^{2}-[E(Y)]^{2})-[E(X)E(Y)]^{2}
=E(X2−[E(X)]2)E(Y2−[E(Y)]2)−[E(X)E(Y)]2
=
[
E
(
X
2
)
−
[
E
(
X
)
]
2
]
[
E
(
Y
2
)
−
[
E
(
Y
)
]
2
]
−
[
E
(
X
)
E
(
Y
)
]
2
=[E(X^{2})-[E(X)]^{2}][E(Y^{2})-[E(Y)]^{2}]-[E(X)E(Y)]^{2}
=[E(X2)−[E(X)]2][E(Y2)−[E(Y)]2]−[E(X)E(Y)]2
=
D
(
X
)
D
(
Y
)
−
[
E
(
X
)
E
(
Y
)
]
2
=D(X)D(Y)-[E(X)E(Y)]^{2}
=D(X)D(Y)−[E(X)E(Y)]2
=
1
∗
1
−
0
∗
0
=1*1-0*0
=1∗1−0∗0
=
1
=1
=1
3、
D
(
Q
K
d
k
)
=
d
k
(
d
k
)
2
=
1
D(\frac{QK}{\sqrt{d_{k}}})=\frac{d_{k}}{(\sqrt{d_{k}})^{2}}=1
D(dkQK)=(dk)2dk=1
需要注意的是，
D
(
Q
K
)
=
D
(
∑
i
=
0
d
k
q
i
k
i
)
=
d
k
∗
1
=
d
k
D(QK)=D(\sum_{i=0}^{d_{k}}q_{i}k_{i})=d_{k}*1=d_{k}
D(QK)=D(∑i=0dkqiki)=dk∗1=dk