Expectation according to joint distribution equals single distribution Expectation [ edit ]
E
p
(
x
1
,
x
2
)
(
X
1
)
=
∑
x
1
,
x
2
p
(
x
1
,
x
2
)
(
x
1
)
=
∑
x
1
∑
x
2
p
(
x
1
)
p
(
x
2
|
x
1
)
x
1
=
{\displaystyle E_{p(x_{1},x_{2})}(X_{1})=\sum _{x_{1},x_{2}}{p(x_{1},x_{2})(x_{1})}=\sum _{x_{1}}\sum _{x_{2}}p(x_{1})p(x_{2}|x_{1})x_{1}=}
∑
x
1
p
(
x
1
)
x
1
∑
x
2
p
(
x
2
|
x
1
)
=
∑
x
1
p
(
x
1
)
x
1
⋅
1
=
E
p
(
x
1
)
(
X
1
)
{\displaystyle \sum _{x_{1}}p(x_{1})x_{1}\sum _{x_{2}}p(x_{2}|x_{1})=\sum _{x_{1}}p(x_{1})x_{1}\cdot 1=E_{p(x_{1})}(X_{1})}
Hence:
E
p
(
x
1
,
x
2
)
(
X
1
)
=
E
p
(
x
1
)
(
X
1
)
{\displaystyle E_{p(x_{1},x_{2})}(X_{1})=E_{p(x_{1})}(X_{1})}
Also:
E
p
(
x
1
,
x
2
)
(
f
(
X
1
)
)
=
E
p
(
x
1
)
(
f
(
X
1
)
)
{\displaystyle E_{p(x_{1},x_{2})}(f(X_{1}))=E_{p(x_{1})}(f(X_{1}))}
E
(
X
1
+
X
2
)
=
∑
x
1
,
x
2
p
(
x
1
,
x
2
)
(
x
1
+
x
2
)
=
{\displaystyle E(X_{1}+X_{2})=\sum _{x_{1},x_{2}}{p(x_{1},x_{2})(x_{1}+x_{2})}=}
∑
x
1
,
x
2
p
(
x
1
,
x
2
)
x
1
+
∑
x
1
,
x
2
p
(
x
1
,
x
2
)
x
2
=
{\displaystyle \sum _{x_{1},x_{2}}{p(x_{1},x_{2})x_{1}}+\sum _{x_{1},x_{2}}{p(x_{1},x_{2})x_{2}}=}
∑
x
1
p
(
x
1
)
x
1
+
∑
x
2
p
(
x
2
)
x
2
=
E
(
X
1
)
+
E
(
X
2
)
{\displaystyle \sum _{x_{1}}{p(x_{1})x_{1}}+\sum _{x_{2}}{p(x_{2})x_{2}}=E(X_{1})+E(X_{2})}
E
(
X
1
+
X
2
)
=
E
(
X
1
)
+
E
(
X
2
)
{\displaystyle E(X_{1}+X_{2})=E(X_{1})+E(X_{2})}
E
(
λ
X
)
=
∑
x
p
(
x
)
λ
x
=
λ
∑
x
p
(
x
)
x
=
λ
E
(
X
)
{\displaystyle E(\lambda X)=\sum _{x}p(x)\lambda x=\lambda \sum _{x}p(x)x=\lambda E(X)}
hence:
E
(
λ
X
)
=
λ
E
(
X
)
{\displaystyle E(\lambda X)=\lambda E(X)}
[ edit ]
V
(
X
)
=
d
e
f
=
E
(
[
X
−
E
(
X
)
]
2
)
{\displaystyle V(X)=def=E([X-E(X)]^{2})}
σ
(
X
)
=
d
e
f
=
V
(
X
)
{\displaystyle \sigma (X)=def={\sqrt {V(X)}}}
The meaning of standard deviation [ edit ] One way to look at standard deviation is as an approximation of the "expected drift" from the expectation. The "expected drift" could be defined as:
E
D
(
X
)
=
d
e
f
?
=
E
(
|
X
−
E
(
X
)
|
)
{\displaystyle ED(X)=def?=E(|X-E(X)|)}
This value is probably not easy to manipulate.
Suppose that X can have only the two values
k
{\displaystyle k}
−
k
{\displaystyle -k}
E
(
X
)
=
0
{\displaystyle E(X)=0}
V
(
X
)
=
d
e
f
=
E
(
[
X
−
E
(
X
)
]
2
)
=
E
(
X
2
)
=
k
2
{\displaystyle V(X)=def=E([X-E(X)]^{2})=E(X^{2})=k^{2}}
and
σ
(
X
)
=
V
(
X
)
=
k
{\displaystyle \sigma (X)={\sqrt {V(X)}}=k}
and
E
D
(
X
)
=
E
(
|
X
−
E
(
X
)
|
)
=
E
(
|
X
|
)
=
E
(
k
)
=
k
=
σ
(
X
)
{\displaystyle ED(X)=E(|X-E(X)|)=E(|X|)=E(k)=k=\sigma (X)}
V
{\displaystyle V}
σ
{\displaystyle \sigma }
E
D
{\displaystyle ED}
X
{\displaystyle X}
k
{\displaystyle k}
σ
(
X
)
=
E
D
(
X
)
{\displaystyle \sigma (X)=ED(X)}
Whenever the drift values are not the same,
σ
{\displaystyle \sigma }
E
D
{\displaystyle ED}
Example :Suppose you are performing the following experiment: you flip a coin, if it's head you go 5 meters to the left, if it is tail, you go 5 meters to the right. The variance in this case is 25 and the standard deviation is 5. The expected drift is also 5 (all the drift values are equal). More on that example, see here .
Alternative definition of variance [ edit ]
V
(
X
)
=
d
e
f
=
E
(
(
X
−
E
(
X
)
)
2
)
=
E
(
X
2
+
E
2
(
X
)
−
2
X
E
(
X
)
)
=
{\displaystyle V(X)=def=E((X-E(X))^{2})=E(X^{2}+E^{2}(X)-2XE(X))=}
E
(
X
2
)
+
E
2
(
X
)
−
2
E
2
(
X
)
=
E
(
X
2
)
−
E
2
(
X
)
{\displaystyle E(X^{2})+E^{2}(X)-2E^{2}(X)=E(X^{2})-E^{2}(X)}
hence:
V
(
X
)
=
E
(
X
2
)
−
E
2
(
X
)
{\displaystyle V(X)=E(X^{2})-E^{2}(X)}
[ edit ]
V
(
X
+
c
)
=
E
(
[
X
+
c
−
E
(
X
+
c
)
]
2
)
=
E
(
[
X
+
c
−
E
(
X
)
−
E
(
c
)
]
2
)
=
E
(
[
X
−
E
(
X
)
]
2
)
=
V
(
X
)
{\displaystyle V(X+c)=E([X+c-E(X+c)]^{2})=E([X+c-E(X)-E(c)]^{2})=E([X-E(X)]^{2})=V(X)}
variance of multiplication [ edit ]
V
(
λ
X
)
=
E
(
(
λ
X
)
2
)
−
E
2
(
λ
X
)
=
λ
2
E
(
X
2
)
−
λ
2
E
2
(
X
)
=
λ
2
(
E
(
X
2
)
−
E
2
(
X
)
)
=
λ
2
V
(
X
)
{\displaystyle V(\lambda X)=E((\lambda X)^{2})-E^{2}(\lambda X)=\lambda ^{2}E(X^{2})-\lambda ^{2}E^{2}(X)=\lambda ^{2}(E(X^{2})-E^{2}(X))=\lambda ^{2}V(X)}
hence:
V
(
λ
X
)
=
λ
2
V
(
X
)
{\displaystyle V(\lambda X)=\lambda ^{2}V(X)}
SD of multiplication [ edit ]
σ
(
λ
X
)
=
V
(
λ
X
)
=
λ
2
V
(
X
)
=
λ
V
(
X
)
=
λ
σ
(
X
)
{\displaystyle \sigma (\lambda X)={\sqrt {V(\lambda X)}}={\sqrt {\lambda ^{2}V(X)}}=\lambda {\sqrt {V(X)}}=\lambda \sigma (X)}
hence:
σ
(
λ
X
)
=
λ
σ
(
X
)
{\displaystyle \sigma (\lambda X)=\lambda \sigma (X)}
Variance of sum of random variables [ edit ]
V
(
X
1
+
X
2
)
=
E
(
(
X
1
+
X
2
)
2
)
−
E
2
(
X
1
+
X
2
)
=
{\displaystyle V(X_{1}+X_{2})=E((X_{1}+X_{2})^{2})-E^{2}(X_{1}+X_{2})=}
E
(
X
1
2
)
+
E
(
X
2
2
)
+
2
E
(
X
1
X
2
)
−
{\displaystyle E(X_{1}^{2})+E(X_{2}^{2})+2E(X_{1}X_{2})-}
(
E
2
(
X
1
)
+
E
2
(
X
2
)
+
2
E
(
X
1
)
E
(
X
2
)
)
=
⋅
{\displaystyle (E^{2}(X_{1})+E^{2}(X_{2})+2E(X_{1})E(X_{2}))=\cdot }
V
(
X
1
)
+
V
(
X
2
)
+
2
C
o
v
(
X
1
,
X
2
)
⋅
{\displaystyle V(X_{1})+V(X_{2})+2Cov(X_{1},X_{2})\cdot }
hence:
V
(
X
1
+
X
2
)
=
V
(
X
1
)
+
V
(
X
2
)
+
2
C
o
v
(
X
1
,
X
2
)
{\displaystyle V(X_{1}+X_{2})=V(X_{1})+V(X_{2})+2Cov(X_{1},X_{2})}
When
X
1
{\displaystyle X_{1}}
X
2
{\displaystyle X_{2}}
C
o
v
(
X
1
,
X
2
)
=
0
{\displaystyle Cov(X_{1},X_{2})=0}
X
1
,
X
2
{\displaystyle X_{1},X_{2}}
⇒
V
(
X
1
+
X
2
)
=
V
(
X
1
)
+
V
(
X
2
)
{\displaystyle \Rightarrow V(X_{1}+X_{2})=V(X_{1})+V(X_{2})}
When
X
1
{\displaystyle X_{1}}
X
2
{\displaystyle X_{2}}
X
1
,
X
2
{\displaystyle X_{1},X_{2}}
⇒
V
(
X
1
+
X
2
)
=
V
(
X
1
)
+
V
(
X
2
)
=
2
V
(
X
1
)
{\displaystyle \Rightarrow V(X_{1}+X_{2})=V(X_{1})+V(X_{2})=2V(X_{1})}
Or more generally:
X
1
,
X
2
,
…
,
X
n
{\displaystyle X_{1},X_{2},\ldots ,X_{n}}
⇒
V
(
∑
i
=
1
n
X
i
)
=
∑
i
=
1
n
V
(
X
i
)
=
n
V
(
X
1
)
{\displaystyle \Rightarrow V(\sum _{i=1}^{n}{X_{i}})=\sum _{i=1}^{n}V(X_{i})=nV(X_{1})}
hence:
X
1
,
X
2
,
…
,
X
n
{\displaystyle X_{1},X_{2},\ldots ,X_{n}}
⇒
σ
(
∑
i
=
1
n
X
i
)
=
V
(
∑
i
=
1
n
X
i
)
=
n
V
(
X
1
)
=
n
⋅
σ
(
X
1
)
{\displaystyle \Rightarrow \sigma (\sum _{i=1}^{n}{X_{i}})={\sqrt {V(\sum _{i=1}^{n}{X_{i}})}}={\sqrt {nV(X_{1})}}={\sqrt {n}}\cdot \sigma (X_{1})}
V
(
X
1
+
X
1
)
=
V
(
2
X
1
)
=
{\displaystyle V(X_{1}+X_{1})=V(2X_{1})=}
4
{\displaystyle 4}
V
(
X
1
)
{\displaystyle V(X_{1})}
and
σ
(
X
1
+
X
1
)
=
σ
(
2
X
1
)
=
{\displaystyle \sigma (X_{1}+X_{1})=\sigma (2X_{1})=}
2
{\displaystyle 2}
σ
(
X
1
)
{\displaystyle \sigma (X_{1})}
more on the last result [ edit ] We've showed that:
X
1
,
X
2
,
…
,
X
n
{\displaystyle X_{1},X_{2},\ldots ,X_{n}}
⇒
σ
(
∑
i
=
1
n
X
i
)
=
n
⋅
σ
(
X
1
)
{\displaystyle \Rightarrow \sigma (\sum _{i=1}^{n}{X_{i}})={\sqrt {n}}\cdot \sigma (X_{1})}
Why is this important?
σ
{\displaystyle \sigma }
mean drift tends to zero:
lim
n
→
∞
σ
(
∑
i
=
1
n
X
i
)
n
=
lim
n
→
∞
n
⋅
σ
(
X
1
)
n
=
0
{\displaystyle \lim _{n\to \infty }{\frac {\sigma (\sum _{i=1}^{n}X_{i})}{n}}=\lim _{n\to \infty }{\frac {{\sqrt {n}}\cdot \sigma (X_{1})}{n}}=0}
Recall the example of the random walk +-5.
Now, suppose You repeat the process
n
{\displaystyle n}
The standard deviation, which can be considered as a measure to that drift is:
n
⋅
5
{\displaystyle {\sqrt {n}}\cdot 5}
The mean drift is:
5
n
n
{\displaystyle 5{\frac {\sqrt {n}}{n}}}
For example, for 10000 iterations, the mean drift is:
5
100000
10000
=
0.05
{\displaystyle 5{\frac {\sqrt {100000}}{10000}}=0.05}
todo:*...example of random walk +-5 gnuplot picture. the relation to the law of big numbers... the fact that frequent ration converges is an assumption in probability theory or a result?..
V
(
X
)
=
0
⇔
E
(
[
X
−
E
(
X
)
]
2
)
=
0
⇔
∀
x
,
x
−
E
(
X
)
=
0
⇔
X
{\displaystyle V(X)=0\Leftrightarrow E([X-E(X)]^{2})=0\Leftrightarrow \forall {x},x-E(X)=0\Leftrightarrow X}
hence:
V
(
X
)
=
0
⇔
X
{\displaystyle V(X)=0\Leftrightarrow X}
Alternative definition [ edit ]
C
o
v
(
X
1
,
X
2
)
=
E
(
(
X
1
−
E
(
X
1
)
)
(
X
2
−
E
(
X
2
)
)
)
=
E
(
X
1
X
2
)
+
E
(
X
1
)
E
(
X
2
)
−
2
E
(
X
1
)
E
(
X
2
)
=
E
(
X
1
X
2
)
−
E
(
X
1
)
E
(
X
2
)
{\displaystyle Cov(X_{1},X_{2})=E((X_{1}-E(X_{1}))(X_{2}-E(X_{2})))=E(X_{1}X_{2})+E(X_{1})E(X_{2})-2E(X_{1})E(X_{2})=E(X_{1}X_{2})-E(X_{1})E(X_{2})}
hence:
C
o
v
(
X
1
,
X
2
)
=
E
(
X
1
X
2
)
−
E
(
X
1
)
E
(
X
2
)
{\displaystyle Cov(X_{1},X_{2})=E(X_{1}X_{2})-E(X_{1})E(X_{2})}
A special case is a covariance of two of the same random variable:
C
o
v
(
X
,
X
)
=
E
(
X
X
)
−
E
(
X
)
E
(
X
)
=
V
(
X
)
{\displaystyle Cov(X,X)=E(XX)-E(X)E(X)=V(X)}
Covariance of independent variables [ edit ] Assume that
X
1
{\displaystyle X_{1}}
X
2
{\displaystyle X_{2}}
E
(
X
1
X
2
)
=
∑
x
1
,
x
2
p
(
x
1
,
x
2
)
x
1
x
2
=
∑
x
1
,
x
2
p
(
x
1
)
p
(
x
2
)
x
1
x
2
{\displaystyle E(X_{1}X_{2})=\sum _{x_{1},x_{2}}{p(x_{1},x_{2})x_{1}x_{2}}=\sum _{x_{1},x_{2}}{p(x_{1})p(x_{2})x_{1}x_{2}}}
∑
x
1
p
(
x
1
)
x
1
∑
x
2
p
(
x
2
)
x
2
=
E
(
X
1
)
E
(
X
2
)
{\displaystyle \sum _{x_{1}}p(x_{1})x_{1}\sum _{x_{2}}p(x_{2})x_{2}=E(X_{1})E(X_{2})}
And hence:
X
1
,
X
2
{\displaystyle X_{1},X_{2}}
⟹
C
o
v
(
X
1
,
X
2
)
=
0
{\displaystyle \implies Cov(X_{1},X_{2})=0}
The contrary is not true, however. For example, if X is a constant random variable then
C
O
V
(
X
,
X
)
=
V
(
X
)
=
0
{\displaystyle COV(X,X)=V(X)=0}
But of course, X and X are very much dependent.
(also known as "Brownian motion")
Let Z be a stochastic process with the following properties:
1. The change
δ
Z
{\displaystyle \delta Z}
δ
t
{\displaystyle \delta t}
δ
Z
=
ϵ
⋅
δ
t
{\displaystyle \delta Z=\epsilon \cdot {\sqrt {\delta t}}}
where:
ϵ
∼
ϕ
(
0
,
1
)
{\displaystyle \epsilon \sim \phi (0,1)}
E
p
(
x
1
,
x
2
)
(
X
1
)
=
E
p
(
x
1
)
(
X
1
)
{\displaystyle E_{p(x_{1},x_{2})}(X_{1})=E_{p(x_{1})}(X_{1})}
E
(
X
1
+
X
2
)
=
E
(
X
1
)
+
E
(
X
2
)
{\displaystyle E(X_{1}+X_{2})=E(X_{1})+E(X_{2})}
E
(
λ
X
)
=
λ
E
(
X
)
{\displaystyle E(\lambda X)=\lambda E(X)}
Variance and standard deviation [ edit ]
V
(
X
)
=
E
(
X
2
)
−
E
2
(
X
)
{\displaystyle V(X)=E(X^{2})-E^{2}(X)}
V
(
λ
X
)
=
λ
2
V
(
X
)
{\displaystyle V(\lambda X)=\lambda ^{2}V(X)}
σ
(
λ
X
)
=
λ
σ
(
X
)
{\displaystyle \sigma (\lambda X)=\lambda \sigma (X)}
V
(
X
1
+
X
2
)
=
V
(
X
1
)
+
V
(
X
2
)
+
2
C
o
v
(
X
1
,
X
2
)
{\displaystyle V(X_{1}+X_{2})=V(X_{1})+V(X_{2})+2Cov(X_{1},X_{2})}
X
1
,
X
2
{\displaystyle X_{1},X_{2}}
⇒
V
(
X
1
+
X
2
)
=
V
(
X
1
)
+
V
(
X
2
)
{\displaystyle \Rightarrow V(X_{1}+X_{2})=V(X_{1})+V(X_{2})}
X
1
,
X
2
,
…
,
X
n
{\displaystyle X_{1},X_{2},\ldots ,X_{n}}
⇒
V
(
∑
i
=
1
n
X
i
)
=
n
V
(
X
1
)
{\displaystyle \Rightarrow V(\sum _{i=1}^{n}{X_{i}})=nV(X_{1})}
X
1
,
X
2
,
…
,
X
n
{\displaystyle X_{1},X_{2},\ldots ,X_{n}}
⇒
σ
(
∑
i
=
1
n
X
i
)
=
n
⋅
σ
(
X
1
)
{\displaystyle \Rightarrow \sigma (\sum _{i=1}^{n}{X_{i}})={\sqrt {n}}\cdot \sigma (X_{1})}
C
o
v
(
X
1
,
X
2
)
=
E
(
X
1
X
2
)
−
E
(
X
1
)
E
(
X
2
)
{\displaystyle Cov(X_{1},X_{2})=E(X_{1}X_{2})-E(X_{1})E(X_{2})}
int main () {
cout << "hello lord \n " ;
}
Determinant is the area of the Parallelogram [ edit ] Let
x
1
=
(
x
1
,
y
1
)
{\displaystyle \mathbf {x_{1}} =(x_{1},y_{1})}
x
2
=
(
x
2
,
y
2
)
{\displaystyle \mathbf {x_{2}} =(x_{2},y_{2})}
R
2
{\displaystyle \mathbb {R} ^{2}}
x
1
⋅
y
2
−
x
2
⋅
y
1
{\displaystyle x_{1}\cdot y_{2}-x_{2}\cdot y_{1}}
let
a
{\displaystyle \mathbf {a} }
x
1
{\displaystyle \mathbf {x_{1}} }
a
=
(
−
y
1
,
x
1
)
‖
x
1
‖
{\displaystyle \mathbf {a} ={\frac {(-y1,x1)}{\|\mathbf {x_{1}} \|}}}
(a word about left/right systems? why we didn't choose
(
y
1
,
−
x
1
)
{\displaystyle (y1,-x1)}
Let
S
{\displaystyle S}
S
=
⟨
a
,
x
2
⟩
⋅
‖
x
1
‖
=
⟨
‖
x
1
‖
a
,
x
2
⟩
=
⟨
(
−
y
1
,
x
1
)
,
(
x
2
,
y
2
)
⟩
=
−
y
1
x
2
+
x
1
y
2
{\displaystyle S=\langle \mathbf {a} ,\mathbf {x_{2}} \rangle \cdot \|\mathbf {x_{1}} \|=\langle \|\mathbf {x_{1}} \|\mathbf {a} ,\mathbf {x_{2}} \rangle =\langle (-y_{1},x_{1}),(x_{2},y_{2})\rangle =-y_{1}x_{2}+x_{1}y_{2}}
Let
N
{\displaystyle \mathbf {N} }
N
=
x
2
−
⟨
x
1
,
x
2
⟩
‖
x
1
‖
x
1
‖
x
1
‖
=
x
2
−
⟨
x
1
,
x
2
⟩
‖
x
1
‖
2
x
1
{\displaystyle \mathbf {N} =\mathbf {x_{2}} -{\frac {\langle \mathbf {x_{1}} ,\mathbf {x_{2}} \rangle }{\|\mathbf {x_{1}} \|}}{\frac {\mathbf {x_{1}} }{\|\mathbf {x_{1}} \|}}=\mathbf {x_{2}} -{\frac {\langle \mathbf {x_{1}} ,\mathbf {x_{2}} \rangle }{\|\mathbf {x_{1}} \|^{2}}}\mathbf {x_{1}} }
Let
S
{\displaystyle S}
S
=
‖
N
‖
⋅
‖
x
1
‖
=
‖
x
2
−
⟨
x
1
,
x
2
⟩
‖
x
1
‖
2
x
1
‖
⋅
‖
x
1
‖
=
‖
‖
x
1
‖
2
x
2
−
⟨
x
1
,
x
2
⟩
x
1
‖
‖
x
1
‖
=
‖
⟨
x
1
,
x
1
⟩
x
2
−
⟨
x
1
,
x
2
⟩
x
1
‖
‖
x
1
‖
{\displaystyle S=\|\mathbf {N} \|\cdot \|\mathbf {x_{1}} \|=\|\mathbf {x_{2}} -{\frac {\langle \mathbf {x_{1}} ,\mathbf {x_{2}} \rangle }{\|\mathbf {x_{1}} \|^{2}}}\mathbf {x_{1}} \|\cdot \|\mathbf {x_{1}} \|={\frac {\left\|\|\mathbf {x_{1}} \|^{2}\mathbf {x_{2}} -\langle \mathbf {x_{1}} ,\mathbf {x_{2}} \rangle \mathbf {x_{1}} \right\|}{\|\mathbf {x_{1}} \|}}={\frac {\left\|\langle \mathbf {x_{1}} ,\mathbf {x_{1}} \rangle \mathbf {x_{2}} -\langle \mathbf {x_{1}} ,\mathbf {x_{2}} \rangle \mathbf {x_{1}} \right\|}{\|\mathbf {x_{1}} \|}}}
S
2
=
‖
⟨
x
1
,
x
1
⟩
x
2
−
⟨
x
1
,
x
2
⟩
x
1
‖
2
‖
x
1
‖
2
=
⟨
⟨
x
1
,
x
1
⟩
x
2
−
⟨
x
1
,
x
2
⟩
x
1
,
⟨
x
1
,
x
1
⟩
x
2
−
⟨
x
1
,
x
2
⟩
x
1
⟩
⟨
x
1
,
x
1
⟩
{\displaystyle S^{2}={\frac {\|\langle \mathbf {x_{1}} ,\mathbf {x_{1}} \rangle \mathbf {x_{2}} -\langle \mathbf {x_{1}} ,\mathbf {x_{2}} \rangle \mathbf {x_{1}} \|^{2}}{\|\mathbf {x_{1}} \|^{2}}}={\frac {\left\langle \langle \mathbf {x_{1}} ,\mathbf {x_{1}} \rangle \mathbf {x_{2}} -\langle \mathbf {x_{1}} ,\mathbf {x_{2}} \rangle \mathbf {x_{1}} ,\langle \mathbf {x_{1}} ,\mathbf {x_{1}} \rangle \mathbf {x_{2}} -\langle \mathbf {x_{1}} ,\mathbf {x_{2}} \rangle \mathbf {x_{1}} \right\rangle }{\langle \mathbf {x_{1}} ,\mathbf {x_{1}} \rangle }}}
=
⟨
x
1
,
x
1
⟩
2
⟨
x
2
,
x
2
⟩
+
⟨
x
1
,
x
2
⟩
2
⟨
x
1
,
x
1
⟩
−
2
⋅
⟨
x
1
,
x
1
⟩
⟨
x
1
,
x
2
⟩
2
⟨
x
1
,
x
1
⟩
{\displaystyle ={\frac {\langle \mathbf {x_{1}} ,\mathbf {x_{1}} \rangle ^{2}\langle \mathbf {x_{2}} ,\mathbf {x_{2}} \rangle +\langle \mathbf {x_{1}} ,\mathbf {x_{2}} \rangle ^{2}\langle \mathbf {x_{1}} ,\mathbf {x_{1}} \rangle -2\cdot \langle \mathbf {x_{1}} ,\mathbf {x_{1}} \rangle \langle \mathbf {x_{1}} ,\mathbf {x_{2}} \rangle ^{2}}{\langle \mathbf {x_{1}} ,\mathbf {x_{1}} \rangle }}}
=
⟨
x
1
,
x
1
⟩
⟨
x
2
,
x
2
⟩
+
⟨
x
1
,
x
2
⟩
2
−
2
⋅
⟨
x
1
,
x
2
⟩
2
=
⟨
x
1
,
x
1
⟩
⟨
x
2
,
x
2
⟩
−
⟨
x
1
,
x
2
⟩
2
{\displaystyle =\langle \mathbf {x_{1}} ,\mathbf {x_{1}} \rangle \langle \mathbf {x_{2}} ,\mathbf {x_{2}} \rangle +\langle \mathbf {x_{1}} ,\mathbf {x_{2}} \rangle ^{2}-2\cdot \langle \mathbf {x_{1}} ,\mathbf {x_{2}} \rangle ^{2}=\langle \mathbf {x_{1}} ,\mathbf {x_{1}} \rangle \langle \mathbf {x_{2}} ,\mathbf {x_{2}} \rangle -\langle \mathbf {x_{1}} ,\mathbf {x_{2}} \rangle ^{2}}
=
(
x
1
2
+
y
1
2
)
(
x
2
2
+
y
2
2
)
−
(
x
1
x
2
+
y
1
y
2
)
2
=
x
1
2
x
2
2
+
y
1
2
y
2
2
+
x
1
2
y
2
2
+
y
1
2
x
2
2
−
(
x
1
2
x
2
2
+
y
1
2
y
2
2
+
2
⋅
x
1
x
2
y
1
y
2
)
{\displaystyle =(x_{1}^{2}+y_{1}^{2})(x_{2}^{2}+y_{2}^{2})-(x_{1}x_{2}+y_{1}y_{2})^{2}=x_{1}^{2}x_{2}^{2}+y_{1}^{2}y_{2}^{2}+x_{1}^{2}y_{2}^{2}+y_{1}^{2}x_{2}^{2}-\left(x_{1}^{2}x_{2}^{2}+y_{1}^{2}y_{2}^{2}+2\cdot x_{1}x_{2}y_{1}y_{2}\right)}
=
x
1
2
y
2
2
+
y
1
2
x
2
2
−
2
⋅
x
1
x
2
y
1
y
2
=
(
x
1
y
2
−
x
2
y
1
)
2
{\displaystyle =x_{1}^{2}y_{2}^{2}+y_{1}^{2}x_{2}^{2}-2\cdot x_{1}x_{2}y_{1}y_{2}=(x_{1}y_{2}-x_{2}y_{1})^{2}}
S
=
x
1
y
2
−
x
2
y
1
{\displaystyle S=x_{1}y_{2}-x_{2}y_{1}}
0
=
0
⋅
(
−
1
)
=
(
1
+
(
−
1
)
)
⋅
(
−
1
)
=
{\displaystyle 0=0\cdot (-1)=(1+(-1))\cdot (-1)=}
=
1
⋅
(
−
1
)
+
(
−
1
)
⋅
(
−
1
)
=
{\displaystyle =1\cdot (-1)+(-1)\cdot (-1)=}
=
(
−
1
)
+
(
−
1
)
⋅
(
−
1
)
{\displaystyle =(-1)+(-1)\cdot (-1)}
לכן:
(
−
1
)
⋅
(
−
1
)
=
1
{\displaystyle (-1)\cdot (-1)=1}
צעד ראשון: 0 כפול כל מספר הוא
0
צעד שני: 0 סכום של הופכיים חיבוריים
צעד שלישי: דיסטריביוטיביות
צעד רביעי: 1 נטרלי כפלי
הסקה: הופכי חיבורי ל - -1-
![{\displaystyle V(X)=def=E([X-E(X)]^{2})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ae1dff4bf1fdb9fdec42b06f0da898521c110cca)
![{\displaystyle V(X)=def=E([X-E(X)]^{2})=E(X^{2})=k^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c1311636108faa39d13273298398bf2fdf9329c5)
![{\displaystyle V(X+c)=E([X+c-E(X+c)]^{2})=E([X+c-E(X)-E(c)]^{2})=E([X-E(X)]^{2})=V(X)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5bdee12489813ae7ab9dbb3a0d121ddaa5e8be49)
![{\displaystyle V(X)=0\Leftrightarrow E([X-E(X)]^{2})=0\Leftrightarrow \forall {x},x-E(X)=0\Leftrightarrow X}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ce8b22b13bd0840cbb79388fd248cb65370323c4)
