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)
