<>1 Moment generating function
Moment generating function is also called moment generating function （Moment Generating Function） Also known as dynamic difference generating function , Is a constructor , It is defined as ：
random variable X X X When it is a continuous random variable , Its moment generating function is ： M X ( t ) = E ( e t X ) = ∫ − ∞ + ∞ e t x f ( x )
d x M_X(t)=\mathrm{E}(e^{tX})=\int_{-\infty}^{+\infty}e^{tx}f(x)dxMX(t)=E(etX)=
∫−∞+∞etxf(x)dx random variable X X X When it is a discrete random variable , Its moment generating function is ： M X ( t ) = E ( e t X ) = ∑ x i e t
x i p ( x i ) M_X(t)=\mathrm{E}(e^{tX})=\sum\limits_{x_i}e^{tx_i}p(x_i)MX(t)=E(
etX)=xi∑etxip(xi)
From Taylor series e x = 1 + x + x 2 2 ! + x 3 3 ! + x 4 4 ! + ⋯ + x n n ! + ⋯
e^{x}=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots+\frac{x^n}{n!}+\cdots
ex=1+x+2!x2+3!x3+4!x4+⋯+n!xn+⋯ obtain ： M X ( t ) = ∫ − ∞ + ∞ ( 1 + x t + x 2 t 2
2 ! + x 3 t 3 3 ! + x 4 t 4 4 ! + ⋯ + x n t n n ! + ⋯ ) f ( x ) d x = ∫ − ∞ +
∞ f ( x ) d x + t ∫ − ∞ + ∞ x f ( x ) d x + t 2 2 ! ∫ − ∞ + ∞ x 2 f ( x ) d x +
⋯ t n n ! ∫ − ∞ + ∞ x n f ( x ) d x + ⋯ = t 0 M 0 + t 1 M 1 + t 2 2 ! M 2 + ⋯ +
t n n ! M n + ⋯
\begin{aligned}M_X(t)&=\int_{-\infty}^{+\infty}(1+xt+\frac{x^2t^2}{2!}+\frac{x^3t^3}{3!}+\frac{x^4t^4}{4!}+\cdots+\frac{x^nt^n}{n!}+\cdots)f(x)dx\\&=\int_{-\infty}^{+\infty}f(x)dx+t\int_{-\infty}^{+\infty}xf(x)dx+\frac{t^2}{2!}\int_{-\infty}^{+\infty}x^2f(x)dx+\cdots
\frac{t^n}{n!}\int_{-\infty}^{+\infty}x^nf(x)dx+\cdots\\&=t^0 M_0 + t^1 M_1
+\frac{t^2}{2!}M_2+\cdots+\frac{t^n}{n!}M_n+\cdots\end{aligned}MX(t)=∫−∞+∞(1+
xt+2!x2t2+3!x3t3+4!x4t4+⋯+n!xntn+⋯)f(x)dx=∫−∞+∞f(x)dx+t∫−∞+∞xf(x)dx+2!t2∫
−∞+∞x2f(x)dx+⋯n!tn∫−∞+∞xnf(x)dx+⋯=t0M0+t1M1+2!t2M2+⋯+n!tnMn+⋯ among , M n
M_nMn mean X X X of n n n Step center distance .
Moment generating function pair t t t seek n n n First derivative available M X ( n ) ( t ) = ( t 0 M 0 ) ( n ) + ( t 1 M 1 ) (
n ) + ( t 2 2 ! M 2 ) ( n ) + ⋯ + ( t n n ! M n ) ( n ) + ( t ( n + 1 ) ( n + 1
) ! M n + 1 ) ( n ) + ⋯ = 0 + 0 + 0 + ⋯ + M n + ( n + 1 ) ⋅ n ⋅ ( n − 1 ) ⋯ 2 ⋅
t ( n + 1 ) ! M n + 1 + ⋯
\begin{aligned}M^{(n)}_X(t)&=(t^0M_0)^{(n)}+(t^1M_1)^{(n)}+\left(\frac{t^2}{2!}M_2\right)^{(n)}+\cdots+\left(\frac{t^n}{n!}M_n\right)^{(n)}+\left(\frac{t^{(n+1)}}{(n+1)!}M_{n+1}\right)^{(n)}+\cdots\\&=0+0+0+\cdots+M_n+\frac{(n+1)\cdot
n \cdot (n-1) \cdots 2 \cdot t}{(n+1)!}M_{n+1}+\cdots\end{aligned}MX(n)(t)=(t0
M0)(n)+(t1M1)(n)+(2!t2M2)(n)+⋯+(n!tnMn)(n)+((n+1)!t(n+1)Mn+1)(n)+⋯=0+0+0
+⋯+Mn+(n+1)!(n+1)⋅n⋅(n−1)⋯2⋅tMn+1+⋯ When t = 0 t=0 t=0 Time , Then there
M X ( n ) ( 0 ) = 0 + 0 + 0 + ⋯ + M n + ( n + 1 ) ⋅ n ⋅ ( n − 1 ) ⋯ 2 ⋅ 0 ( n
+ 1 ) ! M n + 1 + ⋯ 0 + ⋯ = M n
\begin{aligned}M^{(n)}_X(0)&=0+0+0+\cdots+M_n+\frac{(n+1)\cdot n \cdot (n-1)
\cdots 2 \cdot 0}{(n+1)!}M_{n+1}+\cdots 0 +\cdots\\&=M_n\end{aligned}MX(n)(0)=
0+0+0+⋯+Mn+(n+1)!(n+1)⋅n⋅(n−1)⋯2⋅0Mn+1+⋯0+⋯=Mn It can be seen that random variables X X X The mean and variance are ： E (
X ) = M X ( 1 ) ( 0 ) = ∫ − ∞ + ∞ x f ( x ) d x = M 1
\mathrm{E}(X)=M_X^{(1)}(0)=\int_{-\infty}^{+\infty}xf(x)dx=M_1E(X)=MX(1)(0)=∫−∞
+∞xf(x)dx=M1 V a r ( X ) = E ( X 2 ) − [ E ( X ) ] 2 = ∫ − ∞ + ∞ x 2 p ( x )
d x − ( ∫ − ∞ + ∞ x p ( x ) d x ) 2 = M 2 − ( M 1 ) 2
\mathrm{Var}(X)=\mathrm{E}(X^2)-[\mathrm{E}(X)]^2=\int_{-\infty}^{+\infty}x^2p(x)dx-\left(\int_{-\infty}^{+\infty}xp(x)dx\right)^2=M_2-(M_1)^2
Var(X)=E(X2)−[E(X)]2=∫−∞+∞x2p(x)dx−(∫−∞+∞xp(x)dx)2=M2−(M1)2
<>2 Parameter is n n n and p p p Binomial distribution of
Discrete random variable X X X The compliance parameter is n n n and p p p Binomial distribution of , Then its moment generating function is ϕ ( t ) = E [ e t X ] = ∑ k = 0 n
e t k ( n k ) p k ( 1 − p ) n − k = ∑ k = 0 n ( n k ) ( p e t ) k ( 1 − p ) n −
k = ( p e t + 1 − p ) n
\begin{aligned}\phi(t)&=\mathrm{E}[e^{tX}]=\sum\limits_{k=0}^ne^{tk}\left(\begin{array}{c}n\\k\end{array}\right)p^k
(1-p)^{n-k}\\&=\sum\limits_{k=0}^n\left(\begin{array}{c}n\\k\end{array}\right)(pe^t)^k(1-p)^{n-k}\\&=(pe^t+1-p)^n\end{aligned}
ϕ(t)=E[etX]=k=0∑netk(nk)pk(1−p)n−k=k=0∑n(nk)(pet)k(1−p)n−k=(pet+1−p)n therefore ϕ
′ ( t ) = n ( p e t + 1 − p ) n − 1 p e t \phi^{\prime}(t)=n(pe^t+1-p)^{n-1}pe^t
ϕ′(t)=n(pet+1−p)n−1pet So there are E [ X ] = ϕ ′ ( 0 ) = n p
\mathrm{E}[X]=\phi^{\prime}(0)=npE[X]=ϕ′(0)=np Find the second-order Guide ϕ ′ ′ ( t ) = n ( n − 1 )
( p e t + 1 − p ) n − 2 ( p e t ) 2 + n ( p e t + 1 − p ) n − 1 p e t
\phi^{\prime\prime}(t)=n(n-1)(pe^t+1-p)^{n-2}(pe^t)^2+n(pe^t+1-p)^{n-1}pe^tϕ′′(t
)=n(n−1)(pet+1−p)n−2(pet)2+n(pet+1−p)n−1pet therefore E [ X 2 ] = ϕ ′ ′ ( 0 ) = n ( n −
1 ) p 2 + n p \mathrm{E}[X^2]=\phi^{\prime\prime}(0)=n(n-1)p^2+npE[X2]=ϕ′′(0)=n(
n−1)p2+np therefore , X X X The variance of is V a r ( X ) = E [ X 2 ] − ( E [ X ] ) 2 = n ( n − 1 ) p
2 + n p − n 2 p 2 = n p ( 1 − p )
\mathrm{Var}(X)=\mathrm{E}[X^2]-(\mathrm{E}[X])^2=n(n-1)p^2+np-n^2p^2=np(1-p)Var
(X)=E[X2]−(E[X])2=n(n−1)p2+np−n2p2=np(1−p)
<>3 The mean value is λ \lambda λ Poisson distribution of
Discrete random variable X X X Obey the mean value of λ \lambda λ Poisson distribution of , Then its moment generating function is ϕ ( t ) = E [ e t X ] = ∑ k = 0 ∞
e t n e − λ λ n n ! = e − λ ∑ n = 0 ∞ ( λ e t ) n n ! = e − λ e λ e t = exp {
λ ( e t − 1 ) }
\phi(t)=\mathrm{E}[e^{tX}]=\sum\limits_{k=0}^\infty\frac{e^{tn}e^{-\lambda}\lambda^n}{n!}=e^{-\lambda}\sum\limits_{n=0}^{\infty}\frac{(\lambda
e^t)^n}{n!}=e^{-\lambda}e^{\lambda e^t}=\exp\{\lambda(e^t-1)\}ϕ(t)=E[etX]=k=0∑∞
n!etne−λλn=e−λn=0∑∞n!(λet)n=e−λeλet=exp{λ(et−1)} Differential can be obtained ϕ ′ ( t ) = λ e t exp
{ λ ( e t − 1 ) } ϕ ′ ′ ( t ) = ( λ e t ) 2 exp { λ ( e t − 1 ) } + λ e t {
λ ( e t − 1 ) } \begin{aligned}\phi^{\prime}(t)&=\lambda
e^t\exp\{\lambda(e^t-1)\}\\\phi^{\prime\prime}(t)&=(\lambda e^t)^2
\exp\{\lambda(e^t-1)\}+\lambda e^t\{\lambda(e^t-1)\}\end{aligned}ϕ′(t)ϕ′′(t)=λe
texp{λ(et−1)}=(λet)2exp{λ(et−1)}+λet{λ(et−1)} So there are E [ X ] = ϕ ′ ( 0 ) = λ E [
X ] = ϕ ′ ′ ( 0 ) = λ 2 + λ V a r ( X ) = E [ X 2 ] − ( E [ X ] ) 2 = λ
\begin{aligned}\mathrm{E}[X]&=\phi^{\prime}(0)=\lambda\\\mathrm{E}[X]&=\phi^{\prime\prime}(0)=\lambda^2+\lambda\\
\mathrm{Var}(X)&=\mathrm{E}[X^2]-(\mathrm{E}[X])^2=\lambda\end{aligned}E[X]E[X]V
ar(X)=ϕ′(0)=λ=ϕ′′(0)=λ2+λ=E[X2]−(E[X])2=λ therefore , The mean and variance of Poisson distribution are λ \lambda λ.
<>4 Parameter is λ \lambda λ Exponential distribution of
Discrete random variable X X X The compliance parameter is λ \lambda λ Exponential distribution of , Then its moment generating function is ϕ ( t ) = E [ e t X ] = ∫ 0 ∞ e t
x λ e − λ x d x = λ ∫ 0 ∞ e − ( λ − t ) x d x = λ λ − t , t < λ
\phi(t)=\mathrm{E}[e^{tX}]=\int_0^{\infty}e^{tx}\lambda e^{-\lambda
x}dx=\lambda \int_{0}^{\infty}e^{-(\lambda -t)x}dx =\frac{\lambda}{\lambda
-t},\quad t < \lambdaϕ(t)=E[etX]=∫0∞etxλe−λxdx=λ∫0∞e−(λ−t)xdx=λ−tλ,t<λ
From the above derivation, it can be found that , For exponential distribution , ϕ ( t ) \phi(t) ϕ(t) Only for less than λ \lambda λ of t t t Value definition . yes ϕ ( t )
\phi(t)ϕ(t) Differential can be obtained ϕ ′ ( t ) = λ ( λ − t ) 2 , ϕ ′ ′ ( t ) = 2 λ ( λ − t ) 3
\phi^{\prime}(t)=\frac{\lambda}{(\lambda - t)^2},\quad
\phi^{\prime\prime}(t)=\frac{2\lambda}{(\lambda - t)^3}ϕ′(t)=(λ−t)2λ,ϕ′′(t)=(λ−
t)32λ therefore E [ X ] = ϕ ′ ( 0 ) = 1 λ , E [ X 2 ] = ϕ ′ ′ ( 0 ) = 2 λ 2
\mathrm{E}[X]=\phi^{\prime}(0)=\frac{1}{\lambda},\quad
\mathrm{E}[X^2]=\phi^{\prime\prime}(0)=\frac{2}{\lambda^2}E[X]=ϕ′(0)=λ1,E[X2]=ϕ
′′(0)=λ22 therefore X X X The variance of is V a r ( X ) = E [ X 2 ] − ( E [ X ] ) 2 = 1 λ 2
\mathrm{Var}(X)=\mathrm{E}[X^2]-(\mathrm{E}[X])^2=\frac{1}{\lambda^2}Var(X)=E[X2
]−(E[X])2=λ21
<>5 Parameter is μ \mu μ and δ 2 \delta^2 δ2 Normal distribution of
Standard normal random variable X X X The moment generating function of is as follows E [ e t X ] = 1 2 π ∫ − ∞ + ∞ e t x e − x 2 / 2 d x =
1 2 π ∫ − ∞ + ∞ e − ( x 2 − 2 t x ) d x = e t 2 / 2 1 2 π ∫ − ∞ + ∞ e − ( x − t
) 2 / 2 d x = e t 2 / 2
\begin{aligned}\mathrm{E}[e^{tX}]&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}e^{tx}e^{-x^2/2}dx=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}e^{-(x^2-2tx)}dx\\&=e^{t^2/2}\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}e^{-(x-t)^2/2}dx
= e^{t^2/2}\end{aligned}E[etX]=2π 1∫−∞+∞etxe−x2/2dx=2π 1∫−∞+∞e−(x2−2tx)dx
=et2/22π 1∫−∞+∞e−(x−t)2/2dx=et2/2 If X X X Is the standard normal distribution , that Z = σ X + μ Z=\sigma X
+\muZ=σX+μ The parameter is μ \mu μ and σ 2 \sigma^2 σ2 Normal distribution of , Then there ϕ ( t ) = E [ e t Z ] = E [ e
t ( σ X + μ ) ] = e t u E [ e t σ X ] = exp { σ 2 t 2 2 + μ t }
\phi(t)=\mathrm{E}[e^{tZ}]=\mathrm{E}[e^{t(\sigma X +
\mu)}]=e^{tu}\mathrm{E}[e^{t\sigma X}]=\exp\left\{\frac{\sigma^2t^2}{2}+\mu
t\right\}ϕ(t)=E[etZ]=E[et(σX+μ)]=etuE[etσX]=exp{2σ2t2+μt} After differentiation, we can get ϕ ′ ( t ) =
( μ + t σ 2 ) exp { σ 2 t 2 2 + μ t } ϕ ′ ′ ( t ) = ( μ + t σ 2 ) 2 exp { σ
2 t 2 2 + μ t } + σ 2 exp { σ 2 t 2 2 + μ t }
\begin{aligned}\phi^{\prime}(t)&=(\mu+t\sigma^2)\exp\{\frac{\sigma^2
t^2}{2}+\mu t\}\\\phi^{\prime\prime}(t)&=(\mu+ t\sigma^2)^2\exp\{\frac{\sigma^2
t^2}{2}+\mu t\}+\sigma^2 \exp \left\{\frac{\sigma^2 t^2}{2}+\mu
t\right\}\end{aligned}ϕ′(t)ϕ′′(t)=(μ+tσ2)exp{2σ2t2+μt}=(μ+tσ2)2exp{2σ2t2+μt}+
σ2exp{2σ2t2+μt} So there are E [ X ] = ϕ ′ ( 0 ) = μ , E [ X 2 ] = ϕ ′ ′ ( 0 ) = μ 2 +
σ 2 \mathrm{E}[X]=\phi^{\prime}(0)=\mu,\quad
\mathrm{E}[X^2]=\phi^{\prime\prime}(0)=\mu^2+\sigma^2E[X]=ϕ′(0)=μ,E[X2]=ϕ′′(0)=μ
2+σ2 Variance is V a r ( X ) = E [ X 2 ] − ( E [ X ] ) 2 = σ 2
\mathrm{Var(X)}=\mathrm{E}[X^2]-(\mathrm{E}[X])^2=\sigma^2Var(X)=E[X2]−(E[X])2=σ
2
Technology
Daily Recommendation