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I didn't quite understand before moment generating
function（MGF）. Today, I finally dig hard MGF, Make a summary here , And finally helped me understand moment and MGF Role of . If you don't understand, you can take a few minutes to read this article .
first , What is? moment?
moment In statistics, it means expectation , such as E （ X ） , E （ X 2 ） , … , E（X）,E（X^2）,…, E（X）,E（X2）,…,
Each represents a distribution first moment,second moment,…
So what is the use of these expectations in statistics , Relatively simple , Everyone knows first moment It expresses some kind of distributed expectation , and second
moment To some extent, it can reflect the distribution variance, Because we have σ 2 = E （ X 2 ） − E （ X ） 2 \sigma^2 =
E（X^2）- E（X）^2σ2=E（X2）−E（X）2. however third and fourth moment It is also an important parameter reflecting the distribution characteristics , such as third
moment It reflects the asymmetry of the distribution （asymmetry）, It can also be said to be inclination ?（skewness）, The formula is s k e w n e s s = E （
（ x − μ ） 3 ） / σ 3 skewness = E（（x-\mu）^3）/\sigma^3skewness=E（（x−μ）3）/σ3
;fourth moment It reflects kurtosis （kurtosis：how heavy its tails are）, k u r t o s i s = E （
（ x − μ ） 4 ） / σ 4 kurtosis = E（（x-\mu）^4）/\sigma^4kurtosis=E（（x−μ）4）/σ4
. So for us ,momnet It is a very important parameter to reflect the distribution of a sample .
Then we go on to see what is moment generating
function（MGF）, What I understand is MGF Is to build a solution to help you moment An instrumental function of （ Like its name ,to generate
moment）, First, let's see momnet Solution of : E ( X n ) = ∫ − ∞ ∞ π n × p d f ( x ) d x E(X^n)
=\int_{-\infty}^{\infty}{\pi^n}\times pdf(x) dxE(Xn)=∫−∞∞πn×pdf(x)dx
and MGF yes : M G F ( t ) = E ( e t x ) = MGF(t) = E(e^{tx}) = {} MGF(t)=E(etx)=
M G F ( t ) = E ( e t x ) = { ∑ e t x × p ( x ) , x : d i s c r e t e ; p ( x
) : p m f ∫ e t x × f ( x ) , x : c o n t i n o u s ; f ( x ) : p d f
MGF(t)=E(e^{tx})=\left\{ \begin{aligned} & \sum e^{tx}\times p(x), x: discrete;
p(x): pmf \\ & \int e^{tx}\times f(x), x: continous; f(x): pdf \\ \end{aligned}
\right.MGF(t)=E(etx)=⎩⎪⎨⎪⎧∑etx×p(x),x:discrete;p(x):pmf∫etx×f(x),x:continous;f
(x):pdf
At first glance, people may look confused , What I'm asking is moment E （ X n ） E（X^n） E（Xn）, instead of E （ e t x ） E（e^{tx}） E（
etx）. So keep looking , From our past knowledge of Statistics , We know the basis MGF seek moment The best way is ： yes MDF seek n Order derivative , Then bring in t=0, You can get E （ X n
） E（X^n）E（Xn）.
n-th moment:
E ( X n ) = d n d t n M G F ( t ) ∣ t = 0 E(X^n) =
\frac{d^n}{dt^n}MGF(t)|_{t=0}E(Xn)=dtndnMGF(t)∣t=0
But why ? Next, we give the derivation process ：
Using Taylor's expansion, we can get ：
e t x = 1 + t x + ( t x ) 2 2 ! + ( t x ) 3 3 ! + ( t x ) 4 4 ! + . . + ( t
x ) n n ! e^{tx} = 1 + tx + \frac{(tx)^2}{2!}+ \frac{(tx)^3}{3!}+
\frac{(tx)^4}{4!}+...+ \frac{(tx)^n}{n!}etx=1+tx+2!(tx)2+3!(tx)3+4!(tx)4+...+
n!(tx)n
Then we can expect from this formula
E ( e t x ) = E ( 1 ) + t E ( x ) + t 2 2 ! E ( t 2 ) + t 3 3 ! E ( t 3 ) + t
4 4 ! E ( t 4 ) + . . + t n n ! E ( t n ) E(e^{tx}) = E(1) + tE(x) +
\frac{t^2}{2!}E(t^2)+ \frac{t^3}{3!}E(t^3)+ \frac{t^4}{4!}E(t^4)+...+
\frac{t^n}{n!}E(t^n)E(etx)=E(1)+tE(x)+2!t2E(t2)+3!t3E(t3)+4!t4E(t4)+...+n!tn
E(tn)
In the expectation 1 Order derivative , Bring in t=0
d d t E ( e t x ) = d d t E ( 1 ) + d d t t E ( x ) + d d t t 2 2 ! E ( t 2 )
+ d d t t 3 3 ! E ( t 3 ) + d d t t 4 4 ! E ( t 4 ) + . . + d d t t n n ! E (
t n ) \frac{d}{dt}E(e^{tx}) = \frac{d}{dt}E(1) + \frac{d}{dt}tE(x) +
\frac{d}{dt}\frac{t^2}{2!}E(t^2)+ \frac{d}{dt}\frac{t^3}{3!}E(t^3)+
\frac{d}{dt}\frac{t^4}{4!}E(t^4)+...+ \frac{d}{dt}\frac{t^n}{n!}E(t^n)dtdE(etx)
=dtdE(1)+dtdtE(x)+dtd2!t2E(t2)+dtd3!t3E(t3)+dtd4!t4E(t4)+...+dtdn!tnE(
tn)
Substitute in t = 0
= 0 + E ( x ) + 0 + . . + 0 = E ( X ) = 0 +E(x) + 0 +...+ 0 = E(X) =0+E(x)+0
+...+0=E(X)
What we get in the end is X What are your expectations , If we want to second moment, So what are you bringing in t=0 Continue to find a derivative before , Take it again t=0 Can get second
moment. Similarly ,n-th moment It is on this basis n Order derivative .
Then you may have doubts , Why don't we just ask E （ X n ） E（X^n） E（Xn）
, But to pass MGF Come and ask moment And ? The answer is also obvious , Compared with solving higher-order integral , It's much easier to find the derivative .
Let's take an example , Suppose our pdf Is an exponential distribution , So we have
f ( x ) = { λ e − λ x , x > 0 0 , e l s e f(x)=\left\{ \begin{aligned}
&\lambda e^{-\lambda x}, x>0 \\ &0, else \\ \end{aligned} \right.f(x)={λe−λx,x>
00,else
Then his MGF yes
M G F ( t ) = E ( e t x ) = ∫ 0 ∞ e t x × λ e − λ x d x MGF(t) = E(e^{tx})
=\int_{0}^{\infty} e^{tx}\times \lambda e^{-\lambda x}dxMGF(t)=E(etx)=∫0∞etx×λe
−λxdx
= λ ∫ 0 ∞ e ( t − λ ) x d x , ( t − λ < 0 ) =\lambda
\int_{0}^{\infty}e^{(t-\lambda)x}dx, (t-\lambda < 0)=λ∫0∞e(t−λ)xdx,(t−λ<0)
= λ ∣ 1 t − λ × e ( t − λ ) x ∣ 0 ∞ =\lambda|\frac{1}{t-\lambda}\times
e^{(t-\lambda)x}|^\infty_{0}=λ∣t−λ1×e(t−λ)x∣0∞
= λ λ − t =\frac{\lambda}{\lambda-t} =λ−tλ
Is the final result very simple , Just take the derivative of the above simple example moment, For example, ask third moment： E ( x 3 ) = d 3 d t 3
( λ λ − t ) E(x^3) = \frac{d^3}{dt^3}(\frac{\lambda}{\lambda-t})E(x3)=dt3d3(λ−t
λ), And if according to third moment Definition seeking , E ( x 3 ) = ∫ 0 ∞ x 3 λ e − λ x d x E(x^3) =
\int_{0}^{\infty}x^3\lambda e^{-\lambda x}dxE(x3)=∫0∞x3λe−λxdx
So you're doing derivative and integral , Which do you prefer ?
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