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Key review route of probability theory and mathematical statistics examination
2022-07-05 04:23:00 【IOT classmate Huang】
The key review route of probability theory and mathematical statistics exam
List of articles
Preface
Hope to pass a simple route , Achieve accurate and efficient preparation for tomorrow's exam . Don't talk much , To rush !
The content is divided into two parts: probability theory and mathematical statistics , The middle series is the law of large numbers and central limit theorem in Chapter 5 .
MindMap
Probability theory part
Mathematical statistics
probability theory
Basic concepts
The content of this part , My suggestion is to look directly at my previous blog, Or reading books and other online classes ppt And so on. .
I won't enumerate the distribution function of random variables , You can deduce it directly through the distribution law or probability density
discrete
0-1 Distribution
X~b§
Distribution law
P { X = k } = p k ( 1 − p ) 1 − k , k = 1 , 0 P\{X=k \} = p^k(1-p)^{1-k}, \qquad k = 1, 0 P{ X=k}=pk(1−p)1−k,k=1,0
X | 0 | 1 |
---|---|---|
p_k | 1-p | p |
Mathematical expectation
E ( X ) = p E(X) = p E(X)=p
variance
D ( X ) = ( 1 − p ) ⋅ p D(X) = (1-p)\cdot p D(X)=(1−p)⋅p
The binomial distribution
X~b(n, p)
Distribution law
P { X = k } = p k ( 1 − p ) 1 − k P\{X=k \} = p^k(1-p)^{1-k} P{ X=k}=pk(1−p)1−k
Mathematical expectation
E ( X ) = n p E(X) = np E(X)=np
variance
D ( X ) = n ( 1 − p ) ⋅ p D(X) = n(1-p)\cdot p D(X)=n(1−p)⋅p
Poisson distribution
X~π(λ)
Distribution law
P { X = k } = λ k e − λ k ! , k = 0 , 1 , 2... P\{X=k \} = \frac{\lambda^ke^{-\lambda}}{k!}, \qquad k=0,1,2... P{ X=k}=k!λke−λ,k=0,1,2...
Poisson's theorem
Is to use Poisson to approximate binomial ,np=λ
lim n → ∞ C n k ( 1 − p n ) n − k = λ k e − λ k ! \lim_{n\rightarrow \infty}{C_n^k(1-p_n)^{n-k}} = \frac{\lambda^ke^{-\lambda}}{k!} n→∞limCnk(1−pn)n−k=k!λke−λ
Mathematical expectation
E ( X ) = λ E(X) = \lambda E(X)=λ
variance
D ( X ) = λ D(X) = \lambda D(X)=λ
Continuous type
Uniform distribution
X~U(a, b)
Probability density
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expect
E ( X ) = a + b 2 E(X) = \frac {a+b}{2} E(X)=2a+b
variance
D ( X ) = ( b − a ) 2 12 D(X) = \frac{(b-a)^2}{12} D(X)=12(b−a)2
An index distribution
X~E(θ)
Probability density
KaTeX parse error: No such environment: align at position 26: …eft \{ \begin{̲a̲l̲i̲g̲n̲}̲ &\frac{1}{\…
expect
E ( X ) = θ E(X) = \theta E(X)=θ
variance
D ( X ) = θ 2 D(X) = \theta^2 D(X)=θ2
Normal distribution
X~N(μ, σ)
Probability density
f ( x ) = 1 2 π σ e − ( x − u ) 2 2 σ 2 , − ∞ < x < ∞ f(x) = \frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-u)^2}{2\sigma^2}}, \qquad -\infty < x < \infty f(x)=2πσ1e−2σ2(x−u)2,−∞<x<∞
Standard normal distribution
X ∼ N ( 0 , 1 2 ) φ ( x ) = 1 2 π e − x 2 / 2 X\sim N(0, 1^2)\\ \varphi(x) = \frac{1}{\sqrt{2\pi}}e^{-x^2/2} X∼N(0,12)φ(x)=2π1e−x2/2
Expectation and variance , In general, as long as it is converted into a standard normal distribution , Then it can be solved with the variance and expectation of the standard normal distribution .
expect
E ( x ) = μ E(x) = \mu E(x)=μ
variance
D ( X ) = σ 2 D(X) = \sigma^2 D(X)=σ2
In addition to these, there are actually random variable functions in probability theory , Multi dimensional edges and conditions and combinations , There are also covariance and moments in Chapter 4 . But I won't mention these contents , If necessary, you can see blog Or textbooks .
mathematical statistics
It's swinging , Look at this directly . I'm going back to bed .
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