当前位置:网站首页>[the Nine Yang Manual] 2017 Fudan University Applied Statistics real problem + analysis
[the Nine Yang Manual] 2017 Fudan University Applied Statistics real problem + analysis
2022-07-06 13:30:00 【Elder martial brother statistics】
Catalog
The real part
One 、(30 branch ) A term is used to explain
(1)(5 branch ) Sample mean 、 Sample variance ;
(2)(5 branch ) statistic ;
(3)(5 branch ) Order statistics ;
(4)(5 branch ) Median 、 Sample median ;
(5)(5 branch ) Empirical distribution function ;
(6)(5 branch ) Unbiased estimate .
Two 、(20 branch ) X 1 , X 2 , X_{1}, X_{2}, X1,X2, i.i.d ∼ Exp ( 1 ) , \sim \operatorname{Exp}(1), ∼Exp(1), seek
(1)(10 branch ) X 1 X 1 + X 2 \frac{X_{1}}{X_{1}+X_{2}} X1+X2X1 Density function of ;
(2)(10 branch ) X ( 2 ) − X ( 1 ) X_{(2)}-X_{(1)} X(2)−X(1) Density function of .
3、 ... and 、(20 branch ) X 1 , X 2 X_{1}, X_{2} X1,X2 i.i.d. ∼ N ( 0 , 1 ) , \sim N(0,1), ∼N(0,1), seek X 1 X 2 \frac{X_{1}}{X_{2}} X2X1 Probability distribution of .
Four 、(20 branch ) X 1 , X 2 , … , X n X_{1}, X_{2}, \ldots, X_{n} X1,X2,…,Xn i.i.d. ∼ F ( x ) , \sim F(x), ∼F(x), remember Y n ( x ) = ∑ i = 1 n I [ X i ≤ x ] , Y_{n}(x)=\sum_{i=1}^{n} I\left[X_{i} \leq x\right], Yn(x)=∑i=1nI[Xi≤x], seek lim n → ∞ Y n ( x ) n \lim _{n \rightarrow \infty} \frac{Y_{n}(x)}{n} limn→∞nYn(x).
5、 ... and 、(20 branch ) X 0 , X 1 , ⋯ , X 2 n , X_{0}, X_{1}, \cdots, X_{2 n}, X0,X1,⋯,X2n, i.i.d ∼ U ( 0 , 1 ) , X ( 0 ) , X ( 1 ) , ⋯ , X ( 2 n ) \sim U(0,1), \quad X_{(0)}, X_{(1)}, \cdots, X_{(2 n)} ∼U(0,1),X(0),X(1),⋯,X(2n) Is the corresponding order statistic , Try to prove X ( n ) → P 1 2 X_{(n)} \stackrel{P}{\rightarrow} \frac{1}{2} X(n)→P21.
6、 ... and 、(20 branch ) Continuous random variables are known X X X The expectations of the E X E X EX There is , f ( t ) = E ∣ X − t ∣ f(t)=E|X-t| f(t)=E∣X−t∣ stay t = m t=m t=m Take the minimum value , prove P ( X ≤ m ) = 1 2 P(X \leq m)=\frac{1}{2} P(X≤m)=21.
7、 ... and 、(20 branch ) X 1 , X 2 , X 3 X_{1}, X_{2}, X_{3} X1,X2,X3 i.i.d ∼ N ( 0 , 1 ) , \sim N(0,1), ∼N(0,1), seek
(1)(10 branch ) P ( X 1 > X 2 > X 3 ) P\left(X_{1}>X_{2}>X_{3}\right) P(X1>X2>X3);
(2)(10 branch ) P ( X 1 > X 2 , X 1 > X 3 ) P\left(X_{1}>X_{2}, X_{1}>X_{3}\right) P(X1>X2,X1>X3).
The analysis part
One 、(30 branch ) A term is used to explain
(1)(5 branch ) Sample mean 、 Sample variance ;
(2)(5 branch ) statistic ;
(3)(5 branch ) Order statistics ;
(4)(5 branch ) Median 、 Sample median ;
(5)(5 branch ) Empirical distribution function ;
(6)(5 branch ) Unbiased estimate .
Solution: (1) The sample mean is taken from a specific population n n n Independent random samples , Calculated average , Write it down as X ˉ = 1 n ∑ i = 1 n X i \bar{X}=\frac{1}{n} \sum_{i=1}^{n} X_{i} Xˉ=n1∑i=1nXi, If the overall expectation exists , be X ˉ \bar{X} Xˉ It is a strong consistent estimate of the overall expectation . The sample variance is based on this n n n The square of the corrected mean deviation calculated from independent random samples , Correction means that the average is divided by its degrees of freedom n − 1 n-1 n−1 Not the number of data n n n, The sample variance is generally recorded as S 2 = 1 n − 1 ∑ i = 1 n ( X i − X ˉ ) 2 S^{2}=\frac{1}{n-1} \sum_{i=1}^{n}\left(X_{i}-\bar{X}\right)^{2} S2=n−11∑i=1n(Xi−Xˉ)2, If the variance of the population exists , be S 2 S^{2} S2 It is also a strong consistent estimate of the population variance .
(2) A statistic is a function whose expression contains only samples but no unknown parameters , It is essentially a random variable ( vector ), Of course, when the value of random samples is given , Statistics can also be regarded as a known constant ( Constant vector ), At this time, the statistic is “ Observations of Statistics ” For short .
(3) Order statistics refers to random samples X 1 , X 2 , … , X n X_{1}, X_{2}, \ldots, X_{n} X1,X2,…,Xn Rearrange the statistics from small to large , Generally, it is written from small to large X ( 1 ) , X ( 2 ) , … , X ( n ) X_{(1)}, X_{(2)}, \ldots, X_{(n)} X(1),X(2),…,X(n).
(4) If x 0.5 x_{0.5} x0.5 Satisfy P { X ≤ x 0.5 } = 0.5 P\left\{X \leq x_{0.5}\right\}=0.5 P{ X≤x0.5}=0.5 said x 0.5 x_{0.5} x0.5 Is the median . And the median of the sample m 0.5 m_{0.5} m0.5 It refers to the number in the middle of the random sample , If expressed in order statistics, it is m 0.5 = { x ( n + 1 2 ) , n = Odd number , x ( n 2 ) + x ( n 2 + 1 ) 2 , n = even numbers . m_{0.5}= \begin{cases}x_{\left(\frac{n+1}{2}\right)}, & n=\text { Odd number , } \\ \frac{x_{\left(\frac{n}{2}\right)}+x_{\left(\frac{n}{2}+1\right)}}{2}, & n=\text { even numbers . }\end{cases} m0.5=⎩⎨⎧x(2n+1),2x(2n)+x(2n+1),n= Odd number , n= even numbers .
(5) Empirical distribution function is an estimate of the overall distribution function based on sample information , Write it down as F n ( x ) = 1 n ∑ i = 1 n I [ X i ≤ x ] , F_{n}(x)=\frac{1}{n} \sum_{i=1}^{n} I\left[X_{i} \leq x\right], Fn(x)=n1i=1∑nI[Xi≤x], according to Glivenko-Cantelli Theorem , The empirical distribution function is a uniformly strong consistent estimate of the population distribution function , namely sup x ∣ F n ( x ) − F ( x ) ∣ * a . s . 0. \sup _{x}\left|F_{n}(x)-F(x)\right| \stackrel{a . s .}{\longrightarrow} 0 . xsup∣Fn(x)−F(x)∣*a.s.0. (6) If g ^ ( X 1 , … , X n ) \hat{g}\left(X_{1}, \ldots, X_{n}\right) g^(X1,…,Xn) Satisfy E g ^ ( X 1 , … , X n ) = g ( θ ) E \hat{g}\left(X_{1}, \ldots, X_{n}\right)=g(\theta) Eg^(X1,…,Xn)=g(θ), said g ^ \hat{g} g^ yes g g g Unbiased estimation of , Unbiasedness is a A very important good standard , But it's not necessary , As if E X E X EX There is , A random sample X 1 X_{1} X1 Always an unbiased estimate of overall expectations , But you can hardly say that it is a good estimate .
Two 、(20 branch ) X 1 , X 2 , X_{1}, X_{2}, X1,X2, i.i.d ∼ Exp ( 1 ) , \sim \operatorname{Exp}(1), ∼Exp(1), seek
(1)(10 branch ) X 1 X 1 + X 2 \frac{X_{1}}{X_{1}+X_{2}} X1+X2X1 Density function of ;
(2)(10 branch ) X ( 2 ) − X ( 1 ) X_{(2)}-X_{(1)} X(2)−X(1) Density function of .
Solution: (1) According to the meaning , 2 X 1 ∼ χ 2 ( 2 ) , 2 X 2 ∼ χ 2 ( 2 ) 2 X_{1} \sim \chi^{2}(2), 2 X_{2} \sim \chi^{2}(2) 2X1∼χ2(2),2X2∼χ2(2) And independent of each other , so X 1 X 1 + X 2 = 2 X 1 2 X 1 + 2 X 2 ∼ Beta ( 1 , 1 ) \frac{X_{1}}{X_{1}+X_{2}}=\frac{2 X_{1}}{2 X_{1}+2 X_{2}} \sim \operatorname{Beta}(1,1) X1+X2X1=2X1+2X22X1∼Beta(1,1), The density function is f ( x ) = 1 , 0 < x < 1 f(x)=1,0<x<1 f(x)=1,0<x<1, namely U ( 0 , 1 ) U(0,1) U(0,1).
(2) It's easy to see X ( 2 ) − X ( 1 ) = ∣ X 1 − X 2 ∣ X_{(2)}-X_{(1)}=\left|X_{1}-X_{2}\right| X(2)−X(1)=∣X1−X2∣, According to the convolution formula Y = X 1 − X 2 Y=X_{1}-X_{2} Y=X1−X2 The density function of is f ( y ) = { 1 2 e − y , y ≥ 0 1 2 e y , y < 0 f(y)= \begin{cases}\frac{1}{2} e^{-y}, & y \geq 0 \\ \frac{1}{2} e^{y}, & y<0\end{cases} f(y)={ 21e−y,21ey,y≥0y<0 so ∣ Y ∣ |Y| ∣Y∣ The density function of is f ( x ) = e − y , y ≥ 0 f(x)=e^{-y}, y \geq 0 f(x)=e−y,y≥0.
[ notes ] if X ∼ χ 2 ( m ) , Y ∼ χ 2 ( n ) X \sim \chi^{2}(m), Y \sim \chi^{2}(n) X∼χ2(m),Y∼χ2(n), And independent of each other , be X X + Y ∼ Beta ( m 2 , n 2 ) . \frac{X}{X+Y} \sim \operatorname{Beta}\left(\frac{m}{2}, \frac{n}{2}\right). X+YX∼Beta(2m,2n). The analysis is as follows : remember U = X X + Y , V = X + Y U=\frac{X}{X+Y}, V=X+Y U=X+YX,V=X+Y, After inverse solution x = u v , y = v − u v x=u v, y=v-u v x=uv,y=v−uv, ∣ J ∣ = ∣ ∂ ( x , y ) ∂ ( u , v ) ∣ = ∣ v u − v 1 − u ∣ = v , |J|=\left|\frac{\partial(x, y)}{\partial(u, v)}\right|=\left|\begin{array}{cc} v & u \\ -v & 1-u \end{array}\right|=v, ∣J∣=∣∣∣∣∂(u,v)∂(x,y)∣∣∣∣=∣∣∣∣v−vu1−u∣∣∣∣=v, therefore f U , V ( u , v ) = v f X , Y ( u v , ( 1 − u ) v ) = u m 2 − 1 ( 1 − u ) n 2 − 1 Γ ( m 2 ) Γ ( n 2 ) ⋅ ( 1 2 ) m + n 2 v m + n 2 − 1 e − v 2 f_{U, V}(u, v)=v f_{X, Y}(u v,(1-u) v)=\frac{u^{\frac{m}{2}-1}(1-u)^{\frac{n}{2}-1}}{\Gamma\left(\frac{m}{2}\right) \Gamma\left(\frac{n}{2}\right)} \cdot\left(\frac{1}{2}\right)^{\frac{m+n}{2}} v^{\frac{m+n}{2}-1} e^{-\frac{v}{2}} fU,V(u,v)=vfX,Y(uv,(1−u)v)=Γ(2m)Γ(2n)u2m−1(1−u)2n−1⋅(21)2m+nv2m+n−1e−2v, so f U ( u ) = ∫ 0 + ∞ f U , V ( u , v ) d v = Γ ( m + n 2 ) Γ ( m 2 ) Γ ( n 2 ) u m 2 − 1 ( 1 − u ) n 2 − 1 ∼ Beta ( m 2 , n 2 ) . f_{U}(u)=\int_{0}^{+\infty} f_{U, V}(u, v) d v=\frac{\Gamma\left(\frac{m+n}{2}\right)}{\Gamma\left(\frac{m}{2}\right) \Gamma\left(\frac{n}{2}\right)} u^{\frac{m}{2}-1}(1-u)^{\frac{n}{2}-1} \sim \operatorname{Beta}\left(\frac{m}{2}, \frac{n}{2}\right). fU(u)=∫0+∞fU,V(u,v)dv=Γ(2m)Γ(2n)Γ(2m+n)u2m−1(1−u)2n−1∼Beta(2m,2n).
3、 ... and 、(20 branch ) X 1 , X 2 X_{1}, X_{2} X1,X2 i.i.d ∼ N ( 0 , 1 ) , \sim N(0,1), ∼N(0,1), seek X 1 X 2 \frac{X_{1}}{X_{2}} X2X1 Probability distribution of .
Solution: Due to denominator X 2 X_{2} X2 The distribution of is about 0 symmetry , therefore X 1 ∣ X 2 ∣ \frac{X_{1}}{\left|X_{2}\right|} ∣X2∣X1 And X 1 X 2 \frac{X_{1}}{X_{2}} X2X1 Homodistribution , And obviously N ( 0 , 1 ) χ 2 ( 1 ) 1 \frac{N(0,1)}{\sqrt{\frac{\chi^{2}(1)}{1}}} 1χ2(1)N(0,1) Is a degree of freedom 1 Of t t t Distribution , therefore X 1 ∣ X 2 ∣ \frac{X_{1}}{\left|X_{2}\right|} ∣X2∣X1 Also, the degree of freedom is 1 Of t t t Distribution , Its probability density is f ( x ) = Γ ( 1 ) π Γ ( 1 2 ) ( x 2 + 1 ) − 1 = 1 π ⋅ 1 1 + x 2 , − ∞ < x < + ∞ , f(x)=\frac{\Gamma(1)}{\sqrt{\pi} \Gamma\left(\frac{1}{2}\right)}\left(x^{2}+1\right)^{-1}=\frac{1}{\pi} \cdot \frac{1}{1+x^{2}},-\infty<x<+\infty, f(x)=πΓ(21)Γ(1)(x2+1)−1=π1⋅1+x21,−∞<x<+∞, The standard Cauchy distribution .
Four 、(20 branch ) X 1 , X 2 , … , X n X_{1}, X_{2}, \ldots, X_{n} X1,X2,…,Xn i.i.d ∼ F ( x ) , \sim F(x), ∼F(x), remember Y n ( x ) = ∑ i = 1 n I [ X i ≤ x ] , Y_{n}(x)=\sum_{i=1}^{n} I\left[X_{i} \leq x\right], Yn(x)=∑i=1nI[Xi≤x], seek lim n → ∞ Y n ( x ) n \lim _{n \rightarrow \infty} \frac{Y_{n}(x)}{n} limn→∞nYn(x).
Solution: According to the strong law of numbers , lim n → ∞ Y n ( x ) n = E I [ X 1 ≤ x ] = P ( X 1 ≤ x ) = F ( x ) \lim _{n \rightarrow \infty} \frac{Y_{n}(x)}{n}=E I\left[X_{1} \leq x\right]=P\left(X_{1} \leq x\right)=F(x) limn→∞nYn(x)=EI[X1≤x]=P(X1≤x)=F(x), a.s.
5、 ... and 、(20 branch ) X 0 , X 1 , ⋯ , X 2 n , X_{0}, X_{1}, \cdots, X_{2 n}, X0,X1,⋯,X2n, i.i.d ∼ U ( 0 , 1 ) , X ( 0 ) , X ( 1 ) , ⋯ , X ( 2 n ) \sim U(0,1), \quad X_{(0)}, X_{(1)}, \cdots, X_{(2 n)} ∼U(0,1),X(0),X(1),⋯,X(2n) Is the corresponding order statistic , Try to prove X ( n ) → P 1 2 X_{(n)} \stackrel{P}{\rightarrow} \frac{1}{2} X(n)→P21.
Solution: So let's calculate Y = X ( n ) Y=X_{(n)} Y=X(n) Density function of , Thought is : from X 0 , X 1 , ⋯ , X 2 n X_{0}, X_{1}, \cdots, X_{2 n} X0,X1,⋯,X2n Choose one of them as Y Y Y, be left over There should be n n n Ratio Y Y Y Small , n n n Ratio Y Y Y Big , Of course, there are ( 2 n + 1 ) ! n ! ⋅ 1 ! ⋅ n ! \frac{(2 n+1) !}{n ! \cdot 1 ! \cdot n !} n!⋅1!⋅n!(2n+1)! Kind of , therefore
f ( y ) = ( 2 n + 1 ) ! n ! n ! P { X 0 ≤ y , X 1 ≤ y , … , X n − 1 ≤ y } f X n ( y ) P { X n + 1 > y , … , X 2 n > y } P { X 0 ≤ y , X 1 ≤ y , … , X n − 1 ≤ y } = P n { X 0 ≤ y } = y n , P { X n + 1 > y , … , X 2 n > y } = P n { X 2 n > y } = ( 1 − y ) n , \begin{gathered} f(y)=\frac{(2 n+1) !}{n ! n !} P\left\{X_{0} \leq y, X_{1} \leq y, \ldots, X_{n-1} \leq y\right\} f_{X_{n}}(y) P\left\{X_{n+1}>y, \ldots, X_{2 n}>y\right\}\\ P\left\{X_{0} \leq y, X_{1} \leq y, \ldots, X_{n-1} \leq y\right\}=P^{n}\left\{X_{0} \leq y\right\}=y^{n}, \\ P\left\{X_{n+1}>y, \ldots, X_{2 n}>y\right\}=P^{n}\left\{X_{2 n}>y\right\}=(1-y)^{n}, \end{gathered} f(y)=n!n!(2n+1)!P{ X0≤y,X1≤y,…,Xn−1≤y}fXn(y)P{ Xn+1>y,…,X2n>y}P{ X0≤y,X1≤y,…,Xn−1≤y}=Pn{ X0≤y}=yn,P{ Xn+1>y,…,X2n>y}=Pn{ X2n>y}=(1−y)n, so f ( y ) = Γ ( 2 n + 2 ) Γ ( n + 1 ) Γ ( n + 1 ) y n ( 1 − y ) n , 0 < y < 1 f(y)=\frac{\Gamma(2 n+2)}{\Gamma(n+1) \Gamma(n+1)} y^{n}(1-y)^{n}, 0<y<1 f(y)=Γ(n+1)Γ(n+1)Γ(2n+2)yn(1−y)n,0<y<1, This is a Beta ( n + 1 , n + 1 ) \operatorname{Beta}(n+1, n+1) Beta(n+1,n+1) Distribution function of . according to Beta The nature of the distribution , E Y = n + 1 2 n + 2 = 1 2 , Var ( Y ) = ( n + 1 ) 2 ( 2 n + 2 ) 2 ( 2 n + 3 ) → 0 , E Y=\frac{n+1}{2 n+2}=\frac{1}{2},\quad \operatorname{Var}(Y)=\frac{(n+1)^{2}}{(2 n+2)^{2}(2 n+3)} \rightarrow 0, EY=2n+2n+1=21,Var(Y)=(2n+2)2(2n+3)(n+1)2→0, Using Chebyshev inequality , P { ∣ Y − 1 2 ∣ > ε } ≤ Var ( Y ) ε 2 → 0 P\left\{\left|Y-\frac{1}{2}\right|>\varepsilon\right\} \leq \frac{\operatorname{Var}(Y)}{\varepsilon^{2}} \rightarrow 0 P{ ∣∣Y−21∣∣>ε}≤ε2Var(Y)→0, therefore X ( n ) → P 1 2 X_{(n)} \stackrel{P}{\rightarrow} \frac{1}{2} X(n)→P21.
6、 ... and 、(20 branch ) Continuous random variables are known X X X The expectations of the E X E X EX There is , f ( t ) = E ∣ X − t ∣ f(t)=E|X-t| f(t)=E∣X−t∣ stay t = m t=m t=m Take the minimum value , prove P ( X ≤ m ) = 1 2 P(X \leq m)=\frac{1}{2} P(X≤m)=21.
Solution: set up X X X The distribution function of F ( x ) F(x) F(x), The MI degree function is p ( x ) p(x) p(x), be f ( t ) = ∫ − ∞ t ( t − x ) d F ( x ) + ∫ t + ∞ ( x − t ) d F ( x ) = t F ( t ) − ∫ − ∞ t x d F ( x ) + ∫ t + ∞ x d F ( x ) − t ( 1 − F ( t ) ) , f(t)=\int_{-\infty}^{t}(t-x) d F(x)+\int_{t}^{+\infty}(x-t) d F(x)=t F(t)-\int_{-\infty}^{t} x d F(x)+\int_{t}^{+\infty} x d F(x)-t(1-F(t)), f(t)=∫−∞t(t−x)dF(x)+∫t+∞(x−t)dF(x)=tF(t)−∫−∞txdF(x)+∫t+∞xdF(x)−t(1−F(t)), Yes t t t Derivation , have to f ′ ( t ) = F ( t ) + t p ( t ) − t p ( t ) − t p ( t ) − 1 + F ( t ) + t p ( t ) = 2 F ( t ) − 1 , f^{\prime}(t)=F(t)+t p(t)-t p(t)-t p(t)-1+F(t)+t p(t)=2 F(t)-1, f′(t)=F(t)+tp(t)−tp(t)−tp(t)−1+F(t)+tp(t)=2F(t)−1, Make f ′ ( t ) = 0 f^{\prime}(t)=0 f′(t)=0, Solution t = x 0.5 ( x a t=x_{0.5} \left(x_{a}\right. t=x0.5(xa Indicates that it is satisfied F ( x a ) = a ) \left.F\left(x_{a}\right)=a\right) F(xa)=a). The second derivative f ′ ′ ( t ) = 2 p ( t ) ≥ 0 f^{\prime \prime}(t)=2 p(t) \geq 0 f′′(t)=2p(t)≥0, Therefore, the stagnation point is the minimum point . According to the meaning m = x 0.5 m=x_{0.5} m=x0.5, so P ( X ≤ m ) = 1 2 . P(X \leq m)=\frac{1}{2} . P(X≤m)=21.
7、 ... and 、(20 branch ) X 1 , X 2 , X 3 , X_{1}, X_{2}, X_{3}, X1,X2,X3, i.i.d ∼ N ( 0 , 1 ) , \sim N(0,1), ∼N(0,1), seek
(1)(10 branch ) P ( X 1 > X 2 > X 3 ) P\left(X_{1}>X_{2}>X_{3}\right) P(X1>X2>X3);
(2)(10 branch ) P ( X 1 > X 2 , X 1 > X 3 ) P\left(X_{1}>X_{2}, X_{1}>X_{3}\right) P(X1>X2,X1>X3).
Solution: (1) According to the rotation symmetry , P ( X 1 > X 2 > X 3 ) = P ( X 1 > X 3 > X 2 ) = P ( X 2 > X 1 > X 3 ) = P ( X 2 > X 3 > X 1 ) = P ( X 3 > X 1 > X 2 ) = P ( X 3 > X 2 > X 1 ) \begin{aligned} & P\left(X_{1}>X_{2}>X_{3}\right)=P\left(X_{1}>X_{3}>X_{2}\right)=P\left(X_{2}>X_{1}>X_{3}\right) \\ =& P\left(X_{2}>X_{3}>X_{1}\right)=P\left(X_{3}>X_{1}>X_{2}\right)=P\left(X_{3}>X_{2}>X_{1}\right) \end{aligned} =P(X1>X2>X3)=P(X1>X3>X2)=P(X2>X1>X3)P(X2>X3>X1)=P(X3>X1>X2)=P(X3>X2>X1) These events are disjoint , And the probability of these events is 1 Events , so 6 P ( X 1 > X 2 > X 3 ) = 1 ⇒ P ( X 1 > X 2 > X 3 ) = 1 6 . 6 P\left(X_{1}>X_{2}>X_{3}\right)=1 \Rightarrow P\left(X_{1}>X_{2}>X_{3}\right)=\frac{1}{6} . 6P(X1>X2>X3)=1⇒P(X1>X2>X3)=61. (2) We found that :
P ( X 1 > X 2 , X 1 > X 3 ) = P ( X 1 = max { X 1 , X 2 , X 3 } ) , P\left(X_{1}>X_{2}, X_{1}>X_{3}\right)=P\left(X_{1}=\max \left\{X_{1}, X_{2}, X_{3}\right\}\right), P(X1>X2,X1>X3)=P(X1=max{ X1,X2,X3}), According to the rotation symmetry , P ( X 1 = max { X 1 , X 2 , X 3 } ) = P ( X 2 = max { X 1 , X 2 , X 3 } ) = P ( X 3 = max { X 1 , X 2 , X 3 } ) , P\left(X_{1}=\max \left\{X_{1}, X_{2}, X_{3}\right\}\right)=P\left(X_{2}=\max \left\{X_{1}, X_{2}, X_{3}\right\}\right)=P\left(X_{3}=\max \left\{X_{1}, X_{2}, X_{3}\right\}\right), P(X1=max{ X1,X2,X3})=P(X2=max{ X1,X2,X3})=P(X3=max{ X1,X2,X3}),so P ( X 1 > X 2 , X 1 > X 3 ) = 1 3 P\left(X_{1}>X_{2}, X_{1}>X_{3}\right)=\frac{1}{3} P(X1>X2,X1>X3)=31.
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