方差与标准差 方差的定义
方差的数学定义
定义 :设 X X X 是随机变量,E ( X ) = μ E(X) = \mu E ( X ) = μ ,如果 E [ ( X − μ ) 2 ] E[(X - \mu)^2] E [( X − μ ) 2 ] 存在,则称:
D ( X ) = E [ ( X − μ ) 2 ] = E [ ( X − E ( X ) ) 2 ] D(X) = E[(X - \mu)^2] = E[(X - E(X))^2] D ( X ) = E [( X − μ ) 2 ] = E [( X − E ( X ) ) 2 ]
为 X X X 的方差 ,记为 D ( X ) D(X) D ( X ) 或 V a r ( X ) Var(X) Va r ( X ) 。
方差的计算公式
离散型随机变量 :
D ( X ) = ∑ i = 1 ∞ ( x i − μ ) 2 p i D(X) = \sum_{i=1}^{\infty} (x_i - \mu)^2 p_i D ( X ) = ∑ i = 1 ∞ ( x i − μ ) 2 p i
连续型随机变量 :
D ( X ) = ∫ − ∞ + ∞ ( x − μ ) 2 f ( x ) d x D(X) = \int_{-\infty}^{+\infty} (x - \mu)^2 f(x) dx D ( X ) = ∫ − ∞ + ∞ ( x − μ ) 2 f ( x ) d x
方差的简化计算公式
公式 :D ( X ) = E ( X 2 ) − [ E ( X ) ] 2 D(X) = E(X^2) - [E(X)]^2 D ( X ) = E ( X 2 ) − [ E ( X ) ] 2
证明 :
D ( X ) = E [ ( X − μ ) 2 ] = E ( X 2 − 2 μ X + μ 2 ) = E ( X 2 ) − 2 μ E ( X ) + μ 2 = E ( X 2 ) − μ 2 = E ( X 2 ) − [ E ( X ) ] 2 D(X) = E[(X - \mu)^2] = E(X^2 - 2\mu X + \mu^2) = E(X^2) - 2\mu E(X) + \mu^2 = E(X^2) - \mu^2 = E(X^2) - [E(X)]^2 D ( X ) = E [( X − μ ) 2 ] = E ( X 2 − 2 μ X + μ 2 ) = E ( X 2 ) − 2 μ E ( X ) + μ 2 = E ( X 2 ) − μ 2 = E ( X 2 ) − [ E ( X ) ] 2
方差的直观理解
理解 :方差是随机变量取值与其期望的偏离程度的平方的平均值,它反映了随机变量取值的离散程度。
方差的例子
例 1 :掷一颗均匀骰子,用 X X X 表示出现的点数
E ( X ) = 3.5 E(X) = 3.5 E ( X ) = 3.5
E ( X 2 ) = ∑ i = 1 6 i 2 ⋅ 1 6 = 1 + 4 + 9 + 16 + 25 + 36 6 = 91 6 E(X^2) = \sum_{i=1}^6 i^2 \cdot \frac{1}{6} = \frac{1+4+9+16+25+36}{6} = \frac{91}{6} E ( X 2 ) = ∑ i = 1 6 i 2 ⋅ 6 1 = 6 1 + 4 + 9 + 16 + 25 + 36 = 6 91
D ( X ) = E ( X 2 ) − [ E ( X ) ] 2 = 91 6 − ( 7 2 ) 2 = 91 6 − 49 4 = 35 12 D(X) = E(X^2) - [E(X)]^2 = \frac{91}{6} - (\frac{7}{2})^2 = \frac{91}{6} - \frac{49}{4} = \frac{35}{12} D ( X ) = E ( X 2 ) − [ E ( X ) ] 2 = 6 91 − ( 2 7 ) 2 = 6 91 − 4 49 = 12 35
例 2 :在区间 [ 0 , 1 ] [0, 1] [ 0 , 1 ] 上均匀分布的随机变量 X X X
E ( X ) = 1 2 E(X) = \frac{1}{2} E ( X ) = 2 1
E ( X 2 ) = ∫ 0 1 x 2 ⋅ 1 d x = x 3 3 ∣ 0 1 = 1 3 E(X^2) = \int_0^1 x^2 \cdot 1 dx = \frac{x^3}{3}\big|_0^1 = \frac{1}{3} E ( X 2 ) = ∫ 0 1 x 2 ⋅ 1 d x = 3 x 3 0 1 = 3 1
D ( X ) = E ( X 2 ) − [ E ( X ) ] 2 = 1 3 − ( 1 2 ) 2 = 1 3 − 1 4 = 1 12 D(X) = E(X^2) - [E(X)]^2 = \frac{1}{3} - (\frac{1}{2})^2 = \frac{1}{3} - \frac{1}{4} = \frac{1}{12} D ( X ) = E ( X 2 ) − [ E ( X ) ] 2 = 3 1 − ( 2 1 ) 2 = 3 1 − 4 1 = 12 1
标准差的定义
标准差的数学定义
定义 :随机变量 X X X 的标准差 定义为方差的算术平方根:
σ = D ( X ) = E [ ( X − E ( X ) ) 2 ] \sigma = \sqrt{D(X)} = \sqrt{E[(X - E(X))^2]} σ = D ( X ) = E [( X − E ( X ) ) 2 ]
标准差的直观理解
理解 :标准差与随机变量具有相同的量纲,它直接反映了随机变量取值的离散程度。
标准差的例子
例 3 :继续例 1,掷骰子的标准差
σ = D ( X ) = 35 12 ≈ 1.71 \sigma = \sqrt{D(X)} = \sqrt{\frac{35}{12}} \approx 1.71 σ = D ( X ) = 12 35 ≈ 1.71
例 4 :继续例 2,均匀分布的标准差
σ = D ( X ) = 1 12 ≈ 0.289 \sigma = \sqrt{D(X)} = \sqrt{\frac{1}{12}} \approx 0.289 σ = D ( X ) = 12 1 ≈ 0.289
方差的性质
非负性
性质 1 :D ( X ) ≥ 0 D(X) \geq 0 D ( X ) ≥ 0
证明 :D ( X ) = E [ ( X − μ ) 2 ] ≥ 0 D(X) = E[(X - \mu)^2] \geq 0 D ( X ) = E [( X − μ ) 2 ] ≥ 0 ,因为平方项非负。
常数的方差
性质 2 :D ( c ) = 0 D(c) = 0 D ( c ) = 0 ,其中 c c c 是常数。
证明 :D ( c ) = E [ ( c − c ) 2 ] = E ( 0 ) = 0 D(c) = E[(c - c)^2] = E(0) = 0 D ( c ) = E [( c − c ) 2 ] = E ( 0 ) = 0
线性变换的方差
性质 3 :D ( a X + b ) = a 2 D ( X ) D(aX + b) = a^2 D(X) D ( a X + b ) = a 2 D ( X ) ,其中 a , b a, b a , b 是常数。
证明 :
D ( a X + b ) = E [ ( a X + b − E ( a X + b ) ) 2 ] = E [ ( a X + b − a E ( X ) − b ) 2 ] = E [ ( a X − a E ( X ) ) 2 ] = a 2 E [ ( X − E ( X ) ) 2 ] = a 2 D ( X ) D(aX + b) = E[(aX + b - E(aX + b))^2] = E[(aX + b - aE(X) - b)^2] = E[(aX - aE(X))^2] = a^2 E[(X - E(X))^2] = a^2 D(X) D ( a X + b ) = E [( a X + b − E ( a X + b ) ) 2 ] = E [( a X + b − a E ( X ) − b ) 2 ] = E [( a X − a E ( X ) ) 2 ] = a 2 E [( X − E ( X ) ) 2 ] = a 2 D ( X )
独立随机变量的方差
性质 4 :如果 X X X 和 Y Y Y 独立,则 D ( X + Y ) = D ( X ) + D ( Y ) D(X + Y) = D(X) + D(Y) D ( X + Y ) = D ( X ) + D ( Y )
证明 :
D ( X + Y ) = E [ ( X + Y − E ( X + Y ) ) 2 ] = E [ ( X − E ( X ) + Y − E ( Y ) ) 2 ] D(X + Y) = E[(X + Y - E(X + Y))^2] = E[(X - E(X) + Y - E(Y))^2] D ( X + Y ) = E [( X + Y − E ( X + Y ) ) 2 ] = E [( X − E ( X ) + Y − E ( Y ) ) 2 ]
= E [ ( X − E ( X ) ) 2 ] + E [ ( Y − E ( Y ) ) 2 ] + 2 E [ ( X − E ( X ) ) ( Y − E ( Y ) ) ] = E[(X - E(X))^2] + E[(Y - E(Y))^2] + 2E[(X - E(X))(Y - E(Y))] = E [( X − E ( X ) ) 2 ] + E [( Y − E ( Y ) ) 2 ] + 2 E [( X − E ( X )) ( Y − E ( Y ))]
由于 X X X 和 Y Y Y 独立,E [ ( X − E ( X ) ) ( Y − E ( Y ) ) ] = E ( X − E ( X ) ) E ( Y − E ( Y ) ) = 0 E[(X - E(X))(Y - E(Y))] = E(X - E(X))E(Y - E(Y)) = 0 E [( X − E ( X )) ( Y − E ( Y ))] = E ( X − E ( X )) E ( Y − E ( Y )) = 0 ,所以:
D ( X + Y ) = D ( X ) + D ( Y ) D(X + Y) = D(X) + D(Y) D ( X + Y ) = D ( X ) + D ( Y )
方差的可加性
性质 5 :如果 X 1 , X 2 , … , X n X_1, X_2, \dots, X_n X 1 , X 2 , … , X n 两两独立,则:
D ( X 1 + X 2 + ⋯ + X n ) = D ( X 1 ) + D ( X 2 ) + ⋯ + D ( X n ) D(X_1 + X_2 + \dots + X_n) = D(X_1) + D(X_2) + \dots + D(X_n) D ( X 1 + X 2 + ⋯ + X n ) = D ( X 1 ) + D ( X 2 ) + ⋯ + D ( X n )
常见分布的方差
两点分布
分布律 :P ( X = k ) = p k ( 1 − p ) 1 − k P(X = k) = p^k(1-p)^{1-k} P ( X = k ) = p k ( 1 − p ) 1 − k ,k = 0 , 1 k = 0, 1 k = 0 , 1
方差 :D ( X ) = p ( 1 − p ) D(X) = p(1-p) D ( X ) = p ( 1 − p )
证明 :
E ( X ) = p E(X) = p E ( X ) = p
E ( X 2 ) = 0 2 ⋅ ( 1 − p ) + 1 2 ⋅ p = p E(X^2) = 0^2 \cdot (1-p) + 1^2 \cdot p = p E ( X 2 ) = 0 2 ⋅ ( 1 − p ) + 1 2 ⋅ p = p
D ( X ) = E ( X 2 ) − [ E ( X ) ] 2 = p − p 2 = p ( 1 − p ) D(X) = E(X^2) - [E(X)]^2 = p - p^2 = p(1-p) D ( X ) = E ( X 2 ) − [ E ( X ) ] 2 = p − p 2 = p ( 1 − p )
二项分布
分布律 :P ( X = k ) = C n k p k ( 1 − p ) n − k P(X = k) = C_n^k p^k(1-p)^{n-k} P ( X = k ) = C n k p k ( 1 − p ) n − k ,k = 0 , 1 , 2 , … , n k = 0, 1, 2, \dots, n k = 0 , 1 , 2 , … , n
方差 :D ( X ) = n p ( 1 − p ) D(X) = np(1-p) D ( X ) = n p ( 1 − p )
证明 :
E ( X ) = n p E(X) = np E ( X ) = n p
E ( X 2 ) = ∑ k = 0 n k 2 C n k p k ( 1 − p ) n − k E(X^2) = \sum_{k=0}^n k^2 C_n^k p^k(1-p)^{n-k} E ( X 2 ) = ∑ k = 0 n k 2 C n k p k ( 1 − p ) n − k
通过计算可得 E ( X 2 ) = n p + n ( n − 1 ) p 2 E(X^2) = np + n(n-1)p^2 E ( X 2 ) = n p + n ( n − 1 ) p 2
D ( X ) = E ( X 2 ) − [ E ( X ) ] 2 = n p + n ( n − 1 ) p 2 − n 2 p 2 = n p ( 1 − p ) D(X) = E(X^2) - [E(X)]^2 = np + n(n-1)p^2 - n^2p^2 = np(1-p) D ( X ) = E ( X 2 ) − [ E ( X ) ] 2 = n p + n ( n − 1 ) p 2 − n 2 p 2 = n p ( 1 − p )
泊松分布
分布律 :P ( X = k ) = λ k k ! e − λ P(X = k) = \frac{\lambda^k}{k!} e^{-\lambda} P ( X = k ) = k ! λ k e − λ ,k = 0 , 1 , 2 , … k = 0, 1, 2, \dots k = 0 , 1 , 2 , …
方差 :D ( X ) = λ D(X) = \lambda D ( X ) = λ
证明 :
E ( X ) = λ E(X) = \lambda E ( X ) = λ
E ( X 2 ) = ∑ k = 0 ∞ k 2 λ k k ! e − λ = λ + λ 2 E(X^2) = \sum_{k=0}^{\infty} k^2 \frac{\lambda^k}{k!} e^{-\lambda} = \lambda + \lambda^2 E ( X 2 ) = ∑ k = 0 ∞ k 2 k ! λ k e − λ = λ + λ 2
D ( X ) = E ( X 2 ) − [ E ( X ) ] 2 = λ + λ 2 − λ 2 = λ D(X) = E(X^2) - [E(X)]^2 = \lambda + \lambda^2 - \lambda^2 = \lambda D ( X ) = E ( X 2 ) − [ E ( X ) ] 2 = λ + λ 2 − λ 2 = λ
均匀分布
密度函数 :f ( x ) = { 1 b − a , a ≤ x ≤ b 0 , 其他 f(x) = \begin{cases} \frac{1}{b-a}, & a \leq x \leq b \\ 0, & \text{其他} \end{cases} f ( x ) = { b − a 1 , 0 , a ≤ x ≤ b 其他
方差 :D ( X ) = ( b − a ) 2 12 D(X) = \frac{(b-a)^2}{12} D ( X ) = 12 ( b − a ) 2
证明 :
E ( X ) = a + b 2 E(X) = \frac{a+b}{2} E ( X ) = 2 a + b
E ( X 2 ) = ∫ a b x 2 ⋅ 1 b − a d x = b 3 − a 3 3 ( b − a ) = a 2 + a b + b 2 3 E(X^2) = \int_a^b x^2 \cdot \frac{1}{b-a} dx = \frac{b^3-a^3}{3(b-a)} = \frac{a^2+ab+b^2}{3} E ( X 2 ) = ∫ a b x 2 ⋅ b − a 1 d x = 3 ( b − a ) b 3 − a 3 = 3 a 2 + ab + b 2
D ( X ) = E ( X 2 ) − [ E ( X ) ] 2 = a 2 + a b + b 2 3 − ( a + b 2 ) 2 = ( b − a ) 2 12 D(X) = E(X^2) - [E(X)]^2 = \frac{a^2+ab+b^2}{3} - (\frac{a+b}{2})^2 = \frac{(b-a)^2}{12} D ( X ) = E ( X 2 ) − [ E ( X ) ] 2 = 3 a 2 + ab + b 2 − ( 2 a + b ) 2 = 12 ( b − a ) 2
指数分布
密度函数 :f ( x ) = { λ e − λ x , x ≥ 0 0 , x < 0 f(x) = \begin{cases} \lambda e^{-\lambda x}, & x \geq 0 \\ 0, & x < 0 \end{cases} f ( x ) = { λ e − λ x , 0 , x ≥ 0 x < 0
方差 :D ( X ) = 1 λ 2 D(X) = \frac{1}{\lambda^2} D ( X ) = λ 2 1
证明 :
E ( X ) = 1 λ E(X) = \frac{1}{\lambda} E ( X ) = λ 1
E ( X 2 ) = ∫ 0 ∞ x 2 ⋅ λ e − λ x d x = 2 λ 2 E(X^2) = \int_0^{\infty} x^2 \cdot \lambda e^{-\lambda x} dx = \frac{2}{\lambda^2} E ( X 2 ) = ∫ 0 ∞ x 2 ⋅ λ e − λ x d x = λ 2 2
D ( X ) = E ( X 2 ) − [ E ( X ) ] 2 = 2 λ 2 − ( 1 λ ) 2 = 1 λ 2 D(X) = E(X^2) - [E(X)]^2 = \frac{2}{\lambda^2} - (\frac{1}{\lambda})^2 = \frac{1}{\lambda^2} D ( X ) = E ( X 2 ) − [ E ( X ) ] 2 = λ 2 2 − ( λ 1 ) 2 = λ 2 1
正态分布
密度函数 :f ( x ) = 1 2 π σ e − ( x − μ ) 2 2 σ 2 f(x) = \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{(x-\mu)^2}{2\sigma^2}} f ( x ) = 2 π σ 1 e − 2 σ 2 ( x − μ ) 2
方差 :D ( X ) = σ 2 D(X) = \sigma^2 D ( X ) = σ 2
证明 :
E ( X ) = μ E(X) = \mu E ( X ) = μ
通过积分计算可得 E ( X 2 ) = μ 2 + σ 2 E(X^2) = \mu^2 + \sigma^2 E ( X 2 ) = μ 2 + σ 2
D ( X ) = E ( X 2 ) − [ E ( X ) ] 2 = μ 2 + σ 2 − μ 2 = σ 2 D(X) = E(X^2) - [E(X)]^2 = \mu^2 + \sigma^2 - \mu^2 = \sigma^2 D ( X ) = E ( X 2 ) − [ E ( X ) ] 2 = μ 2 + σ 2 − μ 2 = σ 2
切比雪夫不等式
切比雪夫不等式的表述
定理 :设随机变量 X X X 的期望为 μ \mu μ ,方差为 σ 2 \sigma^2 σ 2 ,则对任意正数 ε \varepsilon ε ,有:
P ( ∣ X − μ ∣ ≥ ε ) ≤ σ 2 ε 2 P(|X - \mu| \geq \varepsilon) \leq \frac{\sigma^2}{\varepsilon^2} P ( ∣ X − μ ∣ ≥ ε ) ≤ ε 2 σ 2
切比雪夫不等式的证明
证明 :
P ( ∣ X − μ ∣ ≥ ε ) = ∫ ∣ x − μ ∣ ≥ ε f ( x ) d x ≤ ∫ ∣ x − μ ∣ ≥ ε ( x − μ ) 2 ε 2 f ( x ) d x ≤ 1 ε 2 ∫ − ∞ + ∞ ( x − μ ) 2 f ( x ) d x = σ 2 ε 2 P(|X - \mu| \geq \varepsilon) = \int_{|x-\mu| \geq \varepsilon} f(x) dx \leq \int_{|x-\mu| \geq \varepsilon} \frac{(x-\mu)^2}{\varepsilon^2} f(x) dx \leq \frac{1}{\varepsilon^2} \int_{-\infty}^{+\infty} (x-\mu)^2 f(x) dx = \frac{\sigma^2}{\varepsilon^2} P ( ∣ X − μ ∣ ≥ ε ) = ∫ ∣ x − μ ∣ ≥ ε f ( x ) d x ≤ ∫ ∣ x − μ ∣ ≥ ε ε 2 ( x − μ ) 2 f ( x ) d x ≤ ε 2 1 ∫ − ∞ + ∞ ( x − μ ) 2 f ( x ) d x = ε 2 σ 2
切比雪夫不等式的应用
应用 1 :估计随机变量偏离期望的概率
应用 2 :证明大数定律
应用 3 :在统计推断中的应用
方差的应用
在统计中的应用
应用 1 :样本方差
样本方差是总体方差的无偏估计。
应用 2 :参数估计
方差是衡量估计量好坏的重要指标。
在金融中的应用
应用 3 :风险评估
方差是衡量投资风险的重要指标。
应用 4 :投资组合理论
通过方差来优化投资组合。
练习题
练习 1
已知 E ( X ) = 2 , D ( X ) = 3 E(X)=2, D(X)=3 E ( X ) = 2 , D ( X ) = 3 ,求 D ( 3 X − 1 ) D(3X-1) D ( 3 X − 1 ) 。
参考答案 解题思路 :
使用方差的性质。
详细步骤 :
D ( 3 X − 1 ) = 3 2 D ( X ) D(3X-1) = 3^2 D(X) D ( 3 X − 1 ) = 3 2 D ( X )
= 9 × 3 = 27 = 9 \times 3 = 27 = 9 × 3 = 27
答案 :27 27 27
练习 2
已知随机变量 X X X 服从参数为 λ \lambda λ 的指数分布,求 D ( X ) D(X) D ( X ) 。
参考答案 解题思路 :
使用指数分布的方差公式。
详细步骤 :
指数分布的方差:D ( X ) = 1 λ 2 D(X) = \frac{1}{\lambda^2} D ( X ) = λ 2 1
或者通过计算:
E ( X ) = 1 λ E(X) = \frac{1}{\lambda} E ( X ) = λ 1
E ( X 2 ) = ∫ 0 ∞ x 2 ⋅ λ e − λ x d x = 2 λ 2 E(X^2) = \int_0^{\infty} x^2 \cdot \lambda e^{-\lambda x} dx = \frac{2}{\lambda^2} E ( X 2 ) = ∫ 0 ∞ x 2 ⋅ λ e − λ x d x = λ 2 2
D ( X ) = E ( X 2 ) − [ E ( X ) ] 2 = 2 λ 2 − ( 1 λ ) 2 = 1 λ 2 D(X) = E(X^2) - [E(X)]^2 = \frac{2}{\lambda^2} - (\frac{1}{\lambda})^2 = \frac{1}{\lambda^2} D ( X ) = E ( X 2 ) − [ E ( X ) ] 2 = λ 2 2 − ( λ 1 ) 2 = λ 2 1
答案 :1 λ 2 \frac{1}{\lambda^2} λ 2 1
练习 3
已知随机变量 X X X 服从参数为 n , p n, p n , p 的二项分布,求 D ( X ) D(X) D ( X ) 。
参考答案 解题思路 :
使用二项分布的方差公式。
详细步骤 :
二项分布的方差:D ( X ) = n p ( 1 − p ) D(X) = np(1-p) D ( X ) = n p ( 1 − p )
或者通过计算:
E ( X ) = n p E(X) = np E ( X ) = n p
E ( X 2 ) = n p + n ( n − 1 ) p 2 E(X^2) = np + n(n-1)p^2 E ( X 2 ) = n p + n ( n − 1 ) p 2
D ( X ) = E ( X 2 ) − [ E ( X ) ] 2 = n p ( 1 − p ) D(X) = E(X^2) - [E(X)]^2 = np(1-p) D ( X ) = E ( X 2 ) − [ E ( X ) ] 2 = n p ( 1 − p )
答案 :n p ( 1 − p ) np(1-p) n p ( 1 − p )
练习 4
已知 X X X 的分布律为 P ( X = 1 ) = 0.2 , P ( X = 2 ) = 0.5 , P ( X = 3 ) = 0.3 P(X=1)=0.2, P(X=2)=0.5, P(X=3)=0.3 P ( X = 1 ) = 0.2 , P ( X = 2 ) = 0.5 , P ( X = 3 ) = 0.3 ,求 D ( X ) D(X) D ( X ) 。
参考答案 解题思路 :
使用方差的简化计算公式。
详细步骤 :
E ( X ) = 1 × 0.2 + 2 × 0.5 + 3 × 0.3 = 2.1 E(X) = 1 \times 0.2 + 2 \times 0.5 + 3 \times 0.3 = 2.1 E ( X ) = 1 × 0.2 + 2 × 0.5 + 3 × 0.3 = 2.1
E ( X 2 ) = 1 2 × 0.2 + 2 2 × 0.5 + 3 2 × 0.3 = 0.2 + 2.0 + 2.7 = 4.9 E(X^2) = 1^2 \times 0.2 + 2^2 \times 0.5 + 3^2 \times 0.3 = 0.2 + 2.0 + 2.7 = 4.9 E ( X 2 ) = 1 2 × 0.2 + 2 2 × 0.5 + 3 2 × 0.3 = 0.2 + 2.0 + 2.7 = 4.9
D ( X ) = E ( X 2 ) − [ E ( X ) ] 2 = 4.9 − ( 2.1 ) 2 = 4.9 − 4.41 = 0.49 D(X) = E(X^2) - [E(X)]^2 = 4.9 - (2.1)^2 = 4.9 - 4.41 = 0.49 D ( X ) = E ( X 2 ) − [ E ( X ) ] 2 = 4.9 − ( 2.1 ) 2 = 4.9 − 4.41 = 0.49
答案 :0.49 0.49 0.49
练习 5
已知随机变量 X X X 服从参数为 λ \lambda λ 的泊松分布,求 D ( X ) D(X) D ( X ) 。
参考答案 解题思路 :
使用泊松分布的方差公式。
详细步骤 :
泊松分布的方差:D ( X ) = λ D(X) = \lambda D ( X ) = λ
或者通过计算:
E ( X ) = λ E(X) = \lambda E ( X ) = λ
E ( X 2 ) = λ + λ 2 E(X^2) = \lambda + \lambda^2 E ( X 2 ) = λ + λ 2
D ( X ) = E ( X 2 ) − [ E ( X ) ] 2 = λ + λ 2 − λ 2 = λ D(X) = E(X^2) - [E(X)]^2 = \lambda + \lambda^2 - \lambda^2 = \lambda D ( X ) = E ( X 2 ) − [ E ( X ) ] 2 = λ + λ 2 − λ 2 = λ
答案 :λ \lambda λ