二项分布期望和方差的推导及推广_二项分布的期望和方差推导
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一、二项分布基础内容
如果不想看复杂的推导过程，请点这里。
众所周知，两点分布的分布列为：
| X | 1 | 0 |
|---|---|---|
| P | p | 1-p |
而二项分布可以看为将 n n n 次两点分布的结果累加起来，其分布列为 P ( X = k ) = C n k p k ( 1 − p ) n − k P\left(X=k\right)=C_n^kp^k\left(1-p\right)^{n-k} P(X=k)=Cnkpk(1−p)n−k.
1、期望
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E\left(X\right)=\sum\limits_{k=0}^n kC_n^kp^k\left(1-p\right)^{n-k}
E(X)=k=0∑nkCnkpk(1−p)n−k
累加的式子中存在
k
C
n
k
kC_n^k
kCnk，考虑把
k
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k 变为常数进行运算。
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kC_n^k=k\frac{n!}{k!\left(n-k\right)!}=n\frac{\left(n-1\right)!}{\left(k-1\right)!\left(n-k\right)!}=nC_{n-1}^{k-1}
kCnk=kk!(n−k)!n!=n(k−1)!(n−k)!(n−1)!=nCn−1k−1
这时我们还要考虑
C
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1
k
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1
C_{n-1}^{k-1}
Cn−1k−1 在
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k=0
k=0 时无意义的情况，因此累加时要从
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1
k=1
k=1 开始，并加上
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k=0
k=0 时的第一项
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E\left(X\right)=\sum\limits_{k=1}^n nC_{n-1}^{k-1} p^k\left(1-p\right)^{n-k}
E(X)=k=1∑nnCn−1k−1pk(1−p)n−k
使用二项式定理合并：
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E\left(X\right)=np\sum\limits_{k=1}^nC_{n-1}^{k-1}p^{k-1}\left(1-p\right)^{n-k}
E(X)=npk=1∑nCn−1k−1pk−1(1−p)n−k
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E\left(X\right)=np\sum\limits_{k=0}^{n-1}C_{n-1}^kp^k\left(1-p\right)^{n-1-k}
E(X)=npk=0∑n−1Cn−1kpk(1−p)n−1−k
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E\left(X\right)=np\left(p+1-p\right)^{n-1}
E(X)=np(p+1−p)n−1
E
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p
E\left(X\right)=np
E(X)=np
2、方差
先推导一个计算方差的公式：
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D\left(X\right)=E\left(X^2\right)-E^2\left(X\right)
D(X)=E(X2)−E2(X).
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D\left(X\right)=\sum\limits_{i=1}^n \left[x_i-E\left(X\right)\right]^2p_i
D(X)=i=1∑n[xi−E(X)]2pi
D
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D\left(X\right)=\sum\limits_{i=1}^n \left[x_i^2-2x_iE\left(X\right)+E^2\left(X\right)\right]p_i
D(X)=i=1∑n[xi2−2xiE(X)+E2(X)]pi
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D\left(X\right)=\sum\limits_{i=1}^n \left[x_i^2p_i-2x_ip_iE\left(X\right)+p_iE^2\left(X\right)\right]
D(X)=i=1∑n[xi2pi−2xipiE(X)+piE2(X)]
D
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D\left(X\right)=\sum\limits_{i=1}^n x_i^2p_i -2E\left(X\right)\sum\limits _{i=1}^n x_ip_i +E^2\left(X\right)\sum\limits_{i=1}^n p_i
D(X)=i=1∑nxi2pi−2E(X)i=1∑nxipi+E2(X)i=1∑npi
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D\left(X\right)=E\left(X^2\right)-2E^2\left(X\right)+E^2\left(X\right)
D(X)=E(X2)−2E2(X)+E2(X)
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D\left(X\right)=E\left(X^2\right)-E^2\left(X\right)
D(X)=E(X2)−E2(X)
由于
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{
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}
D\left(X\right)=E\left\{\left[X-E\left(X\right)\right]^2\right\}
D(X)=E{[X−E(X)]2}，也可以这样推导：
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D\left(X\right)=E\left[X^2-2XE\left(X\right)+E^2\left(X\right)\right]
D(X)=E[X2−2XE(X)+E2(X)]
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D\left(X\right)=E\left(X^2\right)-2E\left(X\right)E\left(X\right)+E^2\left(X\right)
D(X)=E(X2)−2E(X)E(X)+E2(X)
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D\left(X\right)=E\left(X^2\right)-E^2\left(X\right)
D(X)=E(X2)−E2(X)
利用这个公式，我们可以先求出
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E\left(X^2\right)
E(X2)，再算出
D
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D\left(X\right)
D(X).
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E\left(X^2\right)=\sum\limits_{k=0}^n k^2C_n^kp^k\left(1-p\right)^{n-k}
E(X2)=k=0∑nk2Cnkpk(1−p)n−k
跟前面一样，我们要把
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k^2
k2 代换掉。
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k^2C_n^k=knC_{n-1}^{k-1}=n\left(k-1\right)C_{n-1}^{k-1}+nC_{n-1}^{k-1}=n\left(n-1\right)C_{n-2}^{k-2}+nC_{n-1}^{k-1}
k2Cnk=knCn−1k−1=n(k−1)Cn−1k−1+nCn−1k−1=n(n−1)Cn−2k−2+nCn−1k−1
同样考虑到
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C_{n-2}^{k-2}
Cn−2k−2 在
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k=0
k=0 和
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1
k=1
k=1 时无意义，我们可以把第一、二项单独拿出来计算。
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E\left(X^2\right)=C_n^1p\left(1-p\right)^{n-1}+\sum\limits_{k=2}^n n\left(n-1\right)C_{n-2}^{k-2}p^k\left(1-p\right)^{n-k} +\sum\limits_{k=2}^n nC_{n-1}^{k-1}p^k\left(1-p\right)^{n-k}
E(X2)=Cn1p(1−p)n−1+k=2∑nn(n−1)Cn−2k−2pk(1−p)n−k+k=2∑nnCn−1k−1pk(1−p)n−k
为了使用二项式定理合并，我们还需要继续拆项。
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E\left(X^2\right)=np\left(1-p\right)^{n-1}+n\left(n-1\right)p^2\sum\limits_{k=2}^n C_{n-2}^{k-2}p^{k-2}\left(1-p\right)^{n-k}+\sum\limits_{k=1}^n nC_{n-1}^{k-1}p^k\left(1-p\right)^{n-k}-nC_n^0p\left(1-p\right)^{n-1}
E(X2)=np(1−p)n−1+n(n−1)p2k=2∑nCn−2k−2pk−2(1−p)n−k+k=1∑nnCn−1k−1pk(1−p)n−k−nCn0p(1−p)n−1
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E\left(X^2\right)=np\left(1-p\right)^{n-1}+n\left(n-1\right)p^2\sum\limits_{k=0}^{n-2} C_{n-2}^kp^k\left(1-p\right)^{n-2-k}+\sum\limits_{k=1}^n nC_{n-1}^{k-1}p^k\left(1-p\right)^{n-k}-np\left(1-p\right)^{n-1}
E(X2)=np(1−p)n−1+n(n−1)p2k=0∑n−2Cn−2kpk(1−p)n−2−k+k=1∑nnCn−1k−1pk(1−p)n−k−np(1−p)n−1
E
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E\left(X^2\right)=n\left(n-1\right)p^2\left(p+1-p\right)^{n-2}+np\left(p+1-p\right)^{n-1}
E(X2)=n(n−1)p2(p+1−p)n−2+np(p+1−p)n−1
E
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E\left(X^2\right)=n\left(n-1\right)p^2+np
E(X2)=n(n−1)p2+np
带入开头的公式中即可得到方差：
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D\left(X\right)=n^2p^2-np^2+np-n^2p^2
D(X)=n2p2−np2+np−n2p2
D
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p
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D\left(X\right)=np\left(1-p\right)
D(X)=np(1−p)
3、较为简洁的推导方法
在推导方差的公式时，我们可以利用期望和方差的性质，十分简洁地把公式推出来。
- 期望的乘法：当 X i X_i Xi 两两相互独立时，若 X = ∏ i = 1 n X i X=\prod\limits_{i=1}^n X_i X=i=1∏nXi，则 E ( X ) = ∏ i = 1 n E ( X i ) E\left(X\right)=\prod\limits_{i=1}^n E\left(X_i\right) E(X)=i=1∏nE(Xi).
- 方差的加法：当 X i X_i Xi 两两相互独立时，若 X = ∑ i = 1 n X i X=\sum\limits_{i=1}^n X_i X=i=1∑nXi，则 D ( X ) = ∑ i = 1 n D ( X i ) D\left(X\right)=\sum\limits_{i=1}^n D\left(X_i\right) D(X)=i=1∑nD(Xi).
由于在二项分布的大前提下， X i X_i Xi 两两相互独立，所以 D ( X ) = n D ( X i ) = n p ( 1 − p ) D\left(X\right)=nD\left(X_i\right)=np\left(1-p\right) D(X)=nD(Xi)=np(1−p).
下面是对两个性质的证明：
证明1
由于通式的证明不太好写，这里我使用数学归纳法来证明，不过道理都是一样的。
先证明
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E\left(X_1X_2\right)=E\left(X_1\right)E\left(X_2\right)
E(X1X2)=E(X1)E(X2)：
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①
E\left(X_1X_2\right)=\sum\limits _{i=1}^{n_1} \sum\limits_{j=1}^{n_2} x_{1_i}x_{2_j}p_{1_i}p_{2_j}.........①
E(X1X2)=i=1∑n1j=1∑n2x1ix2jp1ip2j.........①
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E\left(X_1\right)E\left(X_2\right)=\sum\limits_{i=1}^{n_1} x_{1_i}p_{1_i}\sum\limits_{i=1}^{n_2}x_{2_i}p_{2_i}...②
E(X1)E(X2)=i=1∑n1x1ip1ii=1∑n2x2ip2i...②
①式与②式的右半边是相等的，因此得证。然后再由此进行迭乘，可归纳出原公式。
证明2
D
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D\left(X\right)=E\left(X^2\right)-E^2\left(X\right)
D(X)=E(X2)−E2(X)
D
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D\left(X\right)=E\left[\left(\sum\limits_{i=1}^n X_i\right)^2\right]-\left[\sum\limits_{i=1}^n E\left(X_i\right)\right]^2
D(X)=E
(i=1∑nXi)2
−[i=1∑nE(Xi)]2
D
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D\left(X\right)=E\left(\sum\limits_{i=1}^n \sum\limits_{j=1}^n X_iX_j\right)-\sum\limits_{i=1}^n \sum\limits_{j=1}^n E\left(X_i\right)E\left(X_j\right)
D(X)=E(i=1∑nj=1∑nXiXj)−i=1∑nj=1∑nE(Xi)E(Xj)
D
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D\left(X\right)=\sum\limits_{i=1}^n \sum\limits_{j=1}^n E\left(X_iX_j\right)-\sum\limits_{i=1}^n \sum\limits_{j=1}^n E\left(X_i\right)E\left(X_j\right)
D(X)=i=1∑nj=1∑nE(XiXj)−i=1∑nj=1∑nE(Xi)E(Xj)
D
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D\left(X\right)=\sum\limits_{i=1}^n \sum\limits_{j=1}^n \left[E\left(X_iX_j\right)-E\left(X_i\right)E\left(X_j\right)\right]
D(X)=i=1∑nj=1∑n[E(XiXj)−E(Xi)E(Xj)]
对于
i
=
j
i=j
i=j 的项，我们不能使用期望的乘法公式，而其余情况均可以使用乘法公式将两项合并为
0
0
0 得：
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D\left(X\right)=\sum\limits_{i=1}^n \left[E\left(X_i^2\right)-E^2\left(X_i\right)\right]
D(X)=i=1∑n[E(Xi2)−E2(Xi)]
D
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n
D
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D\left(X\right)=\sum\limits_{i=1}^n D\left(X_i\right)
D(X)=i=1∑nD(Xi)
