作业帮二项式分布问题X~B(n,p),EX=? E(2X)=? E(x^2)=? DX=? D(2x)=? D(x^2)=?
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二项式分布问题X~B(n,p),EX=? E(2X)=? E(x^2)=? DX=? D(2x)=? D(x^2)=?
二项式分布问题
X~B(n,p),EX=? E(2X)=? E(x^2)=?
DX=? D(2x)=? D(x^2)=?
二项式分布问题X~B(n,p),EX=? E(2X)=? E(x^2)=? DX=? D(2x)=? D(x^2)=?
用C(n,k)表示n中选k的组合数,并约定k < 0或k > n时C(n,k) = 0.
用到几个组合恒等式:
① k·C(n,k) = n·C(n-1,k-1);
② k²·C(n,k) = k(k-1)·C(n,k)+k·C(n,k) = n(n-1)·C(n-2,k-2)+n·C(n-1,k-1);
③ k⁴·C(n,k) = k(k-1)(k-2)(k-3)·C(n,k)+6k(k-1)(k-2)·C(n,k)+(7k²-6k)·C(n-1,k-1)
= n(n-1)(n-2)(n-3)·C(n-4,k-4)+6n(n-1)(n-2)·C(n-3,k-3)+7n(n-1)·C(n-2,k-2)+n·C(n-1,k-1).
(易见①是最基本的,②③是在①的基础上向高次的推广).
设q = 1-p,由X B(n,p),有P(X = k) = C(n,k)p^k·q^(n-k).
E(X) = ∑{0 ≤ k ≤ n} k·P(X = k)
= ∑{0 ≤ k ≤ n} k·C(n,k)p^k·q^(n-k)
= ∑{0 ≤ k ≤ n} n·C(n-1,k-1)p^k·q^(n-k) (由①)
= np·∑{0 ≤ k ≤ n} C(n-1,k-1)p^(k-1)·q^(n-k)
= np(p+q)^(n-1) (二项式定理)
= np (q = 1-p).
E(X²) = ∑{0 ≤ k ≤ n} k²·P(X = k)
= ∑{0 ≤ k ≤ n} k²·C(n,k)p^k·q^(n-k)
= ∑{0 ≤ k ≤ n} n(n-1)·C(n-2,k-2)p^k·q^(n-k) + ∑{0 ≤ k ≤ n} n·C(n-1,k-1)p^k·q^(n-k) (由②)
= n(n-1)p²·∑{0 ≤ k ≤ n} C(n-2,k-2)p^(k-2)·q^(n-k) + np·∑{0 ≤ k ≤ n} C(n-1,k-1)p^(k-1)·q^(n-k)
= n(n-1)p²(p+q)^(n-2)+np(p+q)^(n-1) (二项式定理)
= n(n-1)p²+np
= (np)²+np(1-p)
= (np)²+npq (q = 1-p).
E(X⁴) = ∑{0 ≤ k ≤ n} k⁴·P(X = k)
= ∑{0 ≤ k ≤ n} k⁴·C(n,k)p^k·q^(n-k)
= n(n-1)(n-2)(n-3)p⁴+6n(n-1)(n-2)p³+7n(n-1)p²+np (原理同上,
= (np)⁴+6(np)³q+(np)²q(7-11p)+npq(1-6pq).
于是D(X) = E(X²)-E(X)² = npq;
D(X²) = E(X⁴)-E(X²)² = npq(4(np)²+np(6-10p)+(1-6pq)).(不确定有没有更漂亮的写法).
此外易得E(2X) = 2E(X) = 2np,D(2X) = 4D(X) = 4npq.
注:求E(X)和D(X)还有更简单的方法.
首先对Y B(1,p),易得E(Y) = p,D(Y) = pq.
设Y1,Y2,...,Yn B(1,p)彼此独立,则Y1+Y2+...+Yn B(n,p).
故E(X) = E(Y1+Y2+...+Yn) = E(Y1)+E(Y2)+...+E(Yn) = np (随机变量和的期望 = 期望的和),
D(X) = D(Y1+Y2+...+Yn) = D(Y1)+D(Y2)+...+D(Yn) = npq (独立随机变量和的方差 = 方差的和).