题目
已知X,Y的期望分别为-2,2，方差分别为1,4，相关系数为-0.5. 则由契比晓夫不等式得P(|X+Y|≥6)≤A. (1)/(12)B. 1C. (1)/(2)D. (1)/(6)
已知X,Y的期望分别为-2,2，方差分别为1,4，相关系数为-0.5. 则由契比晓夫不等式得P(|X+Y|≥6)≤
A. $\frac{1}{12}$
B. 1
C. $\frac{1}{2}$
D. $\frac{1}{6}$
题目解答
答案
A. $\frac{1}{12}$
解析
步骤 1：定义随机变量Z
设 $ Z = X + Y $，则 $ E(Z) = E(X) + E(Y) = -2 + 2 = 0 $。
步骤 2：计算方差D(Z)
\[ D(Z) = D(X) + D(Y) + 2 \text{Cov}(X, Y) \]
\[ D(Z) = 1 + 4 + 2 \times (-0.5) \times \sqrt{1} \times \sqrt{4} = 3 \]
步骤 3：应用契比晓夫不等式
\[ P(|Z - E(Z)| \geq \epsilon) \leq \frac{D(Z)}{\epsilon^2} \]
\[ P(|X+Y| \geq 6) \leq \frac{3}{6^2} = \frac{1}{12} \]
设 $ Z = X + Y $，则 $ E(Z) = E(X) + E(Y) = -2 + 2 = 0 $。
步骤 2：计算方差D(Z)
\[ D(Z) = D(X) + D(Y) + 2 \text{Cov}(X, Y) \]
\[ D(Z) = 1 + 4 + 2 \times (-0.5) \times \sqrt{1} \times \sqrt{4} = 3 \]
步骤 3：应用契比晓夫不等式
\[ P(|Z - E(Z)| \geq \epsilon) \leq \frac{D(Z)}{\epsilon^2} \]
\[ P(|X+Y| \geq 6) \leq \frac{3}{6^2} = \frac{1}{12} \]