题目
已知随机变量X_(1), X_(2), X_(3)的协方差COV(X_(1), X_(3)) = 2, COV(X_(2), X_(3)) = 1。则COV(X_(1) + X_(2), 2X_(3)) = ____
已知随机变量$X_{1}, X_{2}, X_{3}$的协方差$COV(X_{1}, X_{3}) = 2, COV(X_{2}, X_{3}) = 1$。则$COV(X_{1} + X_{2}, 2X_{3}) = \_\_\_\_$
题目解答
答案
根据协方差的性质，对于随机变量 $X_1, X_2, X_3$，有：
\[
\text{cov}(X_1 + X_2, 2X_3) = \text{cov}(X_1, 2X_3) + \text{cov}(X_2, 2X_3)
\]
由性质 $\text{cov}(X, aY) = a\text{cov}(X, Y)$，可得：
\[
\text{cov}(X_1, 2X_3) = 2\text{cov}(X_1, X_3) = 2 \times 2 = 4
\]
\[
\text{cov}(X_2, 2X_3) = 2\text{cov}(X_2, X_3) = 2 \times 1 = 2
\]
因此：
\[
\text{cov}(X_1 + X_2, 2X_3) = 4 + 2 = 6
\]
答案：6