Пусть E(X)= 36.0, E(Y)= 80.0, D(X)= 43.0, D(Y)= 36.0, Cov(X,Y)= 7.0. Найдите: Cov(3X-4Y+5, 10X+15Y).

Решение:

Используем свойства ковариации:

1. \( \text{Cov}(aX + b, cY + d) = ac \cdot \text{Cov}(X, Y) \)
2. \( \text{Cov}(X, Y) = \text{Cov}(Y, X) \)
3. \( \text{Cov}(X+Y, Z) = \text{Cov}(X, Z) + \text{Cov}(Y, Z) \)
4. \( \text{Cov}(X, X) = \text{Var}(X) \)

В данном случае, нам нужно найти \( \text{Cov}(3X - 4Y + 5, 10X + 15Y) \).

Применим свойства ковариации:

\[ \text{Cov}(3X - 4Y + 5, 10X + 15Y) = \text{Cov}(3X, 10X) + \text{Cov}(3X, 15Y) + \text{Cov}(-4Y, 10X) + \text{Cov}(-4Y, 15Y) + \text{Cov}(5, 10X) + \text{Cov}(5, 15Y) \]\[ = 3 \cdot 10 \cdot \text{Cov}(X, X) + 3 \cdot 15 \cdot \text{Cov}(X, Y) + (-4) \cdot 10 \cdot \text{Cov}(Y, X) + (-4) \cdot 15 \cdot \text{Cov}(Y, Y) + 0 + 0 \]\[ = 30 \cdot \text{Var}(X) + 45 \cdot \text{Cov}(X, Y) - 40 \cdot \text{Cov}(X, Y) - 60 \cdot \text{Var}(Y) \]\[ = 30 \cdot \text{Var}(X) + (45 - 40) \cdot \text{Cov}(X, Y) - 60 \cdot \text{Var}(Y) \]\[ = 30 \cdot \text{Var}(X) + 5 \cdot \text{Cov}(X, Y) - 60 \cdot \text{Var}(Y) \]\[ = 30 \cdot 43.0 + 5 \cdot 7.0 - 60 \cdot 36.0 \]\[ = 1290 + 35 - 2160 \]\[ = 1325 - 2160 \]\[ = -835 \]

Ответ: -835