__-|||-3 随机变量X的密度函数为 f(x)= ) A(x)^2,-1lt xlt 2 0, .-|||-求:(1)常数A;(2) (Xlt 0) : (3)X的分布函数F (x) (4) Y=2X+1 的概率密度-|||-(5.0分)

步骤 1：求常数A
根据概率密度函数的性质，整个区间上的积分应等于1，即
$$
\int_{-1}^{2} A x^{2} \, dx = 1
$$
步骤 2：计算积分
$$
\int_{-1}^{2} A x^{2} \, dx = A \int_{-1}^{2} x^{2} \, dx = A \left[ \frac{x^{3}}{3} \right]_{-1}^{2} = A \left( \frac{2^{3}}{3} - \frac{(-1)^{3}}{3} \right) = A \left( \frac{8}{3} + \frac{1}{3} \right) = A \cdot 3
$$
步骤 3：求解A
$$
A \cdot 3 = 1 \Rightarrow A = \frac{1}{3}
$$
步骤 4：求 $P(X < 0)$
$$
P(X < 0) = \int_{-1}^{0} \frac{1}{3} x^{2} \, dx = \frac{1}{3} \int_{-1}^{0} x^{2} \, dx = \frac{1}{3} \left[ \frac{x^{3}}{3} \right]_{-1}^{0} = \frac{1}{3} \left( 0 - \frac{(-1)^{3}}{3} \right) = \frac{1}{3} \cdot \frac{1}{3} = \frac{1}{9}
$$
步骤 5：求X的分布函数F(x)
$$
F(x) = \int_{-\infty}^{x} f(t) \, dt
$$
当 $x \leq -1$ 时，$F(x) = 0$
当 $-1 < x < 2$ 时，$F(x) = \int_{-1}^{x} \frac{1}{3} t^{2} \, dt = \frac{1}{3} \left[ \frac{t^{3}}{3} \right]_{-1}^{x} = \frac{1}{9} (x^{3} + 1)$
当 $x \geq 2$ 时，$F(x) = 1$
步骤 6：求Y的概率密度函数
$$
Y = 2X + 1
$$
$$
F_{Y}(y) = P(Y \leq y) = P(2X + 1 \leq y) = P(X \leq \frac{y - 1}{2}) = F_{X}(\frac{y - 1}{2})
$$
$$
f_{Y}(y) = \frac{d}{dy} F_{Y}(y) = \frac{d}{dy} F_{X}(\frac{y - 1}{2}) = f_{X}(\frac{y - 1}{2}) \cdot \frac{1}{2}
$$
当 $y \leq -1$ 时，$f_{Y}(y) = 0$
当 $-1 < y < 5$ 时，$f_{Y}(y) = \frac{1}{2} \cdot \frac{1}{3} \left( \frac{y - 1}{2} \right)^{2} = \frac{1}{24} (y - 1)^{2}$
当 $y \geq 5$ 时，$f_{Y}(y) = 0$