题目
题目 设随机变量X的密度函数为 _(x)(x)=dfrac (1)(2)(e)^-|H| -infty lt xlt +infty 求:① (0lt Xlt 1); ②X的分布函数 bigcirc (3)Y=2X+1 的密度函数f(y).
题目
题目解答
答案
解析
步骤 1：计算 $P(0\lt X\lt 1)$
$P(0\lt X\lt 1) = \int_{0}^{1} f(x) dx = \int_{0}^{1} \dfrac {1}{2}{e}^{-|x|}dx = \int_{0}^{1} \dfrac {1}{2}{e}^{-x}dx$
$= \dfrac {1}{2}(-{e}^{-x})|_{0}^{1} = \dfrac {1}{2}(1-\dfrac {1}{e})$
步骤 2：计算X的分布函数 $F(x)$
当 $x \leq 0$ 时，$F(x) = \int_{-\infty}^{x} \dfrac {1}{2}{e}^{-|t|}dt = \int_{-\infty}^{x} \dfrac {1}{2}{e}^{t}dt = \dfrac {1}{2}{e}^{t}|_{-\infty}^{x} = \dfrac {1}{2}{e}^{x}$
当 $x \geq 0$ 时，$F(x) = \int_{-\infty}^{0} \dfrac {1}{2}{e}^{-|t|}dt + \int_{0}^{x} \dfrac {1}{2}{e}^{-|t|}dt = \dfrac {1}{2} + \int_{0}^{x} \dfrac {1}{2}{e}^{-t}dt = \dfrac {1}{2} + \dfrac {1}{2}(-{e}^{-t})|_{0}^{x} = 1 - \dfrac {1}{2}{e}^{-x}$
步骤 3：计算 $Y=2X+1$ 的密度函数 $f(y)$
$y = 2x + 1$，则 $x = \dfrac {y-1}{2}$，$f(y) = \dfrac {1}{2}{e}^{-|\dfrac {y-1}{2}|}$
$P(0\lt X\lt 1) = \int_{0}^{1} f(x) dx = \int_{0}^{1} \dfrac {1}{2}{e}^{-|x|}dx = \int_{0}^{1} \dfrac {1}{2}{e}^{-x}dx$
$= \dfrac {1}{2}(-{e}^{-x})|_{0}^{1} = \dfrac {1}{2}(1-\dfrac {1}{e})$
步骤 2：计算X的分布函数 $F(x)$
当 $x \leq 0$ 时，$F(x) = \int_{-\infty}^{x} \dfrac {1}{2}{e}^{-|t|}dt = \int_{-\infty}^{x} \dfrac {1}{2}{e}^{t}dt = \dfrac {1}{2}{e}^{t}|_{-\infty}^{x} = \dfrac {1}{2}{e}^{x}$
当 $x \geq 0$ 时，$F(x) = \int_{-\infty}^{0} \dfrac {1}{2}{e}^{-|t|}dt + \int_{0}^{x} \dfrac {1}{2}{e}^{-|t|}dt = \dfrac {1}{2} + \int_{0}^{x} \dfrac {1}{2}{e}^{-t}dt = \dfrac {1}{2} + \dfrac {1}{2}(-{e}^{-t})|_{0}^{x} = 1 - \dfrac {1}{2}{e}^{-x}$
步骤 3：计算 $Y=2X+1$ 的密度函数 $f(y)$
$y = 2x + 1$，则 $x = \dfrac {y-1}{2}$，$f(y) = \dfrac {1}{2}{e}^{-|\dfrac {y-1}{2}|}$