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设随机变量x的概率密度函数为f(x),且f(x)=f(-x)则对于任意实数a有F(-a)=1/2-积分0到af(x)dx为什么?
题目内容:
设随机变量x的概率密度函数为f(x),且f(x)=f(-x)
则对于任意实数a
有 F(-a)=1/2-积分0到a f(x)dx
为什么?优质解答
因为f(x)是随机变量x的概率密度函数
所以 ∫f(x)d(x)│(x=- ∞ to +∞)=1
又因为 f(x)=f(-x)
所以 ∫f(x)d(x)│(x=- a to 0)=∫f(x)d(x)│(x=0 to a )
F(0)=∫f(x)d(x)│(x=- ∞ to 0)=∫f(x)d(x)│(x=0 to +∞ )=(1/2)*∫f(x)d(x)│(x=- ∞ to +∞)=1/2
F(-a)=∫f(x)d(x)│(x=- ∞ to -a)=∫f(x)d(x)│(x=- ∞ to 0)-∫f(x)d(x)│(x=- a to 0)=1/2-∫f(x)d(x)│(x=0 to a )
则对于任意实数a
有 F(-a)=1/2-积分0到a f(x)dx
为什么?
优质解答
所以 ∫f(x)d(x)│(x=- ∞ to +∞)=1
又因为 f(x)=f(-x)
所以 ∫f(x)d(x)│(x=- a to 0)=∫f(x)d(x)│(x=0 to a )
F(0)=∫f(x)d(x)│(x=- ∞ to 0)=∫f(x)d(x)│(x=0 to +∞ )=(1/2)*∫f(x)d(x)│(x=- ∞ to +∞)=1/2
F(-a)=∫f(x)d(x)│(x=- ∞ to -a)=∫f(x)d(x)│(x=- ∞ to 0)-∫f(x)d(x)│(x=- a to 0)=1/2-∫f(x)d(x)│(x=0 to a )
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