概率论,第3题,求解
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(1)
F(-a) = ∫(-∞, -a) f(x) dx = ∫(-∞, ∞) f(x) dx - ∫(-a, ∞) f(x) dx
令 t = -x, ∫(-a, ∞) f(x) dx = ∫(-∞, a) f(-t) dt =∫(-∞, a) f(t) dt = F(a)
F(-a) = 1 - F(a)
F(0)=1-F(0), F(0) = 1/2
F(a) = ∫(-∞, a) f(x) dx = ∫(-∞, 0) f(x) dx + ∫(0, a) f(x) dx = F(0) + ∫(0, a) f(x) dx
F(-a) = 1 - F(0) - ∫(0, a) f(x) dx = 1/2 - ∫(0, a) f(x) dx
(2)
P {|X|>a} = P {X < -a} + P {X > a}
P {X < -a} = F(-a) = 1 - F(a)
P {X > a} = 1 - F(a)
P {|X|>a} = 2(1-F(a))
(3)
P{|X|<a} = 1 - P{|X|>a} = 1 - 2 + 2F(a) = 2F(a) - 1
F(-a) = ∫(-∞, -a) f(x) dx = ∫(-∞, ∞) f(x) dx - ∫(-a, ∞) f(x) dx
令 t = -x, ∫(-a, ∞) f(x) dx = ∫(-∞, a) f(-t) dt =∫(-∞, a) f(t) dt = F(a)
F(-a) = 1 - F(a)
F(0)=1-F(0), F(0) = 1/2
F(a) = ∫(-∞, a) f(x) dx = ∫(-∞, 0) f(x) dx + ∫(0, a) f(x) dx = F(0) + ∫(0, a) f(x) dx
F(-a) = 1 - F(0) - ∫(0, a) f(x) dx = 1/2 - ∫(0, a) f(x) dx
(2)
P {|X|>a} = P {X < -a} + P {X > a}
P {X < -a} = F(-a) = 1 - F(a)
P {X > a} = 1 - F(a)
P {|X|>a} = 2(1-F(a))
(3)
P{|X|<a} = 1 - P{|X|>a} = 1 - 2 + 2F(a) = 2F(a) - 1
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