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设随机变量 的概率密度为F(x)={1%|x|,|x|<...

就是找f(x)在所取x值之前一共积分了多少,分段函数就分段考虑,注意累积即可 F(x)=0(x

不会的追问

∫f(x)dx =1 => ∫A/(1+x^2) dx = Aarctan(x) =Api =1, A=1/pi, 积分上下限为正负无穷大 y=2x+3, x=(y-3)/2 P(y

此题有错。f(x) = kcosx, 0

设随机变量(X,Y)的概率密度为 f(x,y) = 3x,0

F(x,y)=C constant. The definition region is a rectangle with area 4. Therefore, C=1/4/

解: E(X)=∫[0,1]x^2dx=1/3x^3|[0,1]=1/3 E(X)=∫[1,2]2x-x^2dx=x^2|[1,2]-1/3x^3[1,2] =3-1/3(8-1)=3 - 7/3 E(X^2)=∫[0,1]x^3dx=1/4x^4|[0.1]=1/4 E(X^2)=∫[1,2]2X^2-X^3=2/3X^3|[1,2] -1/4x^4|[1,2] =2/3(7)-1/4(16-1) =14/3 -15/4

题设不是说了:0

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