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概率理論問題

解決方案:

(1)密度函數在積分區域的積分總是等於1,所以:

定積分(x從0到π) asin(x) dx = -acos(x) | x從0到π = 2a = 1,所以a = 1/2;

(2)直接使用定義,

EX =定積分(x從0到π)axsin(x) dx

=定積分(x從0到π)(-ax) dcos(x)

= -axcos(x) | x從0到π+定積分(x從0到π)acos(x) dx

= aπ+0 = aπ=π/2;

(3)

EX^2 =定積分(x從0到π) ax 2sin (x) dx

=定積分(x從0到π)(-ax ^ 2)dcos(x)

=(-ax ^ 2)cos(x)| x從0到π+定積分(x從0到π)axcos(x) dx

= aπ^2+definite積分(x從0到π)ax dsin(x)

= aπ^2-definite積分(x從0到π)asin(x) dx

= aπ2 = aπ^2-1 = π^2/2-1+0,所以

dx = π^2/2-1-π^2/4 = π^2/4-1;

(4)分布函數的定義:

F(x) =定積分(t從負無窮到x)f(x) dx

= 0,如果x & lt=0;

=定積分(t從0到x) asin (x) dx = (1-cos (x))/2,如果0

= 1,如果x & gtπ.