离散型随机变量方差计算问题第一题

Question

如题所述

Answer 1

离散型随机变量方差计算问题第一题
由题意：

E(x)=-1x(1/2)+0x(1-2q)+1xq^2=q^2-0.5
E(x^2)=(-1)^2x(1/2)+0x(1-2q)+1xq^2=q^2+0.5
因此D(x)=E(x^2)-[E(x)]^2
=q^2+0.5-(q^2-0.5)^2
又：0.5+1-2q+q^2=1
即：q^2-2q+1=1/2
(q-1)^2=1/2
因为：0<q<1
所以：q-1=-(√2)/2
q=1-(√2)/2
q^2=-1/2+2q
所以：E(x)=q^2-0.5
=-1/2+2q-0.5
=2q-1
=2-√2-1
=1-√2
E(x^2)=q^2+0.5
=-1/2+2q+0.5
=2q
=2-√2
D(x)=E(x^2)-[E(x)]^2
=2-√2-(1-√2)^2
=2-√2-(1+2-2√2)
=2-√2-(3-2√2)
=2-√2-3+2√2
=√2-1