设随机变量X
1
，…X
n
，X
n+1
独立同分布，且P(X
1
=1)=p，P(X
1
=0)=1-p，记：
【正确答案】正确答案：EY
i
=P(X
i
+X
i+1
=1)=P(X
i
=0，X
i+1
=1)+P(X
i
=1，X
i+1
=0)=2p(1-p)，i=1，…，n
=2np(1-p)，而E(Y
i
2
)=P(X
i
+X
i+1
=1)=2p(1-p)，∴DY
i
=E(Y
2
2
)-(EY
i
)
2
=2p(1-p)[1-2p(1-p)]，i=1，2，…，n．若l-k≥2，则Y
k
与Y
l
独立，这时cov(Y
k
，Y
l
)=0，而E(Y
k
Y
k+1
)=P(Y
k
=1，Y
k+1
=1)=P(X
k
+X
k+1
=1，X
k+1
+X
k+2
=1)=P(X
k
=0，X
k+1
=1，X
k+2
=0)+P(X
k
=1，X
k+1
=0，X
k+2
=1)=(1-p)
2
p+p
2
(1-p)=p(1-p)，∴cov(Y
k
，Y
k+1
)=E(Y
k
Y
k+1
)-EY
k
．EY
k+1
=p(1-p)-4p
2
(1-p)
2
，故
=2np(1-p)[1-2p(1-p)]+
=2np(1-p)，而E(Y
i
2
)=P(X
i
+X
i+1
=1)=2p(1-p)，∴DY
i
=E(Y
2
2
)-(EY
i
)
2
=2p(1-p)[1-2p(1-p)]，i=1，2，…，n．若l-k≥2，则Y
k
与Y
l
独立，这时cov(Y
k
，Y
l
)=0，而E(Y
k
Y
k+1
)=P(Y
k
=1，Y
k+1
=1)=P(X
k
+X
k+1
=1，X
k+1
+X
k+2
=1)=P(X
k
=0，X
k+1
=1，X
k+2
=0)+P(X
k
=1，X
k+1
=0，X
k+2
=1)=(1-p)
2
p+p
2
(1-p)=p(1-p)，∴cov(Y
k
，Y
k+1
)=E(Y
k
Y
k+1
)-EY
k
．EY
k+1
=p(1-p)-4p
2
(1-p)
2
，故
=2np(1-p)[1-2p(1-p)]+
【答案解析】