就 一視頻 一網(wǎng)友 一彈幕 所提命題 之證明
E(x)
=
Σn(1-p)^(n-1)p
=
pΣn(1-p)^(n-1)
=
p
1·(1-p)^0+2·(1-p)^1+3·(1-p)^2
+
...
+
n(1-p)^(n-1)
設(shè)
Sn
=
1·(1-p)^0+2·(1-p)^1+3·(1-p)^2
+
...
+
n(1-p)^(n-1)
有
(1-p)Sn
=
1·(1-p)^1+2·(1-p)^2+3·(1-p)^3
+
...
+
n(1-p)^n
即
-pSn
=
n(1-p)^n-1
-
(1-p)^1+(1-p)^2+(1-p)^3+...+(1-p)^(n-1)
=
n(1-p)^n-1
-
(1-p)(1-(1-p)^(n-1))/p
即
Sn
=
(1-(1+n)(1-p)^n)/p2
即
E(x)
=
(1-(1+n)(1-p)^n)/p
即
lim(n→+∞)E(x)
=
(1-lim(n→+∞)(1+n)(1-p)^n)/p
=
(1-lim(n→+∞)(1+n)/(1/(1-p)^n))/p
=
(1-lim(n→+∞)1/(-ln(1-p)/(1-p)^n))/p
=
(1-lim(n→+∞)(1-p)^n/(-ln(1-p)))/p
=
1/p
D(x)
=
E(x2)-E2(x)
=
pΣn2(1-p)^(n-1)-1/p2
=
p
12·(1-p)^0+22·(1-p)^1+32·(1-p)^2
+
...
+
n2(1-p)^(n-1)
-
1/p2
設(shè)
Sn
=
12·(1-p)^0+22·(1-p)^1+32·(1-p)^2
+
...
+
n2(1-p)^(n-1)
有
(1-p)Sn
=
12·(1-p)^1+22·(1-p)^2+32·(1-p)^3
+
...
+
n2(1-p)^n
即
-pSn
=
n2(1-p)^n-1
-
3(1-p)^1+5(1-p)^2+7(1-p)^3
+
...
+
(2n-1)(1-p)^(n-1)
設(shè)
Tn
=
3(1-p)^1+5(1-p)^2+7(1-p)^3
+
...
+
(2n-1)(1-p)^(n-1)
有
(1-p)Tn
=
3(1-p)^2+5(1-p)^3+7(1-p)^4
+
...
+
(2n-1)(1-p)^n
即
-pTn
=
(2n-1)(1-p)^n-3(1-p)
-
2
(1-p)^2+(1-p)^3+...+(1-p)^(n-1)
=
(2n-1)(1-p)^n-3(1-p)
-
2
(1-p)2(1-(1-p)^(n-2))/p
=
((2n-1)p(1-p)^n-3(1-p)p)/p
-
(2(1-p)2-2(1-p)^n)/p
=
(2n-1+2/p)(1-p)^n+p+1-2/p
即
Tn
=
((1-2n)/p-2/p2)(1-p)^n-1-1/p+2/p2
即
-pSn
=
n2(1-p)^n-1
-
((1-2n)/p-2/p2)(1-p)^n-1-1/p+2/p2
=
(n2-(1-2n)/p+2/p2)(1-p)^n+1/p-2/p2
即
Sn
=
(-n2/p+(1-2n)/p2-2/p3)(1-p)^n-1/p2+2/p3
即
D(x)
=
(-n2+(1-2n)/p-2/p2)(1-p)^n-1/p+1/p2
即
lim(n→+∞)D(x)
=
lim(n→+∞)
(-n2-2n/p+(p-2)/p2)(1-p)^n+(1-p)/p2
=
lim(n→+∞)
(-n2-2n/p+(p-2)/p2)/(1/(1-p)^n)
+
(1-p)/p2
=
lim(n→+∞)
(-2n-2/p)/(-ln(1-p)/(1-p)^n)
+
(1-p)/p2
=
lim(n→+∞)
-2/(ln2(1-p)/(1-p)^n)
+
(1-p)/p2
=
(1-p)/p2
得證
