First let us find the distribution of
τ
r
{\displaystyle \tau _{r}}
P
(
τ
r
<
x
)
=
P
(
N
1
(
x
)
≥
r
)
=
∑
k
=
r
∞
(
λ
x
)
k
k
!
e
−
λ
x
{\displaystyle P(\tau _{r}<x)=P(N_{1}(x)\geq r)=\sum _{k=r}^{\infty }{\frac {(\lambda x)^{k}}{k!}}e^{-\lambda x}}
Thus by the chain rule, our random variable
τ
r
{\displaystyle \tau _{r}}
p
τ
r
(
x
)
=
∑
k
=
r
∞
k
λ
(
λ
x
)
k
−
1
k
!
e
−
λ
x
+
(
λ
x
)
k
k
!
(
−
λ
)
e
−
λ
x
=
λ
(
λ
x
)
r
−
1
(
r
−
1
)
!
e
−
λ
x
{\displaystyle p_{\tau _{r}}(x)=\sum _{k=r}^{\infty }{\frac {k\lambda (\lambda x)^{k-1}}{k!}}e^{-\lambda x}+{\frac {(\lambda x)^{k}}{k!}}(-\lambda )e^{-\lambda x}=\lambda {\frac {(\lambda x)^{r-1}}{(r-1)!}}e^{-\lambda x}}
So then
E
[
N
2
(
τ
r
2
)
]
=
E
[
E
[
N
2
(
t
)
|
τ
r
2
=
t
]
]
=
∫
0
∞
x
2
λ
2
λ
λ
r
−
1
(
r
−
1
)
!
x
r
−
1
e
−
λ
x
d
x
=
λ
r
+
2
(
r
−
1
)
!
∫
0
∞
x
r
+
1
e
−
λ
x
d
x
{\displaystyle {\begin{aligned}E[N_{2}(\tau _{r}^{2})]=E[E[N_{2}(t)|\tau _{r}^{2}=t]]&=\int _{0}^{\infty }x^{2}\lambda ^{2}\lambda {\frac {\lambda ^{r-1}}{(r-1)!}}x^{r-1}e^{-\lambda x}\,dx\\&={\frac {\lambda ^{r+2}}{(r-1)!}}\int _{0}^{\infty }x^{r+1}e^{-\lambda x}\,dx\end{aligned}}}
Now integrate the remaining integral by parts letting
u
=
x
r
+
1
,
d
v
=
e
−
λ
x
d
x
{\displaystyle u=x^{r+1},dv=e^{-\lambda x}\,dx}
E
[
N
2
(
τ
r
2
)
]
=
λ
r
+
2
(
r
−
1
)
!
[
−
1
λ
e
−
λ
x
x
r
+
1
|
0
∞
+
∫
0
∞
1
λ
r
+
1
x
r
e
−
λ
x
d
x
]
=
λ
r
+
2
(
r
−
1
)
!
[
∫
0
∞
1
λ
r
+
1
x
r
e
−
λ
x
d
x
]
=
λ
r
+
1
(
r
+
1
)
(
r
−
1
)
!
[
∫
0
∞
1
λ
r
+
1
x
r
e
−
λ
x
d
x
]
{\displaystyle {\begin{aligned}E[N_{2}(\tau _{r}^{2})]&={\frac {\lambda ^{r+2}}{(r-1)!}}[-{\frac {1}{\lambda }}e^{-\lambda x}x^{r+1}|_{0}^{\infty }+\int _{0}^{\infty }{\frac {1}{\lambda }}^{r+1}x^{r}e^{-\lambda x}\,dx]\\&={\frac {\lambda ^{r+2}}{(r-1)!}}[\int _{0}^{\infty }{\frac {1}{\lambda }}^{r+1}x^{r}e^{-\lambda x}\,dx]\\&={\frac {\lambda ^{r+1}(r+1)}{(r-1)!}}[\int _{0}^{\infty }{\frac {1}{\lambda }}^{r+1}x^{r}e^{-\lambda x}\,dx]\\\end{aligned}}}
Repeat integration by parts another
r
{\displaystyle r}
E
[
N
2
(
τ
r
2
)
]
=
(
r
+
1
)
r
λ
∫
0
∞
e
−
λ
x
d
x
=
λ
(
r
+
1
)
r
−
1
λ
e
−
λ
x
|
0
∞
=
(
r
+
1
)
r
{\displaystyle E[N_{2}(\tau _{r}^{2})]=(r+1)r\lambda \int _{0}^{\infty }e^{-\lambda x}\,dx=\lambda (r+1)r{\frac {-1}{\lambda }}e^{-\lambda x}|_{0}^{\infty }=(r+1)r}
X
k
n
,
1
≤
k
≤
n
{\displaystyle X_{kn},1\leq k\leq n}
P
(
X
k
n
=
0
)
=
1
−
1
/
n
,
P
(
X
k
n
=
k
2
)
=
1
/
n
{\displaystyle P(X_{kn}=0)=1-1/n,P(X_{kn}=k^{2})=1/n}
(a) Find the characteristic function of
S
−
n
=
∑
k
=
1
n
X
k
n
{\displaystyle S-n=\sum _{k=1}^{n}X_{kn}}
(b) Show that
S
n
/
n
2
{\displaystyle S_{n}/n^{2}}
φ
X
k
n
(
t
)
=
(
1
−
1
/
n
)
+
1
/
n
e
i
t
k
2
{\displaystyle \varphi _{X_{kn}}(t)=(1-1/n)+1/ne^{itk^{2}}}
Then by independence, we have
φ
X
n
(
t
)
=
∏
k
=
1
n
φ
X
k
n
=
∏
k
=
1
n
(
1
−
1
/
n
)
+
1
/
n
e
i
t
k
2
{\displaystyle \varphi _{X_{n}}(t)=\prod _{k=1}^{n}\varphi _{X_{kn}}=\prod _{k=1}^{n}(1-1/n)+1/ne^{itk^{2}}}
![{\displaystyle E[X_{n}]=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5e17c7306cb65535da8562748657ffa596335d0d)
![{\displaystyle \lim _{m\to \infty }E[\tau _{m}]/m^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ce2a62b89d58e773a5016ac82aa07505c18fb396)
![{\displaystyle S_{n}^{2}-cn=E[S_{n+1}^{2}-c(n+1_{|}{\mathcal {F}}_{n}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/352683c7fdceead475aacc9e2f0cddc58a2b230f)
![{\displaystyle {\begin{aligned}S_{n}^{2}-cn=E[S_{n+1}^{2}-c(n+1_{|}{\mathcal {F}}_{n}]&=S_{n}^{2}+E[X_{n+1}^{2}+2\sum _{j=1}^{n}X_{n+1}X_{j}|{\mathcal {F}}_{n}]-cn-c\\&=S_{n}^{2}-cn+E[X_{n+1}^{2}]+2\sum _{j=1}^{n}X_{j}E[X_{n+1}]{\text{ by independence}}\\&=S_{n}^{2}-cn+Var[X_{n+1}]=S_{n}^{2}-cn+\sigma ^{2}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a57cd4cd7097a259dd84a8650d5e744ca9c6039d)
![{\displaystyle \alpha _{r}=E[N_{2}(\tau _{r}^{2})]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f6cbf81ecf94dcc034e0d8bc382e65048e7ae5c8)
![{\displaystyle {\begin{aligned}E[N_{2}(\tau _{r}^{2})]=E[E[N_{2}(t)|\tau _{r}^{2}=t]]&=\int _{0}^{\infty }x^{2}\lambda ^{2}\lambda {\frac {\lambda ^{r-1}}{(r-1)!}}x^{r-1}e^{-\lambda x}\,dx\\&={\frac {\lambda ^{r+2}}{(r-1)!}}\int _{0}^{\infty }x^{r+1}e^{-\lambda x}\,dx\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8fa24d2712917c4a4663bd0cf12d9ba0ecfcdfeb)
![{\displaystyle {\begin{aligned}E[N_{2}(\tau _{r}^{2})]&={\frac {\lambda ^{r+2}}{(r-1)!}}[-{\frac {1}{\lambda }}e^{-\lambda x}x^{r+1}|_{0}^{\infty }+\int _{0}^{\infty }{\frac {1}{\lambda }}^{r+1}x^{r}e^{-\lambda x}\,dx]\\&={\frac {\lambda ^{r+2}}{(r-1)!}}[\int _{0}^{\infty }{\frac {1}{\lambda }}^{r+1}x^{r}e^{-\lambda x}\,dx]\\&={\frac {\lambda ^{r+1}(r+1)}{(r-1)!}}[\int _{0}^{\infty }{\frac {1}{\lambda }}^{r+1}x^{r}e^{-\lambda x}\,dx]\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d7d8c4e61bc73b39dd1703ffff1bb0c80bc36b81)
![{\displaystyle E[N_{2}(\tau _{r}^{2})]=(r+1)r\lambda \int _{0}^{\infty }e^{-\lambda x}\,dx=\lambda (r+1)r{\frac {-1}{\lambda }}e^{-\lambda x}|_{0}^{\infty }=(r+1)r}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8d5fb120c8cbc0934f20f3b6d7987a8811c21fc9)
