設 X 1 , X 2 , . . . {\displaystyle X_{1},X_{2},...} 為 i.i.d. 隨機變數，使得 E [ X n ] = 0 {\displaystyle E[X_{n}]=0} 且 | X n | ≤ 1 {\displaystyle |X_{n}|\leq 1} 幾乎處處成立，設 S n = ∑ k = 1 n X k {\displaystyle S_{n}=\sum _{k=1}^{n}X_{k}} .
(a) 找到一個數字 c {\displaystyle c} ，使得 S n 2 − c n {\displaystyle S_{n}^{2}-cn} 是一個鞅，並證明鞅性質。
(b) 定義 τ m = inf { n ≥ 1 : S n > 2 m or S n < − m } {\displaystyle \tau _{m}=\inf\{n\geq 1:S_{n}>2m{\text{ or }}S_{n}<-m\}} 。計算 lim m → ∞ P ( S τ m > 2 m ) {\displaystyle \lim _{m\to \infty }P(S_{\tau _{m}}>2m)} .
(c) 計算 lim m → ∞ E [ τ m ] / m 2 {\displaystyle \lim _{m\to \infty }E[\tau _{m}]/m^{2}} .
每個 S n {\displaystyle S_{n}} 顯然是 F n {\displaystyle {\mathcal {F}}_{n}} -可測量的，並且幾乎處處有限（因此 L 1 {\displaystyle L^{1}} ）。因此，我們只需要驗證鞅性質。也就是說，我們要證明 S n 2 − c n = E [ S n + 1 2 − c ( n + 1 | F n ] {\displaystyle S_{n}^{2}-cn=E[S_{n+1}^{2}-c(n+1_{|}{\mathcal {F}}_{n}]}
S n 2 − c n = E [ S n + 1 2 − c ( n + 1 | F n ] = S n 2 + E [ X n + 1 2 + 2 ∑ j = 1 n X n + 1 X j | F n ] − c n − c = S n 2 − c n + E [ X n + 1 2 ] + 2 ∑ j = 1 n X j E [ X n + 1 ] by independence = S n 2 − c n + V a r [ X n + 1 ] = S n 2 − c n + σ 2 {\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}}}
我們可以斷言 σ 2 {\displaystyle \sigma ^{2}} 存在且是有限的，因為每個 | X n | ≤ 1 {\displaystyle |X_{n}|\leq 1} 幾乎處處。因此，為了使 S n 2 − c n {\displaystyle S_{n}^{2}-cn} 成為鞅，我們必須有 c = σ 2 {\displaystyle c=\sigma ^{2}} 。
令 N 1 ( t ) , N 2 ( t ) {\displaystyle N_{1}(t),N_{2}(t)} 為獨立泊松過程，分別具有引數 λ , λ 2 {\displaystyle \lambda ,\lambda ^{2}} ，其中 λ {\displaystyle \lambda } 是一個未指定的正實數。對於每個 r ≥ 1 {\displaystyle r\geq 1} ，令 τ r = inf { t > 0 : N 1 ( t ) ≥ r } {\displaystyle \tau _{r}=\inf\{t>0:N_{1}(t)\geq r\}} 。證明 α r = E [ N 2 ( τ r 2 ) ] {\displaystyle \alpha _{r}=E[N_{2}(\tau _{r}^{2})]} 不依賴於 λ {\displaystyle \lambda } ，並明確求出 α r {\displaystyle \alpha _{r}} 。
首先讓我們找到 τ r {\displaystyle \tau _{r}} 的分佈
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}}
因此，根據鏈式法則，我們的隨機變數 τ r {\displaystyle \tau _{r}} 具有機率密度函式
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}}
所以
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}}}
現在用分部積分法求解剩餘積分，令 u = x r + 1 , d v = e − λ x d x {\displaystyle u=x^{r+1},dv=e^{-\lambda x}\,dx} 。我們得到
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}}}
重複分部積分法另外 r {\displaystyle r} 次，我們得到
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}
令 X k n , 1 ≤ k ≤ n {\displaystyle X_{kn},1\leq k\leq n} 是獨立隨機變數，滿足
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) 求 S − n = ∑ k = 1 n X k n {\displaystyle S-n=\sum _{k=1}^{n}X_{kn}} 的特徵函式。
(b) 證明 S n / n 2 {\displaystyle S_{n}/n^{2}} 按分佈收斂於一個非退化的隨機變數。
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}}}
然後由獨立性，我們有 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}}}
顯然是 
-可測量的，並且幾乎處處有限（因此 
）。因此，我們只需要驗證鞅性質。也就是說，我們要證明 ![{\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/a57cd4cd7097a259dd84a8650d5e7444ca9c6039d)
存在且是有限的，因為每個 
幾乎處處。因此，為了使 
成為鞅，我們必須有 
。
的分佈
具有機率密度函式
![{\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)
。我們得到![{\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)
次，我們得到![{\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)
是獨立隨機變數，滿足
的特徵函式。
按分佈收斂於一個非退化的隨機變數。
![{\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 \alpha _{r}=E[N_{2}(\tau _{r}^{2})]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f6cbf81ecf94dcc034e0d8bc382e65048e7ae5c8)
