UA MATH564 概率论 QE练习题 概率极限理论
UA MATH564 概率論 QE練習(xí)題 概率極限理論
- 2015/5/3
- 2016/1/3
這是2015年5月的3題、2016年1月的3題
2015/5/3
這個(gè)題干有點(diǎn)意思，有一列隨機(jī)變量但并不是互相獨(dú)立的，知道第一個(gè)是均勻的，后續(xù)的條件分布給了。
Part a
E[Xn+1r∣Xn]=∫0cXnxr1cXndx=crXnr1+rE[X_{n+1}^r|X_n] = \int_{0}^{cX_n} x^r \frac{1}{cX_n}dx = \frac{c^rX_n^r}{1+r}E[Xn+1r?∣Xn?]=∫0cXn??xrcXn?1?dx=1+rcrXnr??
By the Iteration of expectation,
E[Xnr]=E[E[Xnr∣Xn?1]]=E[crXn?1r1+r]=cr1+rE[Xn?1r]E[X_n^r] = E[E[X_n^r|X_{n-1}]]= E[\frac{c^rX_{n-1}^r}{1+r}] = \frac{c^r}{1+r} E[X_{n-1}^r] E[Xnr?]=E[E[Xnr?∣Xn?1?]]=E[1+rcrXn?1r??]=1+rcr?E[Xn?1r?]
So
E[Xnr]=cr(n?1)(1+r)n?1EX1r=cr(n?1)(1+r)nE[X_n^r] = \frac{c^{r(n-1)}}{(1+r)^{n-1}} EX_1^r = \frac{c^{r(n-1)}}{(1+r)^n}E[Xnr?]=(1+r)n?1cr(n?1)?EX1r?=(1+r)ncr(n?1)?
Part b
Notice 3<c<2\sqrt{3} < c < 23?<c<2 . Let r=1r=1r=1,
EXn=cn?12n=12(c2)n?1→0,asn→∞EX_n = \frac{c^{n-1}}{2^n} = \frac{1}{2} (\frac{c}{2})^{n-1} \to 0,\ as\ n \to \inftyEXn?=2ncn?1?=21?(2c?)n?1→0,?as?n→∞
Let r=2r = 2r=2,
EXn2=(c2)n?13n=13(c23)n?1→∞,asn→∞EX_n^2 = \frac{(c^2)^{n-1}}{3^n} = \frac{1}{3} (\frac{c^2}{3})^{n-1} \to \infty,\ as\ n \to \inftyEXn2?=3n(c2)n?1?=31?(3c2?)n?1→∞,?as?n→∞
Part c （簡(jiǎn)單點(diǎn)的做法是取任意r≤1r\le1r≤1，用Markov不等式即可）
To show ??>0\forall \epsilon>0??>0, ?M>0\exists M>0?M>0 such that
∑n=1∞P(∣Xn?0∣>?)=∑n=1∞P(Xn>?)<M\sum_{n=1}^{\infty} P(|X_n - 0|>\epsilon) =\sum_{n=1}^{\infty} P(X_n >\epsilon) <Mn=1∑∞?P(∣Xn??0∣>?)=n=1∑∞?P(Xn?>?)<M
By Markov inequality
P(Xn>?)≤EXnr?r∑n=1∞P(Xn>?)≤∑n=1∞EXnr?r=12cr?r∑n=1∞(cr)n(1+r)n=12cr?rcr1+r?(cr1+r)∞1?cr1+rP(X_n >\epsilon) \le \frac{EX_n^r}{\epsilon^r} \\ \sum_{n=1}^{\infty}P(X_n >\epsilon) \le \sum_{n=1}^{\infty} \frac{EX_n^r}{\epsilon^r} = \frac{1}{2c^r\epsilon^r} \sum_{n=1}^{\infty} \frac{(c^r)^n}{(1+r)^n} = \frac{1}{2c^r\epsilon^r} \frac{\frac{c^r}{1+r} - (\frac{c^r}{1+r})^{\infty}}{1-\frac{c^r}{1+r}}P(Xn?>?)≤?rEXnr??n=1∑∞?P(Xn?>?)≤n=1∑∞??rEXnr??=2cr?r1?n=1∑∞?(1+r)n(cr)n?=2cr?r1?1?1+rcr?1+rcr??(1+rcr?)∞?
Let cr<1+rc^r < 1+ rcr<1+r holds,
∑n=1∞P(Xn>?)≤12cr?rcr1+r1?cr1+r=12?r(1+r?cr)\sum_{n=1}^{\infty}P(X_n >\epsilon) \le \frac{1}{2c^r\epsilon^r} \frac{\frac{c^r}{1+r} }{1-\frac{c^r}{1+r}} = \frac{1}{2\epsilon^r (1+r - c^r)}n=1∑∞?P(Xn?>?)≤2cr?r1?1?1+rcr?1+rcr??=2?r(1+r?cr)1?
By continuity, ?r?\exists r^*?r? such that cr?<1+r?c^{r^*} < 1+ r^*cr?<1+r?, and for all rrr in the neighborhood of r?r^*r?
δ1<1+r?cr<δ2,?δ2>δ1\delta_1<1+r-c^{r} < \delta_2,\forall \delta_2>\delta_1δ1?<1+r?cr<δ2?,?δ2?>δ1?
Now fix δ1≥1/?r\delta_1 \ge 1/\epsilon^rδ1?≥1/?r, then
12?r?(1+r??cr?)<∞\frac{1}{2\epsilon^{r^*} (1+r^* - c^{r^*})} < \infty2?r?(1+r??cr?)1?<∞
2016/1/3
證明依概率收斂有幾種不同的辦法：用定義證明、證明均方收斂、證明幾乎必然收斂、證明依分布收斂均可以說明依概率收斂，需要我們根據(jù)條件靈活選取方法。這里只有均值、方差的信息，所以顯然是要我們證明均方收斂。
Compute
E(2n(n+1)∑j=1njXj)=2n(n+1)∑j=1njEXj=2n(n+1)n(n+1)2=1Var(2n(n+1)∑j=1njXj)=4n2(n+1)2[∑j=1nj2Var(Xj)]E \left( \frac{2}{n(n+1)}\sum_{j=1}^n jX_j \right) = \frac{2}{n(n+1)} \sum_{j=1}^n jEX_j = \frac{2}{n(n+1)} \frac{n(n+1)}{2} = 1 \\ Var\left( \frac{2}{n(n+1)}\sum_{j=1}^n jX_j \right) = \frac{4}{n^2(n+1)^2}\left[\sum_{j=1}^n j^2Var(X_j) \right]E(n(n+1)2?j=1∑n?jXj?)=n(n+1)2?j=1∑n?jEXj?=n(n+1)2?2n(n+1)?=1Var(n(n+1)2?j=1∑n?jXj?)=n2(n+1)24?[j=1∑n?j2Var(Xj?)]
Note that the variance of population is finite. Denote it as σ2\sigma^2σ2. As n→∞n \to \inftyn→∞
RHS=4σ2n2(n+1)2∑j=1nj2=4σ2n2(n+1)2n(n+1)(2n+1)6=2σ2(2n+1)3n(n+1)→0RHS = \frac{4\sigma^2}{n^2(n+1)^2}\sum_{j=1}^n j^2 =\frac{4\sigma^2}{n^2(n+1)^2} \frac{n(n+1)(2n+1)}{6} = \frac{2\sigma^2 (2n+1)}{3n(n+1)} \to 0RHS=n2(n+1)24σ2?j=1∑n?j2=n2(n+1)24σ2?6n(n+1)(2n+1)?=3n(n+1)2σ2(2n+1)?→0
So
2n(n+1)∑j=1njXj→p1\frac{2}{n(n+1)}\sum_{j=1}^n jX_j \to_p 1n(n+1)2?j=1∑n?jXj?→p?1
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