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\layout Title

Probability A exam solutions
\layout Author

David Rossell i Ribera
\layout Standard

I may have committed a number of errors in writing these solutions, but
 they should be fine for the most part.
 Use them at your own risk!
\layout Section*

Probability January 2003
\layout Section*

Problem 1
\layout Standard

a) Usual definitions
\layout Standard

b) 
\begin_inset Formula $X_{n}=\left\{ \begin{array}{c}
n^{2}wp1/n^{2}\\
0wp1-1/n^{2}\end{array}\right.$
\end_inset 

 converges a.s.
 to zero by Borel-Cantelli and hence also in probability, but not in 
\begin_inset Formula $L_{1}$
\end_inset 

 since 
\begin_inset Formula $E(X_{n}-0)=E(X_{n})=1$
\end_inset 

.
\layout Standard


\begin_inset Formula $X_{n}=\left\{ \begin{array}{c}
1wp1/n\\
0wp1-1/n\end{array}\right.$
\end_inset 

 with 
\begin_inset Formula $X_{n}$
\end_inset 

 independent converges in 
\begin_inset Formula $L_{1}$
\end_inset 

and thus in probability to zero, but not to zero by Borel-Cantelli.
\layout Standard


\begin_inset Formula $X=\left\{ \begin{array}{c}
1wp1/n\\
0wp1-1/n\end{array}\right.$
\end_inset 

, 
\begin_inset Formula $Y=1-X$
\end_inset 

.
 Obviously 
\begin_inset Formula $Y$
\end_inset 

 and 
\begin_inset Formula $X$
\end_inset 

 have the same distribution but 
\begin_inset Formula $P(|X-Y|>0)=1$
\end_inset 

, so we don't have convergence in probability.
\layout Section*

Problem 2
\layout Standard

a) (i) 
\begin_inset Formula $P(A\bigcap B)\geq0)$
\end_inset 

 (ii) 
\begin_inset Formula $A_{1}...A_{n}$
\end_inset 

 disjoint,
\layout Standard


\begin_inset Formula $Q\left(\bigcup_{i=1}^{\infty}A_{i}\right)=P\left(\left(\bigcup_{i=1}^{\infty}A_{i}\right)\bigcap B\right)=$
\end_inset 


\begin_inset Formula $P\left(\bigcup_{i=1}^{\infty}\left(A_{i}\bigcap B\right)\right)=\left\{ disjoint\right\} =\sum_{i=1}^{\infty}P(A_{i}\bigcap B)=\sum_{i=1}^{\infty}Q(A_{i})$
\end_inset 


\layout Standard

For 
\begin_inset Formula $Q$
\end_inset 

 to be a probability measure, we need 
\begin_inset Formula $Q(\Omega)=P(\Omega\bigcap B)=P(B)=1$
\end_inset 

.
\layout Standard

b) Start with indicators
\layout Standard

(i) 
\begin_inset Formula $X=I_{A}$
\end_inset 

.
 Then 
\begin_inset Formula $E(X;B)=\int_{B}I_{A}dP=P(A\bigcap B)=Q(A)=\int_{\Omega}XdQ$
\end_inset 


\layout Standard

(ii) 
\begin_inset Formula $X=\sum_{i=1}^{n}a_{k}I_{A_{k}}$
\end_inset 

.
 Then 
\begin_inset Formula $\int_{\Omega}XdQ=\int_{\Omega}\sum_{i=1}^{n}a_{k}I_{A_{k}}dQ=$
\end_inset 


\begin_inset Formula $\sum_{i=1}^{n}a_{k}\int_{\Omega}I_{A_{k}}dQ=\sum a_{k}P(A_{k}B)$
\end_inset 


\begin_inset Formula $=\int_{B}\sum_{i=1}^{n}a_{k}I_{A_{k}}dP=E(X;B)$
\end_inset 

.
\layout Standard

(iii) 
\begin_inset Formula $X\geq0$
\end_inset 

.
 Let 
\begin_inset Formula $0\leq X_{n}\uparrow X$
\end_inset 

 with 
\begin_inset Formula $X_{n}$
\end_inset 

simple and 
\begin_inset Formula $\int_{\Omega}XdQ=...$
\end_inset 


\begin_inset Formula $=\int_{B}XdP$
\end_inset 


\layout Standard

(iv) General 
\begin_inset Formula $X$
\end_inset 

.
 Proceed as usual.
\layout Section*

Problem 3
\layout Standard

a) It means that (i) 
\begin_inset Formula $Y_{n}$
\end_inset 

 is 
\begin_inset Formula $\sigma(X_{1}...X_{n})$
\end_inset 

 measurable and (ii) 
\begin_inset Formula $E(Y_{n+1}|\sigma(X_{1}...X_{n}))=Y_{n}$
\end_inset 

.
\layout Standard

b) 1) 
\begin_inset Formula $Y_{n}$
\end_inset 

is a function of 
\begin_inset Formula $X_{1}...X_{n}$
\end_inset 

and thus 
\begin_inset Formula $\sigma(X_{1}...X_{n})$
\end_inset 

 measurable, and 
\begin_inset Formula $E(Y_{n+1}|\sigma(X_{1}...X_{n}))=\left\{ independent\right\} =$
\end_inset 


\begin_inset Formula $E(Y_{n+1})=0\neq Y_{n}$
\end_inset 

, so it's not a martingale.
\layout Standard

2) 
\begin_inset Formula $Y_{n}$
\end_inset 

is a function of 
\begin_inset Formula $X_{1}...X_{n}$
\end_inset 

and thus 
\begin_inset Formula $\sigma(X_{1}...X_{n})$
\end_inset 

 measurable, and 
\begin_inset Formula $E(Y_{n+1}|\sigma(X_{1}...X_{n}))=\sum_{k=1}^{n}X_{k}+E(X_{n+1})=$
\end_inset 


\begin_inset Formula $\sum_{k=1}^{n}X_{k}=Y_{n}$
\end_inset 

, so it's a martingale.
\layout Standard

3) 
\begin_inset Formula $Y_{n}$
\end_inset 

is NOT a function of 
\begin_inset Formula $X_{1}...X_{n}$
\end_inset 

and thus NOT 
\begin_inset Formula $\sigma(X_{1}...X_{n})$
\end_inset 

 measurable, so it's not a martingale.
\layout Standard

4) 
\begin_inset Formula $Y_{n}$
\end_inset 

is a function of 
\begin_inset Formula $X_{1}...X_{n}$
\end_inset 

and thus 
\begin_inset Formula $\sigma(X_{1}...X_{n})$
\end_inset 

 measurable and 
\begin_inset Formula $E(Y_{n+1}|\sigma(X_{1}...X_{n}))=\prod_{i=1}^{n}X_{i}E(X_{n+1})=0\neq Y_{n}$
\end_inset 

, so not a martingale.
\layout Standard

5) 
\begin_inset Formula $Y_{n}$
\end_inset 

is a function of 
\begin_inset Formula $X_{1}...X_{n}$
\end_inset 

and thus 
\begin_inset Formula $\sigma(X_{1}...X_{n})$
\end_inset 

 measurable and 
\begin_inset Formula $E(Y_{n+1}|\sigma(X_{1}...X_{n}))=\prod_{i=1}^{n}2^{X_{i}}E(2^{X_{n+1}})$
\end_inset 

, and 
\begin_inset Formula $E(2^{X_{n+1}})=\frac{1}{2}2^{-1}+\frac{1}{2}2\neq1$
\end_inset 

, so it's not a martingale.
\layout Section*

Problem 4
\layout Standard

By the Strong Law of Large Numbers, 
\begin_inset Formula $\sum X_{i}/n$
\end_inset 

 converges to 
\begin_inset Formula $1/2$
\end_inset 

 almost surely.
 For 
\begin_inset Formula $logG_{n}=\frac{1}{n}\sum logX_{i}$
\end_inset 

, since 
\begin_inset Formula $E(logX)=-1$
\end_inset 

 (can be found integrating the pdf as usual) by SLLN again we have that
 
\begin_inset Formula $logG_{n}$
\end_inset 

 converges almost surely to 
\begin_inset Formula $-1$
\end_inset 

.
 Hence by the Continuous Mapping Principle 
\begin_inset Formula $G_{n}$
\end_inset 

 converges almost surely to 
\begin_inset Formula $e^{-1}$
\end_inset 

.
\layout Standard

Now since 
\begin_inset Formula $E(1/X)=\infty$
\end_inset 

, by SLLN 
\begin_inset Formula $\frac{1}{n}\sum\frac{1}{X_{i}}\rightarrow_{a.s.}\infty$
\end_inset 

 and by CMP 
\begin_inset Formula $\frac{n}{\sum1/X_{i}}\rightarrow_{a.s.}0$
\end_inset 

.
 (note that SLLN also applies when the expectation is 
\begin_inset Formula $\infty$
\end_inset 

.
\layout Section*

Problem 5
\layout Standard

1) The chain is irreducible since all states communicate and the period
 of all states is 1 since 
\begin_inset Formula $p_{22}>0$
\end_inset 

 and periodicity is a class property.
 To find 
\begin_inset Formula $E(\tau_{0})$
\end_inset 

 we'll find the stationary distribution first.
 
\begin_inset Formula $\pi=\pi P$
\end_inset 

 gives 
\begin_inset Formula $\pi_{0}=1/4$
\end_inset 

, 
\begin_inset Formula $\pi_{1}=1/4$
\end_inset 

, 
\begin_inset Formula $\pi_{2}=1/2$
\end_inset 

 and thus 
\begin_inset Formula $E(\tau_{0})=1/\pi_{0}=4$
\end_inset 

.
\layout Standard

2) a)This is a branching process, so 
\begin_inset Formula $E(Z_{n})=(2p)^{n}$
\end_inset 

.
\layout Standard

b) We know that 
\begin_inset Formula $\pi=1$
\end_inset 

 iff 
\begin_inset Formula $2p\leq1$
\end_inset 

 so 
\begin_inset Formula $p\leq1/2$
\end_inset 

.
 Hence for 
\begin_inset Formula $p>1/2$
\end_inset 

 we got 
\begin_inset Formula $\pi<1$
\end_inset 

, and 
\begin_inset Formula $\pi$
\end_inset 

 is the smallest solution to 
\begin_inset Formula $\pi=G(\pi)$
\end_inset 

, where 
\begin_inset Formula $G$
\end_inset 

 is the pgf of a binomial(2,p).
 Solve for 
\begin_inset Formula $s$
\end_inset 

 in
\begin_inset Formula $G(s)=(1-p)^{2}+2p(1-p)s+p^{2}s^{2}=s$
\end_inset 

 to get 
\begin_inset Formula $\pi=\left(\frac{1-p}{p}\right)^{2}$
\end_inset 

.
\layout Standard

c) By independence, it's 
\begin_inset Formula $\pi^{N}$
\end_inset 

.
\layout Section*

Problem 6
\layout Standard

b) For 
\begin_inset Formula $X$
\end_inset 

 continuous we know 
\begin_inset Formula $F(X)$
\end_inset 

 is 
\begin_inset Formula $Unif(0,1)$
\end_inset 

 and hence 
\begin_inset Formula $E(X)=1/2$
\end_inset 

.
 For 
\begin_inset Formula $X$
\end_inset 

 categorical consider 
\begin_inset Formula $Y=F_{x}(X)$
\end_inset 

, and let 
\begin_inset Formula $Z$
\end_inset 

 be any continuous random variable such that it's cdf 
\begin_inset Quotes eld
\end_inset 

matches
\begin_inset Quotes erd
\end_inset 

 the cdf of 
\begin_inset Formula $X$
\end_inset 

 in each of its jumps, i.e.
 
\begin_inset Formula $F_{z}(z)=P(Z\leq z)=P(X\leq z)=F_{x}(z)$
\end_inset 

 for all 
\begin_inset Formula $z$
\end_inset 

 discontinuity point.
 Note that 
\begin_inset Formula $F_{z}(z)\geq F_{x}(z)$
\end_inset 

, with strict equality in the discontinuity points (if this is confusing,
 plotting both cdfs may help to see it clearer).
 That gives us that 
\begin_inset Formula $E(F_{X}(z))\geq E(F_{Z}(z))=1/2$
\end_inset 

, the latter equality due to 
\begin_inset Formula $Z$
\end_inset 

 being a continuous random variable.
\layout Standard

c) I don't know how to do it.
\layout Section*

Probability January 2004
\layout Section*

Problem 1
\layout Standard

a) Usual definition
\layout Standard

b) (i) Let 
\begin_inset Formula $x\in(0,1)$
\end_inset 

.
 
\begin_inset Formula $P(m_{n}\leq x)=1-P(m_{n}>x)=1-\left(P(X_{1}>x)\right)^{n}=1-(1-x)^{n}\rightarrow1$
\end_inset 

, i.e.
 
\begin_inset Formula $m_{n}\rightarrow^{D}0$
\end_inset 


\layout Standard

(ii) 
\begin_inset Formula $P(nm_{n}\leq x)=1-\left(P(X_{1}>x/n)\right)^{n}=1-(1-x/n)^{n}\rightarrow1-e^{-x}$
\end_inset 

, which is the cdf of an 
\begin_inset Formula $exp(1)$
\end_inset 

.
\layout Standard

c) Theorem.
 
\begin_inset Formula $\sum P(|X_{n}-X|>\epsilon)<\infty\Rightarrow X_{n}\rightarrow_{a.s.}X$
\end_inset 


\layout Standard

Since 
\begin_inset Formula $\sum P(|m_{n}-0|>\epsilon)=\sum P(m_{n}>\epsilon)=\sum(1-\epsilon)^{n}<\infty$
\end_inset 

, we have 
\begin_inset Formula $m_{n}\rightarrow_{a.s.}0$
\end_inset 

.
 In general 
\begin_inset Formula $nm_{n}$
\end_inset 

 doesn't converge almost surely to 
\begin_inset Formula $X$
\end_inset 

, although by Skorohod's theorem we know that 
\begin_inset Formula $\exists Y_{n}=^{D}nm_{n}$
\end_inset 

, 
\begin_inset Formula $Y=^{D}X$
\end_inset 

 such that 
\begin_inset Formula $Y_{n}\rightarrow Y$
\end_inset 

.
\layout Section*

Problem 2
\layout Standard

a) Delta method.
\layout Standard

b) Let's find first the asymptotic distribution of 
\begin_inset Formula $logG_{n}=\frac{1}{n}\sum logX_{i}$
\end_inset 

.
 It can be seen that 
\begin_inset Formula $E(logX_{i})=-\frac{1}{2}$
\end_inset 

 and 
\begin_inset Formula $V(logX_{i})=\frac{1}{4}$
\end_inset 

, so by the Central Limit Theorem 
\begin_inset Formula $\sqrt{n}\left(logG_{n}+\frac{1}{2}\right)\rightarrow^{D}N(0,1/4)$
\end_inset 

, and since 
\begin_inset Formula $g(x)=e^{x}\Rightarrow g'(x)=e^{x}$
\end_inset 

the delta method gives that 
\begin_inset Formula $\sqrt{n}\left(G_{n}-e^{-1/2}\right)\rightarrow^{D}N(0,e/4)$
\end_inset 


\layout Standard

Now, for 
\begin_inset Formula $H_{n}$
\end_inset 

let's first find the asymptotic distribution of 
\begin_inset Formula $H_{n}^{-1}=\frac{1}{n}\sum\frac{1}{X_{k}}$
\end_inset 

.
 Since 
\begin_inset Formula $E(1/X)=2$
\end_inset 

, 
\begin_inset Formula $V(1/X)=\infty$
\end_inset 

 the CLT doesn't apply to 
\begin_inset Formula $H_{n}^{-1}$
\end_inset 

.
\layout Section*

Problem 3
\layout Standard

a) BCI and BCII
\layout Standard

b) (i) 
\begin_inset Formula $X_{n}=\left\{ \begin{array}{cc}
n^{2} & wp1/n^{2}\\
0 & wp1-1/n^{2}\end{array}\right\} $
\end_inset 

 independent.
\layout Standard

(ii) 
\begin_inset Formula $X_{n}=\left\{ \begin{array}{cc}
n-1 & wp1/n\\
0 & wp1-1/n\end{array}\right\} $
\end_inset 

 doesn't converge in 
\begin_inset Formula $L_{1}$
\end_inset 

 nor a.s., but it does in probability.
\layout Standard

(iii) 
\begin_inset Formula $X_{n}=\left\{ \begin{array}{cc}
n & wp1-1/n^{2}\\
-n(n^{2}-1) & wp1/n^{2}\end{array}\right\} $
\end_inset 

, by BCI 
\begin_inset Formula $X_{n}\rightarrow_{a.s.}-\infty$
\end_inset 

, but 
\begin_inset Formula $E(X_{n})=0$
\end_inset 

.
\layout Standard

(iv) 
\begin_inset Formula $X_{n}=\left\{ \begin{array}{cc}
n & wp1-1/n^{2}\\
-n^{4} & wp1/n^{2}\end{array}\right\} $
\end_inset 

.
 By BCI 
\begin_inset Formula $X_{n}\rightarrow_{a.s.}0$
\end_inset 

 but 
\begin_inset Formula $E(X_{n})\rightarrow-\infty$
\end_inset 

.
\layout Section*

Problem 4
\layout Standard

1) (a) 
\begin_inset Formula $\{0\}$
\end_inset 

 is transient and 
\begin_inset Formula $\{1\},\{2\}$
\end_inset 

 are recurrent.
\layout Standard

(b) We can find the stationary distribution of the Markov Chain defined
 by 
\begin_inset Formula $\{1,2\}$
\end_inset 

, 
\layout Standard


\begin_inset Formula $P=\left(\begin{array}{cc}
\frac{1}{3} & \frac{2}{3}\\
\frac{1}{3} & \frac{2}{3}\end{array}\right)$
\end_inset 

 and 
\begin_inset Formula $\pi=\pi P$
\end_inset 

 gives 
\begin_inset Formula $\pi_{1}=1/3$
\end_inset 

 and 
\begin_inset Formula $\pi_{2}=2/3$
\end_inset 

.
 Hence 
\begin_inset Formula $E(N|X_{0}=2)=3/2$
\end_inset 

.
\layout Standard

(c) Let 
\begin_inset Formula $Y$
\end_inset 

 be the number of transitions until we leave 0.
 
\begin_inset Formula $P(Y=y)=\frac{2}{3}\left(\frac{1}{3}\right)^{y-1}$
\end_inset 

 for 
\begin_inset Formula $y=1,2,3...$
\end_inset 

 i.e.
 
\begin_inset Formula $Y\sim Geom(2/3)$
\end_inset 

.
 Hence 
\begin_inset Formula $E(Y)=3/2$
\end_inset 

.
\layout Standard

2) Not necessary
\layout Section*

Problem 5
\layout Standard

a) State MCT and DCT
\layout Standard

b) (i) 
\begin_inset Formula $\int_{1}^{\infty}I_{[n,n+1]}(x)dx=\int_{n}^{n+1}dx=1$
\end_inset 

 
\begin_inset Formula $\forall n$
\end_inset 


\layout Standard

(ii) Since 
\begin_inset Formula $\left|\frac{sin(nx)}{e^{nx}x^{2}}\right|\leq\frac{1}{e^{x}x^{2}}$
\end_inset 

 whic is integrable, DCT gives 
\begin_inset Formula $\int_{1}^{\infty}\frac{sin(nx)}{e^{nx}}\frac{1}{x^{2}}dx\rightarrow\int_{1}^{\infty}\frac{1}{x^{2}}lim\frac{sin(nx)}{e^{nx}}dx=0$
\end_inset 


\layout Standard

(iii) 
\begin_inset Formula $\int_{1}^{\infty}x/ndx=\frac{1}{n}\int_{1}^{\infty}xdx=\infty$
\end_inset 

 
\begin_inset Formula $\forall n$
\end_inset 


\layout Standard

(iv) For 
\begin_inset Formula $n$
\end_inset 

 odd, 
\begin_inset Formula $-\int_{1}^{\infty}x/n^{2}dx=-\frac{1}{n^{2}}\infty=-\infty$
\end_inset 

 
\begin_inset Formula $\forall n$
\end_inset 

, and for 
\begin_inset Formula $n$
\end_inset 

 even 
\begin_inset Formula $\int_{1}^{\infty}x/n^{2}dx=\frac{1}{n^{2}}\infty=\infty$
\end_inset 

 
\begin_inset Formula $\forall n$
\end_inset 

, so the limit doesn't exist.
\layout Standard

(v) The function inside the integral is positive and decreasing with n,
 with its limit being 0 so for 
\begin_inset Formula $n\geq2$
\end_inset 


\layout Standard


\begin_inset Formula $\left|\frac{(1+nx^{2})}{(1+x^{2})^{n}}\right|<\frac{1+2x^{2}}{(1+x^{2})^{2}}$
\end_inset 

 which is integrable since the degree of the denominator is 4 and for the
 numerator it's 2.
 Hence DCT applies to give 
\begin_inset Formula $\int_{1}^{\infty}\frac{(1+nx^{2})}{(1+x^{2})^{n}}dx\rightarrow\int_{1}^{\infty}lim\frac{(1+nx^{2})}{(1+x^{2})^{n}}dx=0$
\end_inset 

.
\layout Section*

Problem 6
\layout Standard

a) 
\begin_inset Formula $Y_{n}$
\end_inset 

 is a function of 
\begin_inset Formula $X_{1}...X_{n}\Rightarrow\sigma(X_{1}...X_{n})$
\end_inset 

 measurable.
 Some algebra shows that 
\begin_inset Formula $E(S_{n+1}^{4}|\sigma(X_{1}...X_{n}))=S_{n}^{4}+6S_{n}^{2}+1$
\end_inset 

, 
\begin_inset Formula $E(S_{n+1}^{2}|\sigma(X_{1}...X_{n}))=S_{n}^{2}+1$
\end_inset 

.
 Thus 
\begin_inset Formula $E(Y_{n+1}|\sigma(X_{1}...X_{n}))=$
\end_inset 


\begin_inset Formula $E(S_{n+1}^{4}-6(n+1)S_{n+1}^{2}+3(n+1)^{2}+2(n+1)|\sigma(X_{1}...X_{n}))$
\end_inset 


\begin_inset Formula $=...=S_{n}^{4}-6nS_{n}^{2}+3n^{2}+2n=Y_{n}$
\end_inset 

.
\layout Standard

Hence 
\begin_inset Formula $\{ Y_{n}\}$
\end_inset 

is a martingale.
\layout Standard

b) The optional stopping theorem gives that 
\begin_inset Formula $Y_{o}$
\end_inset 

, 
\begin_inset Formula $Y_{\nu}$
\end_inset 

 is a martingale so in particular 
\begin_inset Formula $E(Y_{\nu})=E(Y_{o})=0$
\end_inset 

.
 Since 
\begin_inset Formula $E(Y_{\nu})=E(S_{\nu}^{4}-6\nu S_{\nu}^{2}+3\nu^{2}+2\nu)=...=-5a^{4}+2a^{2}+3E(\nu^{2})$
\end_inset 

, we get that
\layout Standard


\begin_inset Formula $E(\nu^{2})=\frac{5a^{4}-2a^{2}}{3}$
\end_inset 

 and 
\begin_inset Formula $V(\nu)=\frac{2}{3}a^{2}(a^{2}-1)$
\end_inset 

.
\the_end