UMD Probability Qualifying Exams/Jan2015Probability

Solution
(a) Let $$X_n$$ be the value of the nth roll. The number of (6,6)'s in the first 1000 rolls is given by $$ \sum_{n=1}^{999} \mathbb{I}(X_n=6 \text{ and } X_{n+1}=6 )$$

By linearity of expectation,

$$ E[\sum_{n=1}^{999} \mathbb{I}(X_n=6 \text{ and } X_{n+1}=6)]=\sum_{n=1}^{999} E[\mathbb{I}(X_n=6 \text{ and } X_{n+1}=6)]$$

$$=\sum_{n=1}^{999} \mathbb{P}(X_n=6 \text{ and } X_{n+1}=6)=999/36 $$

Similarly, the number of (1,2)'s in the first 1000 rolls is 999/36.

(b) For the general approach to solve such problems see [1]. The expected waiting time till the first (6,6) is 6+36=42, while the expected waiting time till the first (1,2) is 0+36=36.

The argument is as follows. Suppose we continue rolling dice until we hit the desired pattern. At each time n a new gambler begins a game with 1 coin. On a gambler's i-th bet, he bets the i-th character of the pattern will occur. If the gambler loses a bet, he forfeits all his coins. Otherwise he has 6 times as many as he started with. This is a martingale, so at the end of the game, by the optional stopping theorem the expected total sum of all gamblers' coins is equal to the number of gamblers who joined the game, i.e. the expected waiting time for the pattern.

For the pattern (6,6), the second to last gambler has 6 coins and the last gambler has 36 coins. The others have 0.

For the pattern (1,2), the last gambler has 36 coins. The others have 0.


 * Note the optional stopping theorem applies because the expected value of the stopping time is finite and the martingale increments are bounded.
 * A more direct way to do this problem is with absorbing Markov chains

(c) Since the patterns do not overlap, the odds that (6,6) occurs first are given by the expected waiting times, 36 : 42. That is the probability (6,6) occurs first is 36/(42+36)=6/13

For patterns which do overlap, the analysis is slightly more complicated, see Corollary 3.2 in the Shuo-Yen Li paper [1].

[1] A Martingale Approach to the Study of Occurrence of Sequence Patterns in Repeat Experiments, Shuo-Yen Li, Annals of Probability

Solution
(a) True. A random variable is discrete iff its range is countable. The range of the sum of two random variables A, B is a subset of the sum-set range(A)+range(B). The sum-set is countable if range(A), and range(B) are countable.

(b). True. A random variable X is absolutely continuous iff $$\mathbb{P}(\text{X in E})=0$$ for any Lebesgue-null set E.  Let Y be a discrete random variable. $$\mathbb{P}(X+Y\text{ in E}) \le \mathbb{P}(X\text{ in E}-\text{range(Y)})=0$$ since a countable sum of Lebesgue-null sets is Lebesgue-null.

(c). False. For any absolutely continuous random variable X, -X is also absolutely continuous, and X + (-X) =0 which is a discrete random variable.

(d). True, by (a)

(e). True, by (b)

(f). True, the distribution function is given by the convolution theorem.

Solution
(a). Let $$\xi_i=\pm a_i X_i$$ where $$X_i$$ is uniformly distributed in $$[0,1]$$.

$$ \log |\eta_n| = \sum_{i=1}^n{(\log a_i + \log X_i})$$

$$ E[\log X_i] = \int_0^1 \log x dx = -1 $$

By the strong law off large numbers,

$$\frac{1}{n} \sum_{i=1}^n\log X_i $$ converges to -1 almost surely.

$$\lim \sup_{n \rightarrow \infty} a_n \le 2 \implies \lim \sup_{n \rightarrow \infty} \frac{1}{n}\sum_{i=1}^n \log a_i \le \log 2 <1 $$.

So $$\log|\eta_n|$$ converges to $$-\infty$$ a.s. So $$\eta_n$$ converges to 0 a.s.

(b). In this case

$$\lim \sup_{n \rightarrow \infty} \frac{1}{n}\sum_{i=1}^n \log a_i =\infty$$

So $$\log|\eta_n|$$ converges to $$\infty$$ a.s. So $$|\eta_n|$$ converges to $$\infty$$ almost surely. However the sign of $$\eta_n$$ changes infinitely often a.s. (by the second Borel-Cantelli lemma), so $$\eta_n$$ has no limit almost surely.

Solution
(a) Let $$\tau$$ be the stopping time given by $$\min(s,u)$$ where u is the first time $$M_t=\lambda$$. By the optional stopping theorem,

$$M_0=E[M_\tau]=\mathbb{P}(\sup_{0\le t \le s}M_t \ge \lambda | \mathcal{F}_0)\max(\lambda,M_0)+\mathbb{P}(\sup_{0\le t \le s}M_t < \lambda | \mathcal{F}_0)E[M_{s}|\sup_{0\le t \le s}M_t < \lambda, \mathcal{F}_0]$$

Since $$M_t$$ converges to zero a.s., taking the limit as $$s$$ goes to infinity we have $$\mathbb{P}(\sup_{t\ge0} M_t \ge \lambda | \mathcal{F}_0)=M_0/\lambda$$ if $$\lambda>M_0$$.

Clearly, if $$M_0 \ge \lambda$$, $$\mathbb{P}(\sup_{t\ge0} M_t \ge \lambda | \mathcal{F}_0)=1$$.

(b) Wald's martingale $$X_t=e^{aW_t-\frac{1}{2}a^2t}$$ is a well-known martingale.

$$\mathbb{P}(\sup_{t\ge0} (W_t - \frac{1}{2}at) \ge s)=\mathbb{P}(\sup_{t\ge0} (aW_t - \frac{1}{2}a^2t )\ge as)$$

$$=\mathbb{P}(\sup_{t\ge0} e^{(aW_t - \frac{1}{2}a^2t)} \ge e^{as})=\mathbb{P}(\sup_{t\ge0} X_t \ge e^{as})$$

By (a),

$$\mathbb{P}(\sup_{t\ge0} X_t \ge e^{as})=\min(1,e^{-as})=e^{-as}$$

Solution
Suppose $$|\xi(x)|0$$

$$E[\xi^2]=\mathbb{P}(\xi=f_{\epsilon})E[\xi f_{\epsilon}|\xi=f_{\epsilon}]+\mathbb{P}(\xi \ne f_{\epsilon})E[\xi^2|\xi \ne f_{\epsilon}]$$

$$E[\xi f_{\epsilon}]=\mathbb{P}(\xi=f_{\epsilon})E[\xi f_{\epsilon}|\xi=f_{\epsilon}]+\mathbb{P}(\xi \ne f_{\epsilon})E[\xi f_\epsilon|\xi \ne f_{\epsilon}]$$

$$=C-\mathbb{P}(\xi \ne f_{\epsilon})E[\xi^2|\xi \ne f_{\epsilon}]+\mathbb{P}(\xi \ne f_{\epsilon})E[\xi f_\epsilon|\xi \ne f_{\epsilon}]$$

$$>C-2\epsilon B^2$$

Taking $$\epsilon 0$$. Partition the circle into $$\lfloor C n^2 \rfloor$$ non-intersecting arcs of equal length. Assume $$\lfloor C n^2 \rfloor>>n$$ The probability that n points are selected at random and at least two fall into the same arc is given by

$$1-\prod_{i=0}^{n-1}(\lfloor C n^2 \rfloor-i)/(\lfloor C n^2 \rfloor)$$

$$1-\prod_{i=0}^{n-1}(1-i/(\lfloor C n^2 \rfloor)) \approx 1-e^{-1/2C}$$

Taking C sufficiently small, we can ensure

$$e^{-1/2C}<\epsilon$$

If two points fall into the same arc, the distance between them is less than

$$2\pi/\lfloor C n^2 \rfloor$$ and $$n\xi_n<\frac{2\pi n}{\lfloor C n^2 \rfloor}$$

which tends to 0. Since $$\epsilon$$ may be taken to be arbitrarily small, we get that $$n\xi_n$$ tends to zero in probability.