LTTC-mock
Choose four out of the five questions.
Probability spaces and the generation of random variables
Define an explicit function \(\phi: [0,1] \to [0,1] \times \{1,2\}\) such that:
If \(U\) is uniformly distributed in \([0,1]\), then \(\phi(U)\) has the same distribution as \((V,X)\), where \(V\) and \(X\) are independent and uniformly distributed on \([0,1]\) and \(\{1,2\}\), respectively;
There exists a set \(A \subset [0,1]\), with \(\mathbb{P}(U \in A)=1\), such that \(\phi\) is injective on \(A\).
Solution
Let \(b:[0,1] \to \{0,1\}^{\mathbb{Z}^{+}}\) be so that \(b(u)\) is the binary expansion of \(u \in [0,1]\) this is bijective modulo a countable set. We know that \(b(U)\) is a sequence of iid Benroulli \(\tfrac{1}{2}\) random variables, from which we define \[\phi(u) = (b^{-1}[ b(u)_{n=2} ^{\infty}], b(u)_1).\]
Borel-Cantelli
Let \((X_i)_{i=2} ^{\infty}\) be i.i.d. continuous random variables with probability density function given by \[ f(x) = \frac{1}{2} |x|^3 e^{-x^4/4}.\] Find a deterministic sequence \((c_i)_{i=2}^{\infty}\) such that almost surely \[\limsup_{n\to \infty} \frac{X_n}{c_n} = 1.\]
Solution
For \(x\geq0\), we easily see that \[\mathbb{P}(X_n > x) = \frac{1}{2}e^{-x^4/4}.\] Take \[c_n = (4 \log n)^{\tfrac{1}{4}}.\] Will we show using the first Borel-Cantelli lemma that almost surely, \[\limsup_{n\to \infty} \frac{X_n}{c_n} \leq 1\] and then we will show using the second Borel-Cantelli lemma and the independence of the \(X_i\) that almost surely \[ \limsup_{n \to \infty} \frac{X_n}{c_n} \geq 1.\] Let \(\epsilon \geq 0\). We have that \[ \mathbb{P}( X_n/c_n > 1+ \epsilon) = \frac{1}{2 n^{ (1+\epsilon)^4}}=:a_n.\] The sequence \(a_n\) is summable for \(\epsilon>0\), so the upper bound follows by from the first Borel-Cantelli lemma. For the lower bound, if we set \(\epsilon=0\), then \(a_n\) is not summable, and then the lower bound follows from the second Borel-Cantelli lemma.
Markov chains
Prove that for an irreducible aperiodic Markov chain on a finite state space \(S\), we have that for each \(s \in S\), the return time
\[T := \inf\{n \geq 1: X_n=s\}\] has finite expectation, regardless of the starting distribution of the chain.
Solution
We go back to basics; in particular, some observations that were made in connection to the Doeblin coupling. If \(P\) is the transition matrix, then by the assumptions of irreducibility and aperiodicity, we know that \(P^M\) has all positive entries for some \(M>0\); let \(\delta >0\) be the smallest entry. Hence every \(M\) steps, we have a non-zero probability \(\delta >0\) of getting to the state \(s\). Thus \[ \mathbb{P}(T > kM) \leq (1- \delta)^k\] from which we easily deduce that \(T\) has finite expectation.
Poisson processes
Suppose that \(\Pi\) is a Poisson process on \(\mathbb{R}^d\). Let \(t \in \mathbb{R}^d\), and \(c \in (0, \infty)\) Prove that:
The translated point process \(\Gamma:=t + \Pi\) given by translating each point of \(\Pi\) by \(t\) is still a Poisson point process on \(\mathbb{R}^d\).
The scaled point process \(\Sigma := c\Pi\) given by multiplying each point of \(\Pi\) by \(c\) is still a Poisson point process on \(\mathbb{R}^d\).
Solution
Suppose that \(\Pi\) is a Poisson point process of intensity \(\lambda\). Then \(\Pi\) satisfies:
\(\Pi(A)\) is a Poisson random variable with mean \(\lambda |A|\) for every set \(A\) of finite volume.
The random variables \(\Pi(A_1), \ldots, \Pi(A_n)\) are independent for pairwise disjoint sets \(A_1, \dots, A_n\).
It suffices to verify these properties for \(\Gamma\) and \(\Sigma\). For a subset \(A \subset \mathbb{R}^d\), let \(t + A = \{t+a: a \in A\}\) and \(cA = \{ca: a \in A\}\).
Observe that \(\Gamma(A) = \Pi(-t +A)\), so that the two required properties are easily verified.
Similarly \(\Sigma(A) = \Pi(c^{-1}A)\), and \(\Sigma\) is a Poisson point process of intensity \(c^{-1}\lambda\).
Estimating the stationary distribution
Suppose you are given the output of a \(100000\) steps of a irreducible and aperiodic finite state Markov chain. Carefully explain how you could estimate the stationary distribution for this Markov chain, and why you estimator is reasonable.
Import the data from the file markovchain.txt and use this data and your method above to estimate the stationary distribution.
Solution
- Since the chain is on a finite state space and irreducible and aperiodic, it has a unique stationary distribution \(\pi\), and we know that a version of the law of large numbers gives that for every state \(s\), we have
\[\frac{1}{n} \sum_{i=1}^n \mathbf{1}[X_i = s] \to \pi(s).\] Thus for each state, we simply need to count the number of occurrences and divide by \(100000\), to get its approximate probability under \(\pi\).
- A quick scan of the file shows there are \(4\) states: \(1,2,3,4\). With R, we have:
z =read.table("markovchain.txt", sep=",")
z= z[,]
b1 = seq(-1,5, by=1)
hist(z, prob=TRUE, breaks=b1)
sum(z==1)/100000## [1] 0.17023sum(z==2)/100000## [1] 0.24957sum(z==3)/100000## [1] 0.41172sum(z==4)/100000## [1] 0.16849