Suppose R (Python) can only generate uniform random variables. How can you take advantage of this and generate Poisson random variables?
Use the inverse transform method to generate an exponential random variable
Inscribe a circle in a square. Estimate the value of \(\pi\) by computing the ratio of the number of times a uniformly chosen point on the square ends up in the circle.
Using an exponential random variable, generate a normal random variable that is conditioned to be be positive; from here, adjust this result to get a normal random variable.
Let \(X\) and \(Y\) be Poisson random variables with means \(\lambda > \mu\). Show that \[d_{TV}(X, Y) \leq 2 (1-\exp(\mu-\lambda))\]
Let \(P\) be transition matrix on a state space \(S\), and \(\pi\) be a probability measure on \(S\). We say that \(\pi\) is reversible for \(P\) if \(\pi_i p_{ij} = \pi_j p_{ji}\).
Check that the stationary distribution from a random walk on a finite graph is reversible.
Let \(P\) be a transition matrix on a state space \(S\). Check that if \(\pi\) is reversible, then it is stationary.
Let \(\pi\) be a reversible distribution for the transition matrix \(P\) on a state space \(S\). Let \(X\) be Markov chain with transition matrix \(P\), that is started at \(\pi\). Let \(a, b,c,d \in S\). Show that \[ \mathbb{P}(X_0=a,X_1=b, X_2=c, X_3=d) = \mathbb{P}(X_0=d,X_1=c,X_2=b, X_3=a).\]
Suppose that I have \(n=52\) cards, arranged in some initial order. Consider the following procedure: I choose two cards with probability \(1/ {n \choose 2}\), and then change their position; repeat.
Describe this procedure as a Markov chain. Can you code it?
With this procedure, can you get from an initial ordering to any other ordering? Why?
Do you think it has a stationary distribution? If, so, what is it?
Consider a Markov chain on \(15\) states \(1,2,3,4, \ldots, 15\). Suppose that the probability of going from state \(i\) to \(i+1\) is \(0.9\), for all \(1 \leq i \leq 15\) and the probability of going from state \(i\) to \(i-1\) is \(0.1\) for all \(2 \leq i \leq 15\); in the case that \(i =1\), we allow the possibility of going to state \(15\) with probability \(0.1\), and if \(i=15\), the probability of going to state \(1\) is \(0.9\). Do you think this Markov chain has a stationary distribution; if so, what is it? Is this Markov chain reversible? Explain.