Suppose R 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?