In a stochastic process we have a mapping from a variable (time) to a random variable.

Time could be discrete, or continuous.

Temperature over time is a stochastic process, as is the number of cars sold each day.

The state space for temperature is continous, the number of people on the moon is discrete.

We can describe processes by their evolution.

\(p(x_t|x_{t-1}...)\)

\(AC(a,b)=cov(X_a, X_b)\)

The autocorrelation between two time periods is their covariance, normlised by their variances

\(AC(a,b)=\dfrac{E[(X_a-\mu_a)(X_b-\mu_b)]}{\sigma_a \sigma_b}\)

This is also called serial correlation.

For a process with the Markov property, only the current state matters for all probability distributions.

\(P(x_{t+n}|x_t)=P(x_{t+n}|x_t, x_{t-1}...)\)

For a process with the Martingale property, the expected value of all future variables is the current state.

This only restricts expectations.

\(E(X_{n+1}|X_0,...,X_n)=X_n\)

A process can be Markov, Martingale, neither or both.

Unconditional probabilities don’t change over time.

So GDP would not be stationary, but random noise would. A random walk is not stationary, because the variance increases over time.

How many differences to make it stationary?

Mean and autocovarinance don’t change over time.

All moments are the same.

Variables at each time are indepdendent.

How many diffs do you need to do to get a stationary process?

If something is first order integrated it is \(I(1)\).

If we can remove the trend as a function, eg linear or non-linear growth, and the rest is stationary, then the process is trend stationary

Sample moments must converge to generating momements. Not guaranteed.

Eg process with path dependence. 50 Generating average is £50, but sample will only convergen to £100 or £0

This shows the probability for moving between discrete states.

We can show the probability of being in a state by multiplying the vector state by the transition matrix.

\(Mv\)

For time-homogenous Markov chains the transition matrix is independent of time.

For these we can calculate the probability of being in any given state in the future:

\(M^nv\)

This becomes independent of v as we tend to infinity. The initial starting state does not matter for long term probabilities.

How to find steady state probability?

\(Mv=v\)

The eigenvectors! With associated eigenvector \(1\). There is only one eigenvector. We can find it by iteratively multiplying any vector by \(M\).

If we have multiple variables, we can explore the order of integration of linear combinations.

If two series have time trends, a linear combination of them could remove this.

\(Cov(x_{it},u_{it})=0\)

\(Cov (x_{is}, u_{it})=0)\)

This is stronger than contemporeous, all periods.

Shocks don’t affect future outcomes.

Sequential exogeneity: a bit looser than strict exogeneity. only holds when \(s\le t\).

So shocks can affect, but only in future.