Weather Models with Exact Transition Probabilities

From Weather Probabilities to a Chain

A weather model is useful only when its states and probabilities match the question being asked. If the states are “sunny,” “cloudy,” and “rainy,” the model treats those categories as the complete set of possibilities for each day. Its transition probabilities then describe the weather tomorrow conditional on the weather today. Once the conditional probabilities are organized consistently, a matrix calculation gives the forecast even when today’s weather is uncertain.

Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, and let \((X_n)_{n\ge 0}\) be a discrete-time process whose value records the weather on day \(n\). Let \(S\) be a finite set of weather states, and let \(\mathcal{F}_n=\sigma(X_0,\ldots,X_n)\) be the information available through day \(n\). In a homogeneous Markov weather model, the conditional law of tomorrow given the weather history depends only on today and does not change with \(n\). In particular, for a state \(i\) with \(\mathbb{P}(X_n=i)>0\), \(p_{ij}=\mathbb{P}(X_{n+1}=j\mid X_n=i)\). More fully, the Markov assumption says \(\mathbb{P}(X_{n+1}=j\mid\mathcal{F}_n)=p_{X_nj}\) almost surely. Conditional probabilities given an event are defined here only when that event has positive probability; a model may nevertheless specify its transition probabilities for every state, including states that are not reached from its chosen initial law.

State definitions matter. The categories must be mutually exclusive, so a day cannot count as both “sunny” and “rainy” under the model’s coding. They must also cover the outcomes being modelled. If the categories omit a possible weather condition, the transition probabilities need an additional state or a clearly stated convention for assigning that condition. The numerical probabilities are conditional probabilities in a particular direction: the row describes tomorrow given today, not today given tomorrow.

Homogeneity means that the same transition probabilities are used on each day; it does not mean that the weather distribution is unchanged from day to day. In particular, we do not assume the initial distribution is stationary. The immediate task here is a one-step prediction: determine the distribution of \(X_{n+1}\) from the distribution of \(X_n\).

One-Step Prediction

Proof. Fix a state \(j\in S\). The events \(\{X_n=i\}\), for \(i\in S\), are disjoint and together cover the probability space. By the law of total probability and the Markov property,

If \(\mathbb{P}(X_n=i)=0\), its contribution is zero, so no conditional probability on that zero-probability event is needed for this calculation. The \(j\)-th entry of the row-vector product \(\mu_nP\) is precisely \(\sum_{i\in S}\mu_n(i)p_{ij}\). This holds for every \(j\), proving that \(\mu_{n+1}=\mu_nP\). \(\square\)

Proof. For each \(j\), \((\mu P)_j=\sum_i\mu_i p_{ij}\ge0\), because every factor is nonnegative. Summing over all destination states gives

The rearrangement is valid because the state space is finite. The penultimate equality uses the fact that each row of \(P\) sums to \(1\), and the final equality uses that \(\mu\) is a probability distribution. Thus \(\mu P\) has nonnegative entries summing to \(1\). \(\square\)

Worked Weather Models

Reading the Calculation Correctly

Each forecast entry combines all possible weather states today. For example, the probability of sunny weather tomorrow is the sum, over today’s states \(i\), of “probability of \(i\) today” multiplied by “probability of sunny tomorrow given \(i\) today.” This is why a row-vector initial distribution multiplies \(P\) on the left: the \(j\)-th forecast entry collects the contributions \(\mu_i p_{ij}\) from all starting states \(i\).

Common mistake: multiplying in the wrong order. In the uncertain-weather example, treating \(\mu\) as a column vector and calculating \(P\mu\) gives the entries \(\tfrac{13}{36}\), \(\tfrac38\), and \(\tfrac13\). Their sum is \(\tfrac{13}{36}+\tfrac38+\tfrac13=\tfrac{26}{72}+\tfrac{27}{72}+\tfrac{24}{72}=\tfrac{77}{72}\), which is greater than \(1\). That cannot be a probability distribution. The correction is to use the stated row-vector convention: \(\mu P=(\tfrac{13}{40},\tfrac{77}{180},\tfrac{89}{360})\), whose entries sum to \(1\). The row sums of \(P\) are what ensure that multiplying a probability row vector on the left gives another probability row vector.

A second error is to reverse the conditioning. A matrix entry in the sunny row and rainy column means “probability of rain tomorrow given sun today.” It does not mean “probability of sun today given rain tomorrow.” Those are different conditional probabilities and generally have different values. For a one-day weather forecast, identify the state today first, then use that state’s row—or, if today is uncertain, weight each row by today’s probability of being in that state.