Row and Column Conventions for Initial Distributions

Why the Vector Convention Matters

A state model describes the probabilities of occupying its states over time. Those probabilities can be written as a row vector or a column vector. Either choice is valid, but it determines which side of the intensity matrix the distribution multiplies. A mismatch can produce an incorrect rate of change even when every entry of the intensity matrix is correct.

In this tutorial, the intensity matrix uses the row-source convention: row \(i\) describes transitions out of state \(i\), and its off-diagonal entry in column \(j\) is the intensity from state \(i\) to state \(j\). This convention does not change when the distribution is written as a column vector. Instead, the corresponding equation uses the transpose of the intensity matrix.

Definition: For a finite-state continuous-time Markov model with states in a declared order, an initial distribution is a vector of the probabilities of occupying each state at the initial time. Its entries are nonnegative and sum to 1. A row distribution is written horizontally; a column distribution is written vertically. The orientation is a notation choice and does not change the state model.

Let \(Q(t)=(q_{ij}(t))\) be the intensity matrix at time \(t\), with \(q_{ij}(t)\) the intensity of a transition from state \(i\) to state \(j\) for \(i\ne j\). Its diagonal entries are the negatives of the total intensities out of the corresponding states:

q_{ii}(t)=-\sum_{j\ne i}q_{ij}(t).

Thus each row of \(Q(t)\) sums to zero. For a row distribution \(\mathbf p(t)\), the forward equation is \(\mathbf p'(t)=\mathbf p(t)Q(t)\). For a column distribution \(\mathbf c(t)\), the same model is written \(\mathbf c'(t)=Q(t)^{\mathsf T}\mathbf c(t)\). The superscript \({\mathsf T}\) means transpose.

Key convention: With a row-source intensity matrix, a row distribution multiplies \(Q(t)\) on the left: \(\mathbf p'(t)=\mathbf p(t)Q(t)\). A column distribution is multiplied by \(Q(t)^{\mathsf T}\): \(\mathbf c'(t)=Q(t)^{\mathsf T}\mathbf c(t)\). Do not change the source-row interpretation of \(Q(t)\) just because the distribution is a column.

How the Multiplication Gives the Rate of Change

The derivative of a state probability is its instantaneous rate of change, not a probability itself. For example, a negative derivative for one state means that its probability is decreasing at that instant. The component equations express a balance of inflow and outflow.

Identify inflows to state \(j\).
For each other state \(i\), the probability currently in \(i\) contributes at rate \(p_i(t)q_{ij}(t)\) to state \(j\).

Subtract outflow from state \(j\).
The diagonal entry \(q_{jj}(t)\) is the negative total rate of leaving \(j\), so \(p_j(t)q_{jj}(t)\) accounts for that outflow.

Combine the contributions.
The rate for state \(j\) is \(\sum_i p_i(t)q_{ij}(t)\), the \(j\)th component of \(\mathbf p(t)Q(t)\).

This also justifies the column formula. A column vector is the transpose of its corresponding row vector: \(\mathbf c(t)=\mathbf p(t)^{\mathsf T}\). Transposing the row equation gives

\mathbf c'(t) =\big(\mathbf p(t)Q(t)\big)^{\mathsf T} =Q(t)^{\mathsf T}\mathbf p(t)^{\mathsf T} =Q(t)^{\mathsf T}\mathbf c(t).

The transpose appears because transposition reverses the order of a matrix product. It is not a different model or a different definition of the transition intensities.

Worked Calculations With a Row Distribution

Worked Example: Initial Rates in a Three-State Model

Method: Use the row-distribution forward equation, and check that the three component rates sum to zero. Consider a hypothetical model with states \(A\), \(B\), and \(D\), in that order. State \(D\) is absorbing. The constant intensities are per year, and the initial probabilities are \(0.50\), \(0.30\), and \(0.20\), respectively. Find the initial rate of change of each state probability and a first-order estimate of the distribution after \(0.10\) year.

The intensity matrix and initial row distribution are

Q= \begin{pmatrix} -0.12&0.08&0.04\\ 0.03&-0.10&0.07\\ 0&0&0 \end{pmatrix}, \qquad \mathbf p(0)=(0.50,\ 0.30,\ 0.20).

The rows sum to zero: \(-0.12+0.08+0.04=0\), \(0.03-0.10+0.07=0\), and the absorbing row is all zero. Apply \(\mathbf p'(0)=\mathbf p(0)Q\), calculating one component at a time:

\begin{aligned} p_A'(0)&=0.50(-0.12)+0.30(0.03)+0.20(0)\\ &=-0.060+0.009=-0.051,\\ p_B'(0)&=0.50(0.08)+0.30(-0.10)+0.20(0)\\ &=0.040-0.030=0.010,\\ p_D'(0)&=0.50(0.04)+0.30(0.07)+0.20(0)\\ &=0.020+0.021=0.041. \end{aligned}

The initial rate vector is therefore \((-0.051,\ 0.010,\ 0.041)\) per year. Its entries sum to zero: \(-0.051+0.010+0.041=0\), as required because the total probability remains 1. Using the first-order approximation \(\mathbf p(0.10)\approx\mathbf p(0)+0.10\mathbf p'(0)\) gives

\mathbf p(0.10)\approx (0.50,\ 0.30,\ 0.20) +0.10(-0.051,\ 0.010,\ 0.041) =(0.4949,\ 0.3010,\ 0.2041).

These approximate probabilities sum to \(1.0000\). This is a first-order approximation, not an exact calculation of the distribution at \(0.10\) year. Common mistake avoided: multiplying the row distribution by \(Q^{\mathsf T}\) would apply the wrong equation for the declared row-vector convention.

Worked Calculations With a Column Distribution

Worked Example: Applying the Transpose to a Column Distribution

Method: Keep the intensity matrix in row-source form and multiply its transpose by the column distribution. A hypothetical contract model has states \(A\), \(B\), and \(D\), in that order, with \(D\) absorbing. The constant intensities are per year. At time zero the probabilities are \(0.40\), \(0.35\), and \(0.25\). Find the initial rate vector and the initial rate for state \(D\).

The model data are

Q= \begin{pmatrix} -0.25&0.20&0.05\\ 0.10&-0.30&0.20\\ 0&0&0 \end{pmatrix}, \qquad \mathbf c(0)= \begin{pmatrix} 0.40\\ 0.35\\ 0.25 \end{pmatrix}.

Since the distribution is a column, use \(\mathbf c'(0)=Q^{\mathsf T}\mathbf c(0)\), where

Q^{\mathsf T}= \begin{pmatrix} -0.25&0.10&0\\ 0.20&-0.30&0\\ 0.05&0.20&0 \end{pmatrix}.

The components are

\begin{aligned} c_A'(0)&=(-0.25)(0.40)+(0.10)(0.35)+0(0.25)=-0.100+0.035=-0.065,\\ c_B'(0)&=(0.20)(0.40)+(-0.30)(0.35)+0(0.25)=0.080-0.105=-0.025,\\ c_D'(0)&=(0.05)(0.40)+(0.20)(0.35)+0(0.25)=0.020+0.070=0.090. \end{aligned}

Thus \(\mathbf c'(0)=(-0.065,\ -0.025,\ 0.090)^{\mathsf T}\) per year. The rates sum to zero, and the rate of increase for \(D\) is the total inflow \(0.020+0.070=0.090\) per year. Common mistake avoided: using \(Q\mathbf c(0)\) does not match a column distribution with a row-source intensity matrix; the required product is \(Q^{\mathsf T}\mathbf c(0)\).

Changing the State Order

A distribution and its intensity matrix must use the same state order. Reordering states is allowed, but every row, every column, and every component of the distribution must be reordered consistently. If only the distribution is reordered, the resulting multiplication combines probabilities with the wrong states.

Worked Example: Reordering States Without Changing the Model

Method: First compute the rate vector in the original order, then reorder both the states and the matrix to verify the result. A hypothetical model has states \(A\), \(D\), and \(X\), in that original order; \(X\) is absorbing. Intensities are constant per year, and the initial column distribution in that order is \((0.50,\ 0.30,\ 0.20)^{\mathsf T}\). Find the rates in the new order \((X,A,D)\).

In the original order, the model is

Q= \begin{pmatrix} -0.18&0.12&0.06\\ 0.04&-0.14&0.10\\ 0&0&0 \end{pmatrix}, \qquad \mathbf c(0)= \begin{pmatrix} 0.50\\ 0.30\\ 0.20 \end{pmatrix}.

Use the column equation in the original order:

\begin{aligned} \mathbf c'(0)=Q^{\mathsf T}\mathbf c(0) &= \begin{pmatrix} -0.18&0.04&0\\ 0.12&-0.14&0\\ 0.06&0.10&0 \end{pmatrix} \begin{pmatrix} 0.50\\ 0.30\\ 0.20 \end{pmatrix}\\ &= \begin{pmatrix} -0.090+0.012\\ 0.060-0.042\\ 0.030+0.030 \end{pmatrix} = \begin{pmatrix} -0.078\\ 0.018\\ 0.060 \end{pmatrix}. \end{aligned}

The original-order rates are for \(A,D,X\). Therefore, simply expressing these same rates in the requested order \(X,A,D\) gives \((0.060,\ -0.078,\ 0.018)^{\mathsf T}\) per year.

To check by rebuilding the matrix, reorder its rows and columns using the index order \(3,1,2\). The reordered matrix and distribution are

Q_{\mathrm{new}}= \begin{pmatrix} 0&0&0\\ 0.06&-0.18&0.12\\ 0.10&0.04&-0.14 \end{pmatrix}, \qquad \mathbf c_{\mathrm{new}}(0)= \begin{pmatrix} 0.20\\ 0.50\\ 0.30 \end{pmatrix}.

Now \(Q_{\mathrm{new}}^{\mathsf T}\mathbf c_{\mathrm{new}}(0)=(0.060,\ -0.078,\ 0.018)^{\mathsf T}\), which agrees with the reordered original rates. The rates sum to zero. Common mistake avoided: changing the vector order without applying the same reordering to the intensity matrix would not represent the same model.

Time-Varying Intensities and Practical Checks

The orientation rule is the same when intensities vary with time. If the model has an intensity matrix \(Q(t)\), evaluate it at the time of interest and use the distribution at that same time:

\mathbf p'(t)=\mathbf p(t)Q(t), \qquad \mathbf c'(t)=Q(t)^{\mathsf T}\mathbf c(t).

In particular, at the initial time \(t=0\), the rate is found from \(Q(0)\) and the initial distribution. Do not substitute an intensity at a later time merely because the question asks about a later time interval; the instantaneous derivative at time \(0\) uses the rates at time \(0\).

Useful checks include the following:

Match dimensions. With \(n\) states, a row distribution has dimension \(1\times n\), \(Q\) has dimension \(n\times n\), and \(\mathbf pQ\) has dimension \(1\times n\). A column distribution has dimension \(n\times1\), and \(Q^{\mathsf T}\mathbf c\) has dimension \(n\times1\).
Check total probability. The components of the rate vector should sum to zero. For a row distribution this follows from the row sums of \(Q\) being zero; equivalently, the column sums of \(Q^{\mathsf T}\) are zero.
Interpret signs and flows. An absorbing state has no outflow, so its rate of change is its total inflow. A state can still have a negative rate even if it has inflows, when its outflow is larger.
Keep rates and probabilities distinct. A derivative is measured per unit time and may be negative. It is not an occupancy probability.

These conventions are aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus. In calculations, state the order of the states, identify whether the distribution is a row or column, and then write the matching multiplication before evaluating its components.

Check Your Understanding

For each problem, use the declared state order and vector convention. Rates are expressed per year.

A hypothetical model has states \(R,S,D\), with \(D\) absorbing, and \(Q=\begin{pmatrix}-0.20&0.15&0.05\\0.04&-0.14&0.10\\0&0&0\end{pmatrix}\). The initial row distribution is \((0.60,0.25,0.15)\). Find the initial rate vector. (Answer: \((-0.110,\ 0.055,\ 0.055)\) per year.)
For states \(A,B,D\), let \(Q=\begin{pmatrix}-0.16&0.10&0.06\\0.05&-0.20&0.15\\0&0&0\end{pmatrix}\), and let the initial column distribution be \((0.30,0.50,0.20)^{\mathsf T}\). Find the initial rate vector. (Answer: \((-0.023,\ -0.070,\ 0.093)^{\mathsf T}\) per year.)
In state order \(A,B,C\), let \(Q=\begin{pmatrix}-0.30&0.20&0.10\\0.05&-0.20&0.15\\0&0&0\end{pmatrix}\) and the initial row distribution be \((0.20,0.50,0.30)\). What is the initial rate of change of the probability of \(B\)? (Answer: \(-0.060\) per year.)
At time zero, a time-inhomogeneous model in state order \(A,B,D\) has \(Q(0)=\begin{pmatrix}-0.12&0.08&0.04\\0.03&-0.09&0.06\\0&0&0\end{pmatrix}\) and column distribution \((0.25,0.50,0.25)^{\mathsf T}\). Find the initial rate vector. (Answer: \((-0.015,\ -0.025,\ 0.040)^{\mathsf T}\) per year.)
A two-state model has row-source matrix \(Q=\begin{pmatrix}-0.24&0.24\\0.08&-0.08\end{pmatrix}\) and initial row distribution \((0.70,0.30)\). Use the initial rate to find the first-order approximation to the distribution after \(0.05\) year. (Answer: \((0.6928,\ 0.3072)\).)

What You'll Learn