Why the Multiplication Side Matters
For a continuous-time Markov chain, the backward and forward Kolmogorov equations place the generator on opposite sides of the transition matrix. This is not a choice of notation: each side comes from conditioning at a different end of the time interval. With row-stochastic transition matrices, the backward equation is \(P'(t)=QP(t)\), while the forward equation is \(P'(t)=P(t)Q\). The multiplication order matters for general matrices. For a homogeneous finite-state chain, however, these two particular products are equal; understanding why prevents a real distinction in derivation from becoming a false claim that the equations give different transition probabilities.
Let \((X_t)_{t\ge0}\) be a time-homogeneous continuous-time Markov chain on a finite state space \(S\), defined on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Write \(\mathcal{F}_t\) for its natural filtration. Its transition probabilities and transition matrix are \(p_{ij}(t)=\mathbb{P}(X_t=j\mid X_0=i)\) and \(P(t)=(p_{ij}(t))\). Rows are probability distributions, so each row of \(P(t)\) sums to \(1\), and an initial distribution \(\mu\), written as a row vector, evolves as \(\mu P(t)\).
The generator \(Q=(q_{ij})\) has nonnegative off-diagonal entries and rows summing to zero. In a finite-state chain its small-time expansion is \(P(h)=I+hQ+o(h)\) as \(h\downarrow0\), entry by entry. Thus \(q_{ij}\) for \(i\ne j\) is the instantaneous rate of moving from \(i\) to \(j\), and \(q_{ii}=-\sum_{j\ne i}q_{ij}\). The Chapman–Kolmogorov relation is \(P(s+t)=P(s)P(t)\). These definitions and the finite-state small-time expansion are the starting point for both equations.
Two Conditionings, Two Products
For the backward equation, split the elapsed time \(t+h\) into a short initial interval of length \(h\) followed by an interval of length \(t\). Chapman–Kolmogorov gives \(P(t+h)=P(h)P(t)\). Substituting the small-time expansion into this product gives \(P(t+h)=P(t)+hQP(t)+o(h)\). The generator therefore multiplies on the left.
For the forward equation, split the same elapsed time into an interval of length \(t\) followed by a short final interval of length \(h\). Then \(P(t+h)=P(t)P(h)\), so \(P(t+h)=P(t)+hP(t)Q+o(h)\). Now the generator multiplies on the right. Both factorizations are valid because the chain is time-homogeneous and has the semigroup property; the conditioning step determines which factor appears first.
Proof. For fixed \(t\), the two Chapman–Kolmogorov factorizations and the expansion \(P(h)=I+hQ+o(h)\) give the right-hand limits \((P(t+h)-P(t))/h\to QP(t)\) and \((P(t+h)-P(t))/h\to P(t)Q\). These limits are equal, so \(QP(t)=P(t)Q\). To verify differentiability from the left when \(t>0\), write \((P(t)-P(t-h))/h=P(t-h)(P(h)-I)/h\), which tends to \(P(t)Q\) by continuity of the finite-state transition semigroup and the small-time expansion. Thus the derivative exists and equals both products. At \(t=0\), the right derivative is \(Q\), since \(P(0)=I\). \(\square\)
The same conclusion follows directly from the two derivations: each product is the derivative of the same matrix-valued function \(P(t)\). The useful distinction is therefore about how each equation is obtained and which indices it sums over, not about two different derivatives for the same homogeneous chain.
Reading the Indices
Matrix multiplication makes the two conditioning arguments explicit. For states \(i,j\in S\), the backward equation reads \((QP(t))_{ij}=\sum_{k\in S}q_{ik}p_{kj}(t)\). It sums over the state \(k\) reached during the initial short interval: first take a rate out of \(i\), then a transition probability from \(k\) to \(j\) during the remaining time. The forward equation reads \((P(t)Q)_{ij}=\sum_{k\in S}p_{ik}(t)q_{kj}\). It sums over the state \(k\) occupied just before the final short interval: first reach \(k\), then take a rate from \(k\) to \(j\).
These expressions are not interchangeable for arbitrary matrices. Their equality here follows from the homogeneous semigroup, as the proof showed. In particular, \(Q\) need not be symmetric, and \(P(t)\) need not be symmetric. Symmetry is not the reason the two products agree.
Worked Example: A Row Distribution’s Instantaneous Change
Consider the state space \(S=\{1,2,3\}\) and generator $$ Q=\begin{bmatrix} -1&1&0\\ 0&-2&2\\ 3&0&-3 \end{bmatrix}. $$ Each row sums to zero: \(-1+1+0=0\), \(0-2+2=0\), and \(3+0-3=0\). The off-diagonal rates are nonnegative, so this is a valid finite-state generator. Suppose the initial distribution is \(\mu=(\tfrac12,\tfrac13,\tfrac16)\). Find its instantaneous change at time zero.
Since the distribution at time \(t\) is \(\mu P(t)\), the forward equation gives derivative \(\mu P'(0)=\mu Q\). Calculate each entry by multiplying the row vector by the corresponding column of \(Q\): $$ \begin{aligned} (\mu Q)_1&=\tfrac12(-1)+\tfrac13(0)+\tfrac16(3)=-\tfrac12+\tfrac12=0,\\ (\mu Q)_2&=\tfrac12(1)+\tfrac13(-2)+\tfrac16(0)=\tfrac12-\tfrac23=-\tfrac16,\\ (\mu Q)_3&=\tfrac12(0)+\tfrac13(2)+\tfrac16(-3)=\tfrac23-\tfrac12=\tfrac16. \end{aligned} $$ Thus the derivative is \((0,-\tfrac16,\tfrac16)\).
Check: its entries sum to \(0-\tfrac16+\tfrac16=0\), as required for the instantaneous change of a probability distribution to preserve total mass. The first-order distribution is \(\mu+t(0,-\tfrac16,\tfrac16)+o(t)\); the initial mass in state \(1\) has zero net instantaneous change because its outflow \(\tfrac12\cdot1\) is balanced by inflow \(\tfrac16\cdot3\).
Worked Example: A Generator and a Stochastic Matrix That Do Not Commute
Use the same valid generator \(Q\) and the stochastic matrix $$ R=\begin{bmatrix} 0&1&0\\ 1&0&0\\ 0&0&1 \end{bmatrix}. $$ Each row of \(R\) sums to \(1\): \(0+1+0=1\), \(1+0+0=1\), and \(0+0+1=1\). This matrix swaps states \(1\) and \(2\) and leaves state \(3\) fixed. It is a stochastic matrix, but it is not being asserted to be \(P(t)\) for the generator \(Q\). Compute both product orders.
Multiplying \(Q\) by \(R\), whose columns reorder the first two columns of \(Q\), gives $$ QR=\begin{bmatrix} 1&-1&0\\ -2&0&2\\ 0&3&-3 \end{bmatrix}. $$ Multiplying \(R\) by \(Q\) swaps the first two rows of \(Q\), giving $$ RQ=\begin{bmatrix} 0&-2&2\\ -1&1&0\\ 3&0&-3 \end{bmatrix}. $$ The two products differ; for example, their \((1,1)\) entries are \(1\) and \(0\). Each row of both products still sums to zero: the row sums of \(QR\) are \(1-1+0=0\), \(-2+0+2=0\), and \(0+3-3=0\); those of \(RQ\) are \(0-2+2=0\), \(-1+1+0=0\), and \(3+0-3=0\).
Check: \(QR\ne RQ\), so the order of multiplication cannot be discarded as a general matrix rule. There is no conflict with the Kolmogorov equations: the theorem says \(Q\) commutes with the transition matrix \(P(t)\) of its own homogeneous chain, not with every stochastic matrix \(R\).
Worked Example: A Single Entry in Both Equations
For the same three-state chain, consider the transition probability \(p_{12}(t)\). From the first row of \(Q\), the backward equation gives $$ p'_{12}(t)=(QP(t))_{12}=-p_{12}(t)+p_{22}(t). $$ This combines the rate \(-1\) on the diagonal with the rate \(1\) from state \(1\) to state \(2\). From the second column of \(Q\), the forward equation gives $$ p'_{12}(t)=(P(t)Q)_{12}=p_{11}(t)-2p_{12}(t). $$ Here the contributions are from being in state \(1\), which has rate \(1\) to state \(2\), and being in state \(2\), whose diagonal rate is \(-2\).
At time zero, \(P(0)=I\), so \(p_{12}(0)=0\), \(p_{22}(0)=1\), and \(p_{11}(0)=1\). The backward expression gives \(-0+1=1\); the forward expression gives \(1-2(0)=1\). Both calculate \(p'_{12}(0)=1\), matching \(q_{12}=1\). At later times, the theorem ensures the two expressions remain equal for this chain, even though their terms are organized differently.
Check: Substitution at zero gives \(p'_{12}(0)=1\) from each equation. This agrees with the generator’s rate of moving from state \(1\) to state \(2\).
Common Mistake and Practical Use
A common mistake is to say that the backward and forward equations are identical because matrix multiplication commutes. The calculation with \(Q\) and \(R\) disproves that general claim. Another mistake is to use the wrong factorization: writing \(P(t+h)=P(h)P(t)\) leads to \(QP(t)\), not \(P(t)Q\); writing \(P(t+h)=P(t)P(h)\) leads to \(P(t)Q\), not \(QP(t)\). Both are valid factorizations here, but each must be multiplied in its stated order.
For a nonhomogeneous chain, or for two different generators acting in sequence, the commutation argument need not apply; one cannot infer \(QP(t)=P(t)Q\) just from knowing that products of matrices are involved. For the homogeneous finite-state chain considered here, the semigroup property and small-time expansion establish both equations and their equality. This distinction is useful when checking calculations: row distributions evolve by multiplication on the right, while the forward equation describes how probability flows into each destination state. The backward equation instead describes the effect of the initial state's possible short-time moves on future transition probabilities.
Check Your Understanding
For each question, keep the row-vector convention and multiplication order explicit.
- Which short interval is used to derive the backward equation, and on which side does \(Q\) multiply?
- For the forward equation, what intermediate state is summed over in \((P(t)Q)_{ij}\)?
- Why does \(QP(t)=P(t)Q\) hold for a homogeneous finite-state chain even though \(QR\ne RQ\) in the worked example?
- For a row distribution \(\mu\), which product gives its instantaneous derivative at time zero?
- Using the displayed \(Q\), calculate \((QP(0))_{12}\) and \((P(0)Q)_{12}\) from \(P(0)=I\), and check that they agree.