Holding-Time Survival and No-Jump Probabilities

Introduction

In a continuous-time state model, an individual can remain in a state for some period and then make a transition out of it. The probability of remaining in that state without any transition depends on the total intensity of all possible exits during the period. This is a holding-time survival probability: it describes the time until the first exit from the state, not necessarily the state occupied at a later endpoint.

The key calculation is to add the exit intensities and integrate that total rate over the duration. When the total intensity is constant, the result reduces to a familiar exponential expression. When the intensity changes over time, the integral preserves the effect of that changing rate.

Definition: Suppose a process is in state \(i\) at time \(t\). Let \(T_i\) be the time from \(t\) until its first transition out of state \(i\). The holding-time survival probability for duration \(u\) is \(P(T_i>u\mid X(t)=i)\), interpreted as the probability that no transition out of state \(i\) occurs during the next \(u\) time units.

Total Exit Intensity and the No-Transition Formula

Write \(\mu_{ij}(r)\) for the transition intensity from state \(i\) to a different state \(j\) at time \(r\). For the formula below, assume these intensities describe the process while it remains in state \(i\), are nonnegative and integrable over the interval, and the process has no infinite accumulation of transitions in a finite time. The total exit intensity from \(i\) at time \(r\) is

\lambda_i(r)=\sum_{j\ne i}\mu_{ij}(r).

Only transitions out of \(i\) are included in this sum. The intensity of a transition from a state to itself is not an exit and is not included. If there is only one possible exit, the total exit intensity is just the intensity for that exit.

Theorem: Conditional on being in state \(i\) at time \(t\), the probability of no transition out of \(i\) over the next \(u\) time units is

P(T_i>u\mid X(t)=i) = \exp\left(-\int_t^{t+u}\lambda_i(r)\,dr\right) = \exp\left(-\int_0^u\lambda_i(t+s)\,ds\right).

The second form changes the integration variable to elapsed time \(s\). It is often convenient when the intensity is specified as a function of time since the starting time. If the process starts at time zero, the formula becomes \(\exp(-\int_0^u\lambda_i(s)\,ds)\).

Why the Exponential Formula Holds

Let \(S(u)=P(T_i>u\mid X(t)=i)\). At elapsed time \(u\), conditional on having remained in state \(i\) so far, the probability of an exit during a short additional interval of length \(h\) is \(\lambda_i(t+u)h+o(h)\). Therefore the probability of no exit during that interval is \(1-\lambda_i(t+u)h+o(h)\). Multiplying by the probability of surviving to elapsed time \(u\) gives

S(u+h)=S(u)\left[1-\lambda_i(t+u)h+o(h)\right].

Subtract \(S(u)\), divide by \(h\), and let \(h\) decrease to zero. This gives the differential equation and initial condition

S'(u)=-\lambda_i(t+u)S(u),\qquad S(0)=1.

Separating variables and integrating from \(0\) to \(u\) yields

\int_1^{S(u)}\frac{dz}{z} = -\int_0^u\lambda_i(t+s)\,ds, \qquad \log S(u)=-\int_0^u\lambda_i(t+s)\,ds.

Exponentiating proves the stated formula. The argument also explains why the intensities for different exits are added: over a short interval, each possible exit contributes to the probability of leaving state \(i\), and the probability of two or more transitions contributes only to the higher-order remainder under the stated intensity assumptions.

Worked Example: Constant Total Exit Intensity

A hypothetical income-protection model starts a covered person in the active state at time zero. While active, the transition intensity to disability is \(0.03\) per year and the intensity to death is \(0.01\) per year. Assuming both intensities remain constant, find the probability of no transition from active during the next five years.

Method. Use the total exit intensity, then apply the constant-intensity form of the holding-time survival formula. This avoids treating either possible exit as the only way to leave the state.

The total intensity is

\lambda_{\text{active}}=0.03+0.01=0.04\text{ per year}.

The duration is five years, so the integrated intensity is \(0.04\times5=0.20\). The no-transition probability is

P(T_{\text{active}}>5) =\exp(-0.20) =0.8187308\ldots \approx 0.8187.

Thus the probability is 0.8187, or 81.87% to two decimal places as a percentage. As a check, the probability is less than one and is below the probability that would result from using only the smaller exit intensity, since either exit ends the period of continuous residence in the active state.

Constant and Changing Intensities

If the total exit intensity is a constant \(\lambda\) per time unit, the integral is simply \(\lambda u\), giving

P(T_i>u\mid X(t)=i)=e^{-\lambda u}.

The product \(\lambda u\) must be dimensionless: a rate per year must be multiplied by a duration in years. If the intensity varies with time, integrate its rate over the interval instead of multiplying a single rate by the entire duration. A higher rate near the end of the interval still affects the probability of surviving the whole interval.

The intensity may vary with calendar time or with elapsed time since the start of the holding period. In either case, the formula uses the intensity along the interval on which the individual remains in state \(i\). If a model specifies different rates in different time periods, split the integral at each point where the rate changes. The resulting probabilities multiply because exponentials of summed integrals multiply.

Worked Example: A Total Intensity That Increases With Duration

In a hypothetical rehabilitation model, a person enters a monitored state at time zero. Over the next three years, the intensity of leaving that state for one destination is \(\mu_1(s)=0.01+0.004s\) per year, and the intensity for another destination is \(\mu_2(s)=0.02+0.006s\) per year, where \(s\) is measured in years since entry. Find the probability of no transition during those three years.

Method. Add the intensities at each elapsed time, integrate the resulting total rate, and exponentiate the negative integral. The common mistake is to use only the rate at entry and treat it as constant.

The total intensity at elapsed time \(s\) is

\lambda(s)=\mu_1(s)+\mu_2(s) =(0.01+0.004s)+(0.02+0.006s) =0.03+0.01s.

Integrating from entry to year three gives

\int_0^3(0.03+0.01s)\,ds = \left[0.03s+0.005s^2\right]_0^3 =0.09+0.045 =0.135.

Therefore,

P(T_i>3)=e^{-0.135} =0.8737159\ldots \approx 0.8737.

The probability of remaining without a transition for all three years is 0.8737. As a check on the arithmetic, the total rate rises from \(0.03\) per year to \(0.06\) per year; the integrated intensity \(0.135\) is greater than \(0.03\times3=0.09\), which it would have been if the entry rate had incorrectly been held constant.

Worked Example: Piecewise-Constant Rates and Time Units

A hypothetical long-term-care state has a total exit intensity of \(0.18\) per year during the first four months after entry and \(0.06\) per year during the following eight months. Assuming these rates are constant within their respective periods, calculate the probability of no transition throughout the full year.

Method. Integrate each rate over the duration for which it applies, using years as the time unit in both pieces. This avoids multiplying a yearly rate by a number of months as if those months were years.

Four months is \(4/12=1/3\) of a year, and eight months is \(8/12=2/3\) of a year. The integrated intensity is

0.18\left(\frac{1}{3}\right) + 0.06\left(\frac{2}{3}\right) = 0.06+0.04 = 0.10.

Thus the probability of no transition for the full year is

P(T_i>1)=e^{-0.10} =0.9048374\ldots \approx 0.9048.

The answer is 0.9048. The integrated intensity is dimensionless, as required. A useful check is that the average rate over the year is \(0.10\) per year, since the two rate-duration products sum to \(0.10\).

No Transition Is Not Always the Same as Being in the Same State

A no-transition probability concerns the entire path over the interval: the individual must stay in state \(i\) continuously. By contrast, a probability of occupying state \(i\) at the endpoint only asks where the individual is at that one time. If a model permits leaving \(i\) and later returning, the endpoint state can be \(i\) even though a transition occurred in between. Consequently, a same-state endpoint probability can exceed the no-transition probability.

For a state with no possible route back after exit, the two events may coincide. Otherwise, do not substitute an endpoint transition probability for the holding-time survival probability. The formula in this tutorial answers only the no-exit question; it does not count paths that leave and return.

Common Exam Traps

Using just one exit intensity. If several destinations leave the current state, add all their intensities before integrating.
Using the wrong time unit. Convert months to years when rates are specified per year, or convert the rate consistently before calculating.
Using a varying rate as if constant. Integrate the specified rate function over the interval; do not automatically use the initial or final value for the whole duration.
Forgetting the interval's starting time. With calendar-time intensity \(\lambda_i(r)\), integrate from \(t\) to \(t+u\), not necessarily from zero to \(u\).
Confusing no transition with the endpoint state. The holding-time formula excludes every path with an exit, including paths that could later return.
Applying the formula outside its assumptions. The integrated intensity must describe the process while it remains in the state, under the conditioning used. If rates depend on additional information, use the appropriate conditional intensity path.

Key takeaway: For continuous residence in a state, add all exit intensities to obtain the total rate and calculate \(e^{-\text{integrated total intensity}}\). For a constant rate this is \(e^{-\lambda u}\); for varying or piecewise rates, integrate over the actual interval with consistent time units.

Check Your Understanding

For each question, calculate the probability of no transition over the stated interval and round the probability to four decimal places.

A state has two constant exit intensities, \(0.04\) and \(0.03\) per year. What is the probability of no transition over two years? (Answer: \(e^{-0.14}=0.8694\))
A state has two constant exit intensities, \(0.015\) and \(0.025\) per month. What is the probability of no transition over nine months? (Answer: \(e^{-0.36}=0.6977\))
Starting at time zero, a state's total intensity is \(\lambda(s)=0.02+0.005s\) per year, where \(s\) is in years. Find the probability of no transition for two years. (Answer: \(e^{-0.05}=0.9512\))
A state's total intensity is \(0.12\) per year for the first six months and \(0.24\) per year for the next six months. Find the probability of no transition over the full year. (Answer: \(e^{-0.18}=0.8353\))
At calendar time \(t=2\), a state has total intensity \(\lambda(r)=0.03+0.01r\) per year, where \(r\) is calendar time in years. What is the probability of no transition from time 2 through time 4? (Answer: \(e^{-0.12}=0.8869\))

What You'll Learn