I'm trying to understand the staleness property of the detector in my work. I have simplified the question as follows:
Support the state of a machine changes from $0$ to $1,2,3......$, and the change interval between two states follows an exponential distribution with parameter $\lambda$. For every T second, we check the current state of the machine and record it on a board beside the machine.
My question is, what is the expected ratio of the whole time during which the recorded state is correct, i.e., the recorded state on the board is exactly the current state of the machine?
Any ideas or related materials are welcomed.
See if I understand the question correctly.
For each time interval of length $T$ between two successive checking time, the recorded state is considered to be correct at a given time moment if there is no change in state since the last recorded time.
Let $X$ be the total time that the recorded state is correct for a given time interval. So $X$ is the time for the next change of state since last recorded time, if it is less than $T$, and equal to $T$ otherwise.
Since exponential time is memoryless, if we let $W$ to be the time for the next change of state since last recorded time, we have $W \sim \text{Exponential}(\lambda)$ regardless of the last state change time, and $$ X = \min\{W, T\} $$
Therefore $$ \begin{align} E[X] &= E[\min\{W, T\}] \\ &= \int_0^T w \lambda e^{-\lambda w} dw + \int_T^{+\infty} T \lambda e^{-\lambda w} dw \\ &= -\int_0^T w d e^{-\lambda w} + T[e^{-\lambda T} - 0] \\ &= -Te^{-\lambda T} + 0 + \int_0^T e^{-\lambda w} dw + Te^{-\lambda T} \\ &= \frac {1} {\lambda} [1 - e^{-\lambda T}] \end{align} $$
As a result the expected ratio is $$ E\left[\frac {X} {T}\right] = \frac {1} {\lambda T} [1 - e^{-\lambda T}]$$
If the time between the two successive state change is not exponential (a more general renewal process), then it will be more complicated as we need to condition on the time between the last state change and the last recorded time since memoryless property no longer hold. We should be able to use renewal theory to deduce the long run ratio also.