Mathematical Rules of Thumb, illustrated reader · Chapter 14

14Stochastic Processes

Rates, Waiting, and Dependence

5 demonstrations follow the chapter's rules. Choose a value, watch the figure and the numbers change, and check your prediction. Every choice is precomputed from the notebook calculations.

Ask the chapter skill

“Help me use Chapter 14 for my question. Choose a rule, check its assumptions, and show how the result changes when an input changes.”

Use math-thumb-stochastic-processes from the companion's skill package. The demonstrations below also work on their own.

Examples use constructed inputs or the book's own values, disclosed in each panel. A picture illustrates a rule; its assumptions set its scope.

1Demonstration 1 of 5

Approach the queue capacity boundary

Can you report a negative wait when λ exceeds μ?

A small demand increase near capacity creates a large delay increase. The invalid boundary is shown explicitly.

W=1μ−λ,λ<μ W=\frac1{\mu-\lambda},\quad\lambda<\mu

Arrivals per hour λ. Stationary M/M/1 queue, exponential service, Poisson arrivals, μ=10/hour.

Predict first. Can you report a negative wait when λ exceeds μ?

Choose an example

Approach the queue capacity boundary. At service rate 10/hour and arrivals 9.5/hour, mean system time is 2 hours under the M/M/1 assumptions.
Arrivals per hour λ: 9.5
Constructed teaching inputs; calculations executed locally. Supported menu choices are precomputed.

Calculated values

Utilization
0.95
Mean system time
2
Mean customers in system
19

At service rate 10/hour and arrivals 9.5/hour, mean system time is 2 hours under the M/M/1 assumptions.

Use the idea

Use rule 14.3.3 when its stated conditions fit. Compare the calculation with your own decision threshold; retain the relevant error or uncertainty.

Where the conclusion applies

Stationary M/M/1 queue, exponential service, Poisson arrivals, μ=10/hour.

Check your understanding: Can you report a negative wait when λ exceeds μ?
No. The stationary formula is invalid at or above capacity; a negative output would be a model-use error.

Book source: Rule 14.3.3: M/M/1 Delay Blows Up Near Capacity. Demonstration C14-D01. Worked illustration.

2Demonstration 2 of 5

Follow a correlated shock through time

Does a higher φ make the same shock vanish sooner?

Plot the decay of one shock and compare its half-life with long-run variance amplification.

ϕh=12,h=log⁡(.5)log⁡ϕ \phi^h=\frac12,\quad h=\frac{\log(.5)}{\log\phi}

AR(1) coefficient φ. Stationary AR(1), 0<φ<1, independent innovation model for the variance factor.

Predict first. Does a higher φ make the same shock vanish sooner?

Choose an example

Follow a correlated shock through time. A shock is multiplied by φ=0.8 each step (it keeps 80% per step), with half-life 3.11 steps. A long average is 9 times as noisy (in variance) as an average of the same number of independent readings. Lingering shocks make neighboring readings repeat each other, so persistence costs precision.
AR(1) coefficient φ: 0.8
Constructed teaching inputs; calculations executed locally. Supported menu choices are precomputed.

Calculated values

AR coefficient
0.8
Half-life in steps
3.10628
Long-run variance factor
9

A shock is multiplied by φ=0.8 each step (it keeps 80% per step), with half-life 3.11 steps. A long average is 9 times as noisy (in variance) as an average of the same number of independent readings. Lingering shocks make neighboring readings repeat each other, so persistence costs precision.

Use the idea

Use rule 14.2.3 when its stated conditions fit. Compare the calculation with your own decision threshold; retain the relevant error or uncertainty.

Where the conclusion applies

Stationary AR(1), 0<φ<1, independent innovation model for the variance factor.

Check your understanding: Does a higher φ make the same shock vanish sooner?
No. A higher positive φ retains more of the shock each step and increases its half-life.

Book source: Rule 14.2.3: AR(1) Shock Half-Life. Demonstration C14-D02. Worked illustration.

3Demonstration 3 of 5

Turn exposure into a Poisson count

What mean count do you expect over two hours?

The count distribution and probability of any arrival change with rate times exposure.

N(t)∼Poisson⁡(λt) N(t)\sim\operatorname{Poisson}(\lambda t)

Arrival rate per hour λ. Constant Poisson rate over one hour; independent increments.

Predict first. What mean count do you expect over two hours?

Choose an example

Turn exposure into a Poisson count. A constant Poisson rate 5/hour gives mean count 5 in one hour and mean exponential wait 0.2 hours. Changing or dependent arrival rates need a different model.
Arrival rate per hour λ: 5
Constructed teaching inputs; calculations executed locally. Supported menu choices are precomputed.

Calculated values

Hourly arrival rate
5
Mean count in one hour
5
Probability of any arrival
0.993262
Mean wait (hours)
0.2

A constant Poisson rate 5/hour gives mean count 5 in one hour and mean exponential wait 0.2 hours. Changing or dependent arrival rates need a different model.

Use the idea

Use rule 14.1.1 when its stated conditions fit. Compare the calculation with your own decision threshold; retain the relevant error or uncertainty.

Where the conclusion applies

Constant Poisson rate over one hour; independent increments.

Check your understanding: What mean count do you expect over two hours?
Twice the hourly rate. The exposure duration must accompany the rate.

Book source: Rule 14.1.1: Poisson Count Equals Rate Times Exposure. Demonstration C14-D03. Worked illustration.

4Demonstration 4 of 5

Watch a random walk spread like a square root

If a walker takes 100 times more steps, how much farther does it typically get?

Five sample walks wander around the ±√n curves. The root-mean-square distance of 2000 walks tracks √n.

E[Sn2]=n \sqrt{E[S_n^2]}=\sqrt n

Steps n. Fair independent ±1 steps, seed 1421. RMS is a typical distance, not a bound on any one walk.

Predict first. If a walker takes 100 times more steps, how much farther does it typically get?

Choose an example

Watch a random walk spread like a square root. After 1,000 fair ±1 steps, the typical distance from start is about √1,000=31.62 steps (simulation: 31.83). That is only 3.2% of the steps taken. Ten times more steps would move the walker only about 3.16 times farther.
Steps n: 1000
Constructed teaching inputs; calculations executed locally. Supported menu choices are precomputed.

Calculated values

Steps
1000
Theory √n
31.6228
Simulated RMS distance
31.8311
Typical distance / steps
0.0318311
Seed
1421

After 1,000 fair ±1 steps, the typical distance from start is about √1,000=31.62 steps (simulation: 31.83). That is only 3.2% of the steps taken. Ten times more steps would move the walker only about 3.16 times farther.

Use the idea

Use rule 14.2.1 when its stated conditions fit. Compare the calculation with your own decision threshold; retain the relevant error or uncertainty.

Where the conclusion applies

Fair independent ±1 steps, seed 1421. RMS is a typical distance, not a bound on any one walk.

Check your understanding: If a walker takes 100 times more steps, how much farther does it typically get?
Ten times farther, since √100=10.

Book source: Rule 14.2.1: Random-Walk Displacement Grows as Root N. Demonstration C14-D04. Worked illustration.

5Demonstration 5 of 5

Check whether waiting makes the end nearer

For an exponential wait with mean 10 minutes, after 15 minutes of waiting, what is the expected extra wait?

Compare the remaining-wait curve of a memoryless exponential wait with an aging wait that has the same 10-minute mean.

P(T>s+t∣T>s)=P(T>t)=e−t/μ P(T>s+t\mid T>s)=P(T>t)=e^{-t/\mu}

Minutes already waited s. Exponential with mean 10 minutes versus Weibull shape 2 with mean 10 minutes. Constructed models, not observed data.

Predict first. For an exponential wait with mean 10 minutes, after 15 minutes of waiting, what is the expected extra wait?

Choose an example

Check whether waiting makes the end nearer. After 5 minutes already waited, a memoryless (exponential) wait still has mean 10 more minutes and a 36.8% chance of lasting 10 more: the past wait is forgotten. An aging wait with the same 10-minute mean (Weibull shape 2) now expects only 6.46 more minutes. Whether waiting longer makes the end nearer depends on the model, not on intuition.
Minutes already waited s: 5
Constructed teaching inputs; calculations executed locally. Supported menu choices are precomputed.

Calculated values

Minutes already waited
5
Exponential: expected extra wait
10
Aging model: expected extra wait
6.46061
Exponential: P(over 10 more)
0.367879
Aging model: P(over 10 more)
0.20788

After 5 minutes already waited, a memoryless (exponential) wait still has mean 10 more minutes and a 36.8% chance of lasting 10 more: the past wait is forgotten. An aging wait with the same 10-minute mean (Weibull shape 2) now expects only 6.46 more minutes. Whether waiting longer makes the end nearer depends on the model, not on intuition.

Use the idea

Use rule 14.1.4 when its stated conditions fit. Compare the calculation with your own decision threshold; retain the relevant error or uncertainty.

Where the conclusion applies

Exponential with mean 10 minutes versus Weibull shape 2 with mean 10 minutes. Constructed models, not observed data.

Check your understanding: For an exponential wait with mean 10 minutes, after 15 minutes of waiting, what is the expected extra wait?
Still 10 minutes. The exponential forgets the past; only an aging model shortens the remaining wait.

Book source: Rule 14.1.4: Memorylessness Means No Aging. Demonstration C14-D05. Worked illustration.

Bring the idea to a question of your own

Choose the relationship that answers your question, check its conditions, and compare the result with the accuracy or decision threshold you need.

The chapter skill can adapt these calculations to your inputs. It should name the assumptions, explain what the result supports, and say what still needs evidence. The chapter workbook adds a lab and three exercises with answers.