Exponential distribution and memorylessness

Objectives and prerequisites

Use the rate parameter of an exponential distribution; calculate waiting-time probabilities; derive mean, variance and median; apply memorylessness with the correct event; and relate a first-arrival wait to a homogeneous Poisson process. Prerequisites: PDFs, CDFs, conditional probability, integration and basic Poisson counts. Exponential distributions are explicitly listed in the official GATE 2026 CS syllabus. Reliability examples are mathematical models, not claims that all real lifetimes are exponential.

1. Rate and scale conventions

Write T~Exp(λ) with rate λ>0. Its support is t≥0 and its PDF is f(t)=λe^(−λt) there, zero for t<0. Normalization follows from ∫₀∞λe^(−λt)dt=1. The rate has units reciprocal to time. If λ=2 per minute, then λt is dimensionless when t is in minutes.

Some sources parameterize with scale θ=1/λ, writing f(t)=(1/θ)e^(−t/θ). The scale is the mean and has time units. Always read the convention. “Exponential parameter two” is ambiguous unless rate or scale is specified. This lesson uses the rate convention throughout.

There is no atom at zero: P(T=0)=0 even though f(0)=λ. Nonnegative support includes zero as an endpoint; the continuous density model assigns it no point mass. The parameter λ=0 does not define a normalized exponential density and is excluded.

2. CDF and survival function

For t≥0, F(t)=P(T≤t)=∫₀ᵗλe^(−λu)du=1−e^(−λt). For t<0, F(t)=0. The survival probability for t≥0 is S(t)=P(T>t)=e^(−λt). For t<0, S(t)=1. The density describes probability per unit time; F and S are probabilities.

If 0≤a<b, P(a<T≤b)=F(b)−F(a)=e^(−λa)−e^(−λb). Endpoints do not change the answer in this continuous model. A tail is not the density height λe^(−λt).

Worked example: T~Exp(2 per minute). P(T>1 minute)=e^(−2)≈.135335. P(T≤.5 minute)=1−e^(−1)≈.632121. P(.5<T≤1)=e^(−1)−e^(−2)≈.232544. The density at t=.5 is 2e^(−1) per minute; this is neither the CDF nor the survival probability. Converting .5 minute to 30 seconds while keeping λ=2 per minute in the exponent would mix units; convert λ to 1/30 per second as well.

3. Mean, second moment and variance

Integration by parts gives E[T]=∫₀∞tλe^(−λt)dt=1/λ. A second application yields E[T²]=2/λ². Hence Var(T)=2/λ²−(1/λ)²=1/λ², and standard deviation=1/λ. Boundary terms vanish because polynomial factors times e^(−λt) tend to zero as t→∞ when λ>0.

For λ=2 per minute, mean=.5 minute, variance=.25 minute² and standard deviation=.5 minute. A higher rate means a shorter expected waiting time. Rate and mean are reciprocals; equating them confuses units and behavior.

The median m solves 1−e^(−λm)=1/2, giving m=(ln 2)/λ. It is smaller than the mean since ln 2<1. For 0<q<1, the q-quantile is −ln(1−q)/λ. With λ=2, the 90th percentile is (ln 10)/2≈1.151293 minutes. The density decreases from its maximum at zero, so the usual density representation has mode zero. Mean, median and mode differ.

4. Memorylessness: the remaining wait

For s,t≥0,

P(T>s+t | T>s)=P(T>s+t)/P(T>s) =e^(−λ(s+t))/e^(−λs)=e^(−λt)=P(T>t).

The intersection simplifies because {T>s+t} is a subset of {T>s}. Having already waited s without the event, the additional wait has the same exponential law as a fresh wait. This statement is about the conditional remaining duration, not the total duration measured from the beginning.

Example: T~Exp(.2 per hour). Given survival beyond ten hours, the probability of surviving at least three more hours is e^(−.2×3)=e^(−.6). The unconditional probability of total lifetime exceeding thirteen hours is e^(−2.6), a different quantity. With zero-probability endpoints irrelevant here, “beyond” and “at least” give the same value.

The conditional mean remaining lifetime is E[T−s|T>s]=1/λ. The conditional mean total lifetime is E[T|T>s]=s+1/λ. In the example these are five and fifteen hours respectively. Do not report the unchanged remaining mean as the total lifetime.

Memorylessness is an assumption-dependent consequence of the exponential law; most real aging components need not satisfy it. A uniform waiting time on [0,10], for example, has a remaining upper limit that shrinks after surviving longer, so it is not memoryless.

5. Connection with Poisson arrivals

If events follow a homogeneous Poisson process of rate λ, let T be the time from the start to the first arrival. T>t exactly when no arrivals occur in [0,t]. Since the count has Poisson mean λt,

P(T>t)=P(N(t)=0)=e^(−λt).

Thus T~Exp(λ). The count is discrete and its parameter for an interval is λt; the waiting time is continuous and its rate is λ. Keep these two random variables and parameters distinct. An average arrival rate alone does not establish either process model.

Example: a homogeneous Poisson stream has rate three arrivals per hour. The probability the first arrival takes more than twenty minutes is e^(−3×(1/3))=e^(−1). It is not e^(−60) or e^(−20/3); convert twenty minutes to one-third hour before multiplying by the hourly rate.

6. Minimum of independent exponential waits

If T₁ and T₂ are independent exponential waits with rates λ₁,λ₂, then M=min(T₁,T₂)>t exactly when both exceed t. Independence gives P(M>t)=e^(−λ₁t)e^(−λ₂t)=e^(−(λ₁+λ₂)t), so M~Exp(λ₁+λ₂). Its mean is 1/(λ₁+λ₂), not the sum of the individual means. The first of two independent opportunities arrives sooner on average.

This does not say the sum T₁+T₂ is exponential. For two equal independent rates, the sum waits for two events, a different law outside this lesson's main scope. Without independence, the minimum-rate-addition argument also fails.

Checklist: state rate versus scale; convert time units; choose CDF, survival or interval difference; use memorylessness only for the conditional remaining wait; and justify independence before multiplying survival probabilities. Check that longer time gives smaller survival probability and that a larger rate gives a smaller mean.

Original practice: six single-correct MCQs

These are original questions, not previous-year questions. Choose exactly one option.

EX01. T has an exponential distribution with rate 2 per minute. Find E[T].

A. 2 minutes B. 1/2 minute C. 4 minutes D. 1/4 minute

Answer: B

A: The rate is the reciprocal of the mean, not the mean.

B: The mean is 1/λ=1/2 minute.

C: This squares the rate instead of taking its reciprocal.

D: This is the numerical variance in minute², not the mean in minutes.

EX02. T~Exp(2) with time in minutes and rate per minute. Find P(0.5<T≤1).

A. e^(−2) B. 1−e^(−1) C. e^(−1)−e^(−2) D. 2e^(−1)

Answer: C

A: This is survival beyond one minute.

B: This is arrival by half a minute.

C: Subtract the two survival values at .5 and one minute.

D: This is density at half a minute, not interval probability.

EX03. T~Exp(0.2 per hour). Given T>10 hours, what is P(T>13 hours|T>10 hours)?

A. e^(−2.6) B. e^(−2) C. 1−e^(−0.6) D. e^(−0.6)

Answer: D

A: This ignores the condition and gives unconditional survival to thirteen.

B: This is survival to ten, the condition probability.

C: This is failure within the next three hours, not survival.

D: Memorylessness uses the additional three hours: e^(−.2×3).

EX04. T~Exp(λ), λ>0. What is its median?

A. (ln 2)/λ B. 1/λ C. λ ln 2 D. 1/λ²

Answer: A

A: Solve 1−e^(−λm)=1/2.

B: This is the mean, which is larger than the median.

C: The rate belongs in the denominator; this also has the wrong time units.

D: This is the variance, in squared time units.

EX05. A homogeneous Poisson process has rate three per hour. What is the probability its first arrival takes more than twenty minutes?

A. e^(−60) B. e^(−1) C. 1−e^(−1) D. e^(−20/3)

Answer: B

A: This multiplies hourly rate by minutes without converting units.

B: Twenty minutes is one-third hour, so λt=1.

C: This is arrival within twenty minutes.

D: The exponent must be rate times time expressed in matching units.

EX06. T₁,T₂ are independent exponential waits with rates 2 and 3 per minute. What is E[min(T₁,T₂)]?

A. 5/6 minute B. 1/2 minute C. 1/5 minute D. 5 minutes

Answer: C

A: This adds individual means and describes the mean of the sum.

B: This ignores the second independent opportunity.

C: The minimum is exponential with rate 2+3=5, giving mean 1/5.

D: This uses the combined rate as though it were the mean.

Source and scope

Exponential distributions are named in Section 1 Probability and Statistics of the official GATE 2026 CS syllabus. The Poisson-process connection explains a common model; it is not a claim that additional stochastic-process topics are separately listed. Original teaching and questions; partial syllabus coverage.

Official GATE 2026 CS syllabus (verified 2026-10-14).