Exponential & Poisson
Two distributions that always travel together: the exponential times the wait until an event, the Poisson counts how many events land. Both lean on one number, lambda.
What you'll learn
- Exponential(lambda): density, CDF, mean = 1/lambda, variance = 1/lambda squared
- The memoryless property: a used exponential clock is as good as new
- Poisson(lambda) PMF, and its signature fact — mean equals variance equals lambda
- Solving for lambda from a mean/variance condition (a real 2024 NAT)
Before you start
Stand at a bus stop. Buses come at random — sometimes two within a minute,
sometimes a long quiet stretch. Two natural questions follow you around. How
long until the next bus? That answer is a continuous waiting time — the
exponential. How many buses arrive in the next hour? That answer is a
count — the Poisson. Both questions share a single dial called lambda,
the rate of events per unit time. Learn them together and a whole family of
exam questions falls open — and the same pair models server-request arrivals,
time-between-failures, and rare-event counts you will meet in real systems work.
Exponential(lambda)
A continuous random variable with density
- Density:
f(x) = lambda · e^(−lambda·x)forx ≥ 0(and 0 otherwise). - CDF:
F(x) = 1 − e^(−lambda·x)— soP(X > x) = e^(−lambda·x). - Mean:
E(X) = 1/lambda. - Variance:
Var(X) = 1/lambda².
A bigger rate lambda means events come faster, so the average wait 1/lambda
shrinks — exactly what you’d expect.
The memoryless property
The exponential forgets the past. If you’ve already waited s minutes, the chance
of waiting at least t more is the same as if you’d just started:
P(X > s + t | X > s) = P(X > t)A bus that hasn’t come in 10 minutes is no “more due” than a fresh one — the remaining wait has the same exponential distribution. This is the only continuous distribution with this property, and GATE leans on it constantly.
Poisson(lambda)
The discrete partner. If events occur at rate lambda per interval, the number X
that land in one interval has PMF
P(X = k) = e^(−lambda) · lambda^k / k! , k = 0, 1, 2, …Its signature fact, and a GATE favourite:
- Mean:
E(X) = lambda. - Variance:
Var(X) = lambda— equal to the mean.
So a Poisson random variable has equal mean and variance (a true and frequently-tested fact, confirmed again in GATE DA 2024).
Open the Exponential tab below, drag the rate slider, and watch the curve
steepen. Bracket a region (say 0 < X < 1) and the shaded area is P(X < 1) —
exactly 1 − e^(−lambda), no integration needed. Try a small lambda (slow
events, long waits) versus a large one (fast events, short waits) and see the
mean 1/lambda track the shaded mass.
How GATE asks this
Most often a NAT. The classic move: give you a relationship between the mean and variance and ask you to solve for lambda, which forces you to know the formulas cold. Or supply a rate and ask for a single Poisson/exponential probability. MCQs test whether you can tell the two distributions’ formulas apart.
Worked example — a real 2024 question
Let
X ~ Exponential(lambda). If5 · E(X) = Var(X), findlambda.
This is an actual GATE DA 2024 problem. Substitute the two formulas and solve:
E(X) = 1/lambda
Var(X) = 1/lambda^2
5 · E(X) = Var(X)
5 · (1/lambda) = 1/lambda^2
5/lambda = 1/lambda^2
Multiply both sides by lambda^2:
5·lambda = 1
lambda = 0.2So lambda = 0.2. The whole question is just “do you know E(X) = 1/lambda and
Var(X) = 1/lambda²?” — answer them confidently and it falls out in two lines.
Quick check
Quick check
Practice this in an interview
All questionsBinomial counts successes in a fixed number of independent trials with a fixed success probability. Poisson counts events in a continuous interval when events are rare and arrive independently at a constant average rate. Poisson is the limiting case of Binomial as n → ∞ and p → 0 with np = λ fixed.
Each distribution has a natural generative story: Bernoulli is a single coin flip; Binomial sums Bernoullis; Poisson counts rare arrivals; Normal emerges from sums of many small effects; Exponential models waiting times between Poisson events; Uniform assigns equal probability across a range. Choosing correctly comes from matching that story to the data-generating process.
An Exponential random variable satisfies P(X > s+t | X > s) = P(X > t): the remaining waiting time has the same distribution regardless of how long you have already waited. This is the only continuous distribution with this property, and it directly corresponds to a constant hazard rate — neither ageing nor improving over time.
A Bernoulli(p) trial is the atomic unit: a single experiment with success probability p. Binomial(n, p) is the sum of n independent, identically distributed Bernoulli(p) trials, counting total successes. Because Binomial is a sum of independent random variables, its mean and variance are n times those of a single Bernoulli.