Sign in to save

Bookmark this page so you can find it later.

Sign in to save

Bookmark this page so you can find it later.

The exponential distribution models the waiting time until the next event in a process where events occur continuously and independently at a constant average rate. This cheat sheet helps students recognize when the model applies, compute probabilities, and interpret the rate parameter. It is especially useful in reliability, queueing, survival analysis, and Poisson process problems.

The main parameter is the rate λ>0\lambda > 0, which controls how quickly events tend to occur. The density is f(x)=λeλxf(x)=\lambda e^{-\lambda x} for x0x \ge 0, and the cumulative distribution is F(x)=1eλxF(x)=1-e^{-\lambda x}. The mean is E[X]=1λE[X]=\frac{1}{\lambda}, the variance is Var(X)=1λ2\operatorname{Var}(X)=\frac{1}{\lambda^2}, and the distribution has the important memoryless property P(X>s+tX>s)=P(X>t)P(X>s+t\mid X>s)=P(X>t).

Key Facts

  • If XExp(λ)X\sim \operatorname{Exp}(\lambda), then the probability density function is f(x)=λeλxf(x)=\lambda e^{-\lambda x} for x0x\ge 0 and f(x)=0f(x)=0 for x<0x<0.
  • The cumulative distribution function is F(x)=P(Xx)=1eλxF(x)=P(X\le x)=1-e^{-\lambda x} for x0x\ge 0.
  • The survival function is P(X>x)=1F(x)=eλxP(X>x)=1-F(x)=e^{-\lambda x} for x0x\ge 0.
  • The mean waiting time is E[X]=1λE[X]=\frac{1}{\lambda}, so larger λ\lambda means shorter average waiting time.
  • The variance and standard deviation are Var(X)=1λ2\operatorname{Var}(X)=\frac{1}{\lambda^2} and σ=1λ\sigma=\frac{1}{\lambda}.
  • The memoryless property is P(X>s+tX>s)=P(X>t)=eλtP(X>s+t\mid X>s)=P(X>t)=e^{-\lambda t} for s,t0s,t\ge 0.
  • The median is m=ln2λm=\frac{\ln 2}{\lambda} because P(Xm)=0.5P(X\le m)=0.5.
  • If events follow a Poisson process with rate λ\lambda, then the waiting time until the next event follows Exp(λ)\operatorname{Exp}(\lambda).

Vocabulary

Exponential distribution
A continuous probability distribution used to model waiting time until the next event in a constant-rate process.
Rate parameter
The parameter λ\lambda represents the average number of events per unit time and must satisfy λ>0\lambda>0.
Probability density function
The function f(x)=λeλxf(x)=\lambda e^{-\lambda x} describes relative likelihood for possible waiting times x0x\ge 0.
Cumulative distribution function
The function F(x)=1eλxF(x)=1-e^{-\lambda x} gives the probability that the waiting time is at most xx.
Survival function
The function S(x)=P(X>x)=eλxS(x)=P(X>x)=e^{-\lambda x} gives the probability that the waiting time exceeds xx.
Memoryless property
The rule P(X>s+tX>s)=P(X>t)P(X>s+t\mid X>s)=P(X>t) means that the remaining waiting time does not depend on time already waited.

Common Mistakes to Avoid

  • Using the mean as the rate is wrong because λ\lambda is the event rate and E[X]=1λE[X]=\frac{1}{\lambda}, not λ\lambda.
  • Forgetting the support x0x\ge 0 is wrong because exponential waiting times cannot be negative, so f(x)=0f(x)=0 for x<0x<0.
  • Confusing P(X>x)P(X>x) with P(Xx)P(X\le x) is wrong because P(X>x)=eλxP(X>x)=e^{-\lambda x} while P(Xx)=1eλxP(X\le x)=1-e^{-\lambda x}.
  • Mixing time units is wrong because λ\lambda and xx must use compatible units, such as events per hour with time measured in hours.
  • Assuming every waiting-time problem is exponential is wrong because the model requires independent events occurring at a constant average rate.

Practice Questions

  1. 1 If XExp(0.25)X\sim \operatorname{Exp}(0.25), find P(X>6)P(X>6).
  2. 2 The average lifetime of a component is 500500 hours. Assuming an exponential model, find λ\lambda and P(X200)P(X\le 200).
  3. 3 Calls arrive according to a Poisson process at a rate of 33 calls per hour. Find the probability that the next call arrives within 1010 minutes.
  4. 4 Explain why the memoryless property is reasonable for some electronic components but not for a human lifespan.