Interpreting an exponential waiting time density function

Expert intro: This problem uses the exponential distribution to model waiting times between website hits. The key is to recognize the pdf, verify it is a valid density, and use the distribution’s memoryless properties when answering follow-up parts.

Detailed walkthrough

Identify the distribution

The given density is

$f(x)=e^{-x}, \quad x>0.$

This is the pdf of an exponential random variable with rate parameter $\lambda=1$. In standard form, an exponential pdf is

$f(x)=\lambda e^{-\lambda x}, \quad x>0.$

Matching terms gives $\lambda=1$. That means the waiting time between hits is modeled as an exponential distribution with mean

$E[X]=\frac{1}{\lambda}=1$

minute and variance

$\mathrm{Var}(X)=\frac{1}{\lambda^2}=1.$

How to use the pdf

Because this is a continuous distribution, probabilities come from integrating the density over an interval. For example,

$P(a<X<b)=\int_a^b e^{-x}\,dx = e^{-a}-e^{-b}$

for any $0<a<b$. The cumulative distribution function is

$F(x)=P(X\le x)=\int_0^x e^{-t}\,dt=1-e^{-x}, \quad x>0.$

So tail probabilities are especially easy:

$P(X>x)=1-F(x)=e^{-x}.$

Key properties to remember

The exponential distribution has a memoryless property:

$P(X>s+t\mid X>s)=P(X>t).$

That means once you have already waited $s$ minutes, the additional waiting time behaves the same as if you were starting from zero. This is a major reason the exponential model is used for random arrival processes.

If the question asks for the probability of waiting more than a certain time, use the survival function $e^{-x}$. If it asks for the probability of waiting between two times, subtract CDF values or integrate directly.

Common computations

If a part asks for the probability that the next hit occurs within 2 minutes, compute

$P(X\le 2)=1-e^{-2}.$

If it asks for the probability that the waiting time exceeds 3 minutes, compute

$P(X>3)=e^{-3}.$

If it asks for the median waiting time, solve

$1-e^{-m}=0.5 \Rightarrow e^{-m}=0.5 \Rightarrow m=\ln 2.$

These formulas all come from recognizing the exponential pdf correctly.

💡 Pitfall guide

A common mistake is to treat $f(x)=e^{-x}$ as though it were valid for all real numbers. It is only a pdf on $x>0$, because negative waiting times are impossible and the integral over the support must equal 1. Another frequent error is forgetting the exponential rate parameter. Here $\lambda=1$, so the mean is 1 minute; if the coefficient had been different, the answer would change. Students also sometimes use $F(x)=e^{-x}$, but that is the survival function $P(X>x)$, not the CDF. Keeping pdf, CDF, and tail probability distinct avoids most calculation errors.

🔄 Real-world variant

If the density were changed to $f(x)=2e^{-2x},\ x>0$, the same method would apply, but now the rate would be $\lambda=2$. That would change the mean to $1/2$ minute and the CDF to $F(x)=1-e^{-2x}$. If the problem instead asked for the waiting time between website hits in seconds rather than minutes, the form of the distribution would stay exponential, but the parameter interpretation would need to match the new time unit. Always identify the support and the rate before computing probabilities or expectations.

🔍 Related terms

exponential distribution, memoryless property, survival function