Exponential Distribution

Concept

A random time \(T\) is said to be exponentially distributed with parameter \(\lambda > 0\) if its probability density function (PDF) is:

\[ f(x; \lambda) = \begin{cases} \lambda e^{-\lambda t} & \text{for } t \geq 0 \\ 0 & \text{for } t < 0 \end{cases} \]

where,

• Parameter \(\lambda\): Called the rate parameter
  • Determines how quickly events are expected to happen.
  • Higher \(\lambda\) → more frequent events.

It is used to model: Time between rare or random events, e.g.:

• Time until a lightbulb fails
• Time until radioactive decay
• Waiting time at the DMV

Descriptive Statistics

Type	Formula
Support	\(t \in [0, \infty)\)
Rate Parameter	\(\lambda > 0\)
Mean	\(\mathbb{E}[T] = \frac{1}{\lambda}\)
Variance	\(\text{Var}(T) = \frac{1}{\lambda^2}\)
Standard Deviation	\(\sigma = \frac{1}{\lambda}\)
Probability Density Function (PDF)	\(f(t) = \lambda e^{-\lambda t} \quad for \quad t\geq 0\)
Cumulative Distribution Function (CDF)	\(F(t) = \mathbb{P}(T \leq t) = 1 - e^{-\lambda t}\)

Probability Calculations

• Infinitesimal probability between \(t\) and \(t + dt\):

  \[ P(t < T \leq t + dt) = \lambda e^{-\lambda t} dt \]

• Probability between any two times aa and bb:

  \[ P(a \leq T \leq b) = \int_a^b \lambda e^{-\lambda t} dt = e^{-\lambda a} - e^{-\lambda b} \]

Cumulative Distribution Function (CDF)

Defined as:

\[ F(t) = \int_0^t \lambda e^{-\lambda x} \, dx= \left[ -e^{-\lambda x} \right]_0^t= -e^{-\lambda t} + e^{0}= 1 - e^{-\lambda t} \]

\[ \text{Total Probability} = \int_0^\infty \lambda e^{-\lambda x} \, dx= \left[ -e^{-\lambda x} \right]_0^\infty= 0 + 1 = 1 \]

Example Use-Cases

• Lightbulb lifetime: Good for modeling how long a lightbulb might last before it fails.
• Radioactive decay: Time until a single atom decays.
• Queueing scenarios: Time until the next customer is seen.

Example 1: Lightbulb Lifetime

Average lifetime: 100 hours

Find: Probability lightbulb lasts at least 50 hours

• Given: \(\lambda = \frac{1}{100}\)
• Compute \(P(T \geq 50)\):

\[ P(T \geq 50) = 1 - P(T < 50) = 1 - F(50) = e^{-\lambda \cdot 50} = e^{-0.5} \approx 0.606 \]

Example 2: Radioactive Decay (Polonium-210)

Halflife = 138 days

Find: Value of \(\lambda\)

• Half-life definition: \(F(t) = 0.5\) when \(t = 138\)

\[ F(138) = 1 - e^{-138\lambda} = 0.5 \Rightarrow e^{-138\lambda} = 0.5 \Rightarrow -138\lambda = \ln(0.5) \Rightarrow \lambda = \frac{\ln(2)}{138} \]

Memoryless Property

One of the most profound properties of the exponential distribution:

\[ P(T > t + s \mid T > s) = P(T > t) \]

• Interpretation:
  • The probability that the event lasts at least an additional \(t\) time units given that \(s\) time units have already passed is still \(P(T > t)\)
  • Past time has no effect on the future probability.
  • The clock resets back to \(t\) if the time \(s\) has already passed
  • Only memoryless continuous distribution
• Example:
  • Lightbulb lasts 100 hours on average
  • If 100 hours have passed with no failure, the expected lifetime from that point onward is still 100 hours

Intuition Conflict

“If it was supposed to fail around 100 hours, shouldn't it be close to failure at 100?”

• No. The exponential distribution says no aging: surviving longer does not increase the risk of near failure.

The Hazard Rate

The hazard rate tells us the chance of something failing right now, assuming it has survived up to time \(t\).

Think of it like this:

If a lightbulb has already lasted 100 hours, what is the chance it fails in the next tiny moment?

For the exponential distribution, this hazard rate is constant:

\[ \text{Hazard rate} = \lambda \]

Suppose a lightbulb’s lifetime follows an exponential distribution with \(\lambda = 0.01\) (so average lifetime is 100 hours).

Even if the bulb has lasted:

• 0 hours: chance of failure in next instant = 0.01
• 100 hours: still 0.01
• 500 hours: still 0.01

The risk of failure never changes with age. That is why the exponential distribution is called memoryless.

Proof

\[ h(t) = \lim_{\Delta t \to 0} \frac{\mathbb{P}(t \leq T < t + \Delta t \mid T > t)}{\Delta t} \]

For exponential distribution:

\[ \mathbb{P}(t \leq T < t + \Delta t \mid T > t) = 1 - e^{-\lambda \Delta t} \]

Expand using Taylor series:

\[ e^{-\lambda \Delta t} \approx 1 - \lambda \Delta t + \frac{1}{2} \lambda^2 \Delta t^2 - \cdots \]

So, Probability of dying/failure in next \(\Delta t\):

\[ 1 - e^{-\lambda \Delta t} \approx \lambda \Delta t \]

Thus, hazard rate is:

\[ h(t) = \frac{\lambda \Delta t}{\Delta t} = \lambda \]

Interpretation

• The hazard rate is constant: \(h(t) = \lambda\)
• Meaning: The risk of failure per unit time is always the same, no matter how long it has survived.

Derived Formula

\[ \mathbb{P}(T \in [t, t + \Delta t] \mid T > t) \approx \lambda \Delta t \]

Or equivalently:

\[ \mathbb{P}(T \in [t, t + \Delta t]) = \lambda \cdot \mathbb{P}(T > t) \cdot \Delta t \]

Connection to the PDF

\[ f(t) = \lambda e^{-\lambda t} = \lambda \cdot \mathbb{P}(T > t) \]

This confirms:

• The PDF is the hazard rate times survival function where \(S(t)=P(T>t)\)
• A core identity in survival analysis: \(f(t) = h(t) \cdot S(t)\)
• The exponential distribution's constant hazard rate is unique real-world systems often do not behave this way (aging, wear, etc.).

• We can write down a differential equation for \(f(t)\) using the hazard rate:

  \[ \frac{d}{dt} \mathbb{P}(T > t) = -\lambda \mathbb{P}(T > t) \Rightarrow \mathbb{P}(T > t) = e^{-\lambda t} \]