Problem 60 - Calculus for Biology and Medicine

Question

Suppose the lifetime of an organism is exponentially distributed with hazard rate function \(\lambda(x)=2 /\) day. (a) Find the probability that an individual of this species lives for more than three days. (b) What is the expected lifetime?

Step-by-Step Solution

Answer

(a) The probability is approximately 0.0025. (b) The expected lifetime is 0.5 days.

Step 1: Understand the exponential distribution

The hazard rate function given is constant: \( \lambda(x) = 2 \). This implies that the lifetime of the organism follows an exponential distribution with rate parameter \( \lambda = 2 \). The probability density function (PDF) for an exponential distribution is given by: \( f(x) = \lambda e^{-\lambda x} \).

Step 2: Calculate the probability of living more than three days

To find the probability that an individual lives more than three days, we calculate \( P(X > 3) \). For an exponential distribution, this is given by: \( P(X > t) = e^{-\lambda t} \). Substitute \( \lambda = 2 \) and \( t = 3 \): \( P(X > 3) = e^{-2 \times 3} = e^{-6} \).

Step 3: Simplify the expression for the probability

The expression \( e^{-6} \) is the exact probability that an organism lives for more than three days. We can compute this value using a calculator: \( e^{-6} \approx 0.0025 \).

Step 4: Determine the expected lifetime

The expected value for an exponential distribution with rate parameter \( \lambda \) is given by: \( E(X) = \frac{1}{\lambda} \). For \( \lambda = 2 \), the expected lifetime \( E(X) \) is \( \frac{1}{2} = 0.5 \) days.

Key Concepts

Hazard Rate Function

In an exponential distribution, the hazard rate function, often denoted as \( \lambda(x) \), represents the instantaneous rate at which events occur. For the lifetime of an organism, this function gives us an instantaneous risk of failure per unit time. In this specific problem, the hazard rate is constant, \( \lambda(x) = 2 \), meaning every day, the organism has the same chance of dying as any other day.

This constant hazard rate indicates that the process is memoryless. This property implies that the future probability of survival is not dependent on how long an organism has already lived. Each day stands equal in the risk perspective, and this characteristic is crucial for understanding the exponential distribution.

For practical purposes, a constant hazard rate simplifies calculations, as it allows us to use the straightforward formula for probabilities or expected values, without adjusting for time passed.

Probability Density Function

The Probability Density Function (PDF) for an exponential distribution describes the likelihood of various outcomes for continuous random variables, particularly useful for modeling time until an event occurs.

For an exponential distribution with a hazard rate \( \lambda \), the PDF is expressed as \( f(x) = \lambda e^{-\lambda x} \), \( x \geq 0 \). In this scenario, with \( \lambda = 2 \), the PDF becomes \( f(x) = 2e^{-2x} \).

This PDF indicates that:

• The most probable values for the time until an event are when \( x \) is small, as the probability diminishes as \( x \) increases due to the \( e^{-\lambda x} \) term.
• The PDF function gradually slopes down, indicating quickly diminishing probabilities for larger lifespans when the rate is high, such as our example where \( \lambda = 2 \).

By integrating the PDF over a range (ex: day 3 onward), you find the probability that the organism survives past certain times, using its complement for calculations like \( P(X > 3) \).

Expected Value

The expected value of a random variable gives us an average outcome over multiple trials of an experiment or process. In the context of an exponential distribution, the expected value signifies the average time until the event occurs.

For an exponential distribution with hazard rate \( \lambda \), the expected value, \( E(X) \), is calculated using the formula \( E(X) = \frac{1}{\lambda} \). This relation emerges from the integral properties of the PDF.

In our given exercise, with \( \lambda = 2 \), the expected lifetime of the organism is \( E(X) = \frac{1}{2} = 0.5 \) days. This indicates that, on average, the organism lives for half a day given the constant rate. It's essential to note:

• The expectation inversely depends on the rate \( \lambda \), meaning higher rates lead to shorter expected lifetimes.
• Expectation provides a means to compare lifetimes across various organisms or processes given different rates.

Understanding this concept is fundamental in predicting outcomes for processes following an exponential distribution.