Problem 8 - Calculus for Biology and Medicine

Question

Let \(X\) be a continuous random variable with $$ P(X>x)=e^{-a x}, \quad x \geq 0 $$ where \(a\) is a positive constant. Find \(E(X)\) and \(\operatorname{var}(X)\).

Step-by-Step Solution

Answer

The expected value is \( \frac{1}{a} \) and the variance is \( \frac{1}{a^2} \).

Step 1: Identify the Cumulative Distribution Function (CDF)

Given the survival function (tail probability) as \( P(X > x) = e^{-ax} \) for \( x \geq 0 \), the CDF \( F(x) \) can be found using \( F(x) = 1 - P(X > x) \). Therefore, \( F(x) = 1 - e^{-ax} \).

Step 2: Find the Probability Density Function (PDF)

Differentiate the CDF \( F(x) = 1 - e^{-ax} \) with respect to \( x \) to obtain the PDF. This gives \( f(x) = \frac{d}{dx}[1-e^{-ax}] = ae^{-ax} \). So, the PDF is \( f(x) = ae^{-ax} \) for \( x \geq 0 \).

Step 3: Calculate the Expected Value \( E(X) \)

The expected value of \( X \) is given by \( E(X) = \int_{0}^{\infty} x f(x) \, dx \). Substituting the PDF \( f(x) = ae^{-ax} \), we have:\[ E(X) = \int_{0}^{\infty} x ae^{-ax} \, dx = \frac{1}{a} \].

Step 4: Calculate the Second Moment \( E(X^2) \)

The second moment is calculated as \( E(X^2) = \int_{0}^{\infty} x^2 f(x) \, dx \). Using the same PDF, calculate:\[ E(X^2) = \int_{0}^{\infty} x^2 ae^{-ax} \, dx = \frac{2}{a^2} \].

Step 5: Find the Variance \( \operatorname{var}(X) \)

Variance is calculated by \( \operatorname{var}(X) = E(X^2) - [E(X)]^2 \). Using previously calculated values: \[ \operatorname{var}(X) = \frac{2}{a^2} - \left(\frac{1}{a}\right)^2 = \frac{1}{a^2} \].

Key Concepts

Understanding the Cumulative Distribution Function

When dealing with continuous random variables, the cumulative distribution function (CDF) is a powerful tool. It gives us the probability that a random variable, say \( X \), will take a value less than or equal to \( x \). This is expressed as \( F(x) = P(X \leq x) \).

For the random variable in our exercise, using the given formula for the survival function \( P(X > x) = e^{-ax} \), we derive the CDF as:

• \( F(x) = 1 - P(X > x) = 1 - e^{-ax} \)

This formulation ensures the CDF ranges from 0 to 1 as \( x \) progresses from 0 to infinity. The CDF is non-decreasing, meaning it never decreases as \( x \) increases. This characteristic is crucial in defining the underlying behavior of the random variable.

Exploring the Probability Density Function

The probability density function (PDF) is closely tied to the CDF. It provides the density of probabilities at each distinct point. For our random variable, after differentiating the CDF \( F(x) = 1 - e^{-ax} \) with respect to \( x \), we find the PDF:

• \( f(x) = \frac{d}{dx}[1-e^{-ax}] = ae^{-ax} \)

Here \( f(x) \) indicates how the probability is distributed across different values of \( X \). Importantly, while probabilities for continuous distributions are measured over intervals, the PDF shows how dense these probabilities are at each point. The area under the curve within any range of \( x \) provides the probability for that range. For example, \( \int_{a}^{b} f(x) \, dx \) gives the probability that \( X \) falls between \( a \) and \( b \).

Calculating Expected Value

The expected value of a continuous random variable is akin to its mean. It gives us a measure of the central tendency or the average outcome expected from a distribution. For our exercise's random variable, using the PDF \( f(x) = ae^{-ax} \), the expected value \( E(X) \) is calculated by the integral:

\[ E(X) = \int_{0}^{\infty} x ae^{-ax} \, dx = \frac{1}{a} \]
This is essentially the mean of an exponential distribution characterized by a parameter \( a \). The integral computes this by weighing each outcome \( x \) by its probability density and summing over all possible outcomes, providing a comprehensive view of where the data clusters on average.

Understanding Variance

Variance is a critical measure of spread in a distribution, indicating how much the values deviate from the expected value. It helps us understand the variability or consistency of a random variable. For the exercise at hand, variance \( \operatorname{var}(X) \) of \( X \) is derived using:

• The formula \( \operatorname{var}(X) = E(X^2) - [E(X)]^2 \).

Using the calculations:

• \( E(X^2) = \int_{0}^{\infty} x^2 ae^{-ax} \, dx = \frac{2}{a^2} \)
• \( E(X) = \frac{1}{a} \)

We find:
\[ \operatorname{var}(X) = \frac{2}{a^2} - \left(\frac{1}{a}\right)^2 = \frac{1}{a^2} \]
This indicates how much the values of \( X \) vary around the mean and can thus provide valuable insights into the data's predictability and consistency.