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 \(E(X) = \frac{1}{a}\) and the variance is \(\operatorname{var}(X) = \frac{1}{a^2}\).

Step 1: Identify the cumulative distribution function (CDF)

We know that the probability that the random variable is greater than a certain value is given by the survival function, \(P(X > x) = e^{-ax}\). The cumulative distribution function (CDF) \(F(x) = P(X \leq x)\) can be calculated as \(F(x) = 1 - P(X>x) = 1 - e^{-ax}\) for \(x \geq 0\).

Step 2: Identify the probability density function (PDF)

The probability density function \(f(x)\) is the derivative of the cumulative distribution function \(F(x)\). Therefore, \(f(x) = \frac{d}{dx}F(x) = \frac{d}{dx}(1 - e^{-ax}) = ae^{-ax}\) for \(x \geq 0\).

Step 3: Calculate the expected value \(E(X)\)

The expected value \(E(X)\) for a continuous random variable is given by \(E(X) = \int_{0}^{\infty} x f(x) \, dx\). Substituting \(f(x) = ae^{-ax}\), we have:\[E(X) = \int_{0}^{\infty} x ae^{-ax} \, dx\]This is an integral that can be solved using integration by parts. Let \(u = x\) and \(dv = ae^{-ax} \, dx\), then \(du = dx\) and \(v = -e^{-ax}\). Applying integration by parts:\[E(X) = \left[ -xe^{-ax} \right]_{0}^{\infty} + \int_{0}^{\infty} e^{-ax} \, dx\]The first term evaluates to zero at both limits, and the second integral evaluates to \(\frac{1}{a}\). Hence:\[E(X) = \frac{1}{a}\]

Step 4: Calculate \(E(X^2)\) for variance

To calculate the variance, we also need \(E(X^2)\). The formula is:\[E(X^2) = \int_{0}^{\infty} x^2 f(x) \, dx = \int_{0}^{\infty} x^2 ae^{-ax} \, dx\]Using integration by parts again, let \(u = x^2\) and \(dv = ae^{-ax} \, dx\). This results in:\[E(X^2) = +\int_{0}^{\infty} (2xe^{-ax}) \, dx = \frac{2}{a^2}\]

Step 5: Calculate the variance \(\operatorname{var}(X)\)

The variance is calculated using \(\operatorname{var}(X) = E(X^2) - (E(X))^2\). We have \(E(X) = \frac{1}{a}\) and \(E(X^2) = \frac{2}{a^2}\), so:\[\operatorname{var}(X) = \frac{2}{a^2} - \left(\frac{1}{a}\right)^2 = \frac{2}{a^2} - \frac{1}{a^2} = \frac{1}{a^2}\]

Step 6: Summary

We have determined that \(E(X) = \frac{1}{a}\) and \(\operatorname{var}(X) = \frac{1}{a^2}\). These results are consistent with expectations for an exponential distribution, which is what we have in this case with parameter \(a\).

Key Concepts

Continuous Random Variable

A continuous random variable is essentially a variable that can take on an infinite number of values within a given range. Unlike a discrete random variable, which is countable, continuous variables can take any value between two endpoints. This characteristic is crucial when working with distributions like the exponential distribution.
When working with continuous variables, we often use functions like the Probability Density Function (PDF) and Cumulative Distribution Function (CDF) to respectively describe the likelihood of specific outcomes and the probability of a variable being less than or equal to a certain value.
In the exercise example, the continuous random variable is represented by \(X\), and it is described by an exponential distribution with a survival function \(P(X > x) = e^{-ax}\), where \(a\) is a positive constant. This shows that for larger values of \(x\), the probability that \(X\) is greater than \(x\) decreases exponentially.

Expected Value

The expected value of a continuous random variable, often denoted as \(E(X)\), is a measure of the central tendency of the variable. It represents the long-term average or mean of the random variable if the experiment were repeated many times.
For a continuous random variable, the expected value is calculated using the probability density function. Integration is used to weigh all possible values of the variable by their likelihood of occurring.

• The formula for calculating the expected value for a continuous random variable is \(E(X) = \int_{0}^{\infty} x f(x) \, dx\).
• In the exercise, the probability density function \(f(x)\) is \(ae^{-ax}\). Plugging this into the formula and solving gives \(E(X) = \frac{1}{a}\), showcasing that the expected value of an exponentially distributed random variable is the reciprocal of the rate parameter \(a\).

Variance

Variance is a statistical measure that describes the spread of a set of data points around their mean value. For a continuous random variable, it helps us understand how much the variable can deviate from the expected value.
To find the variance, two calculations are necessary:

• The expected value of the random variable squared, \(E(X^2)\).
• Subtract the square of the expected value, \( (E(X))^2 \), from \(E(X^2)\).

For the exponential distribution in the exercise, \(E(X^2)\) is found using further integration by parts, resulting in \(\frac{2}{a^2}\). The variance \(\operatorname{var}(X)\) is then calculated with the formula \(\operatorname{var}(X) = E(X^2) - (E(X))^2\).
Thus, for our variable \(X\), \(\operatorname{var}(X) = \frac{2}{a^2} - \left(\frac{1}{a}\right)^2 = \frac{1}{a^2}\). This indicates that the standard deviation is the reciprocal of the parameter \(a\), signaling how data points spread around the mean.

Probability Density Function

The Probability Density Function (PDF) is a fundamental concept for continuous random variables. Unlike discrete distributions, which use probabilities for individual outcomes, PDFs describe the density of the probability across a range of values.
For a given value of \(x\), the PDF \(f(x)\) shows how densely packed the outcomes are around \(x\). However, note that the density itself is not the probability. Instead, the area under the PDF curve over a given range gives the probability that the variable falls within that range.
From the exercise, the PDF for our exponential distribution is derived as \(f(x) = ae^{-ax}\), which is the derivative of the CDF \(F(x)\). This function characterizes how the likelihood of different values of \(X\) are distributed over the interval \([0, \infty)\), scaling with the parameter \(a\). Understanding the shape and behavior of the PDF helps in assessing where the variable values are most likely to be concentrated and is an essential tool in probabilistic modeling and statistical analysis.