Problem 9 - Calculus for Biology and Medicine

Question

Let \(X\) be a continuous random variable with density function $$ f(x)=\left\\{\begin{array}{cl} (a-1) x^{-a} & \text { for } x>1 \\ 0 & \text { for } x \leq 1 \end{array}\right. $$ (a) Show that \(E(X)=\infty\) when \(a \leq 2\). (b) Compute \(E(X)\) when \(a>2\).

Step-by-Step Solution

Answer

(a) \(E(X) = \infty\) for \(a \leq 2\). (b) \(E(X) = \frac{a-1}{a-2}\) for \(a > 2\).

Step 1: Define Expected Value

The expected value of a continuous random variable \(X\) with density function \(f(x)\) is given by the integral \(E(X) = \int_{-\infty}^{\infty} x f(x) \, dx\). In this exercise, since the density function is only non-zero for \(x > 1\), we need to calculate \(E(X) = \int_{1}^{\infty} x f(x) \, dx\).

Step 2: Substitute and Simplify

For \(x > 1\), we have \(f(x) = (a-1)x^{-a}\). Substitute \(f(x)\) into the integral to find \(E(X)\):\[E(X) = \int_{1}^{\infty} x (a-1) x^{-a} \, dx = (a-1) \int_{1}^{\infty} x^{1-a} \, dx.\]

Step 3: Integrate

Calculate the integral \((a-1) \int_{1}^{\infty} x^{1-a} \, dx\). The antiderivative of \(x^{1-a}\) is \(\frac{x^{2-a}}{2-a}\), valid for \(a eq 2\).\[E(X) = (a-1) \left[ \frac{x^{2-a}}{2-a} \right]_{1}^{\infty},\] which evaluates to:\[= (a-1) \left( \lim_{b \to \infty} \frac{b^{2-a}}{2-a} - \frac{1}{2-a} \right).\]

Step 4: Evaluate Limit and Consider Cases

Evaluate \(\lim_{b \to \infty} b^{2-a}\):- If \(a \leq 2\), then \(2-a \geq 0\) and \(b^{2-a} \to \infty\), so \(E(X) = \infty\).- If \(a > 2\), then \(2-a < 0\) and \(b^{2-a} \to 0\), so \(E(X)\) simplifies to \(\frac{a-1}{2-a}.\)

Step 5: Show Results

- For \(a \leq 2\), \(E(X) = \infty\), confirming the statement for part (a).- For \(a > 2\), compute the simplified expression: \[E(X) = \frac{a-1}{a-2}.\] This confirms part (b) of the question.

Key Concepts

Expected Value

The expected value, often denoted as \(E(X)\), is a measure of the center or average of a probability distribution for a random variable \(X\). When \(X\) is a continuous random variable, the expected value is calculated using an integral of the form:

• \(E(X) = \int_{-inf}^{inf} x f(x) \, dx\)

Here, \(f(x)\) is the probability density function (PDF) of the random variable. This integral represents a sort of weighted average, where each possible value of \(x\) is weighted by the probability (density) of that value.
In our problem, the integral simplifies to:
\(E(X) = \int_{1}^{inf} x (a-1) x^{-a} \, dx\)
This integral helps us determine the expected value based on the density function provided. The specific outcome of this integral can give us insights into the behavior of \(X\), such as when certain parameters like \(a\) are varied. The results are different depending on whether \(a \leq 2\) or \(a > 2\), which highlights the sensitivity of the expected value to the distribution's characteristics.

Density Function

A density function, or probability density function (PDF), is a function that describes the relative likelihood for a continuous random variable \(X\) to take on a given value. For our exercise, the density function is defined as:

• \(f(x)=\left\{\begin{array}{cl}(a-1) x^{-a} & \text { for } x>1 \ 0 & \text { for } x \leq 1 \end{array}\right.\)

This function shows that the probability is concentrated in the region where \(x > 1\). The PDF gives us crucial information about how \(X\) behaves over different intervals.
Key properties of density functions include:

• The area under the curve of the PDF over its range must equal 1, ensuring that \(X\) has a probability distribution.
• The PDF value \(f(x)\) can't be negative, since probabilities are non-negative.

The density function we've provided allows us to compute the expected value by integrating the function and exploring how \(X\) behaves with respect to the parameter \(a\). Such computations give us meaningful insights into the distribution's characteristics and behavior over its range.

Integral Calculus

Integral Calculus is one of the two main branches of calculus, and it focuses on the accumulation of quantities. In the context of probability and statistics, it's crucial for calculating things like expected values for continuous random variables.
Here's how it works for this exercise:

• We use the integral \(\int_{1}^{\infty} x (a-1) x^{-a} \, dx\) to find \(E(X)\).
• This involves computing the antiderivative of the function, which can require substitution and algebraic manipulation.

The primary goal here is to use calculus to find accumulations over an interval, which is exactly what expected value represents.
A key part of solving the exercise is evaluating the limit of the antiderivative as \(x\) approaches infinity, potentially leading to different results based on the exponent \(2-a\). Using the antiderivative \(\frac{x^{2-a}}{2-a}\), we plug in the limits and subtract, which reveals whether the expected value converges or diverges.
This process, known as evaluating improper integrals, is critical in handling cases where a PDF does not naturally terminate at a finite point. Understanding these operations is essential for interpreting probabilities in continuous settings.