Problem 2 - Calculus for Biology and Medicine

Question

Show that \(f(x)=\left\\{\begin{array}{ll}\frac{1}{2} & \text { for } 0

Step-by-Step Solution

Answer

The function is a density function, and the distribution function is \( F(x) \) as defined.

Step 1: Check Non-negativity

To be a density function, \(f(x)\) must be non-negative for all \(x\). From the given function, \(f(x) = \frac{1}{2}\) for \(0 < x < 2\) and \(f(x) = 0\) otherwise. Clearly, \(f(x)\) is non-negative for all values of \(x\).

Step 2: Check Integral over Whole Space

The integral of a density function over the entire real line must equal 1. Calculate the integral of \(f(x)\) over its support:\[\int_{-\infty}^{\infty} f(x) \,dx = \int_{0}^{2} \frac{1}{2} \,dx.\]Evaluating this, we find:\[\int_{0}^{2} \frac{1}{2} \,dx = \frac{1}{2} \times [x]_{0}^{2} = \frac{1}{2} \times (2 - 0) = 1.\]Therefore, the integral is 1.

Step 3: Define Distribution Function

The distribution function \(F(x)\) is obtained by integrating the density function from \(-\infty\) to \(x\). Specifically:1. For \(x < 0\), \(F(x) = \int_{-\infty}^{x} f(t) \, dt = 0\) because \(f(x) = 0\).2. For \(0 \leq x < 2\), \(F(x) = \int_{0}^{x} \frac{1}{2} \, dt = \frac{1}{2}x\).3. For \(x \geq 2\), \(F(x) = \int_{0}^{2} \frac{1}{2} \, dt = 1\).Thus, \(F(x)\) is given by:\[F(x)=\begin{cases} 0 & \text{if } x < 0 \ \frac{1}{2}x & \text{if } 0 \leq x < 2 \ 1 & \text{if } x \geq 2 \end{cases}\]

Key Concepts

Understanding Non-negativity

In probability density functions, non-negativity is a crucial property. This means that the function should never take on negative values. After all, probabilities represent the likelihood of an event, and you can't have a negative chance of something occurring! In the provided function, we have:

• For values of \(x\) between 0 and 2, \(f(x) = \frac{1}{2}\). Clearly, \(\frac{1}{2}\) is greater than 0, so these values are non-negative.
• For values of \(x\) outside the range 0 to 2, \(f(x) = 0\). Zero is non-negative, consistent with the rule.

So, the function is non-negative for all values of \(x\), satisfying this important requirement of a probability density function.

Evaluating the Integral

A key task when working with density functions is evaluating the integral over the entire real line to ensure it sums to 1. This step validates the function as a true probability density function because the total probability across the entire space should equal 1. For our function, we only need to consider the interval from 0 to 2, as \(f(x) = 0\) elsewhere.To perform the integral: \[\int_{-fty}^{fty} f(x) \,dx = \int_{0}^{2} \frac{1}{2} \,dx\]The calculation within this interval is quite simple:\[\int_{0}^{2} \frac{1}{2} \,dx = \frac{1}{2} \times \left[ x \right]^{2}_{0} = \frac{1}{2} \times (2 - 0) = 1.\] This integral evaluates to 1, confirming that the function correctly describes a probability density distribution. Ensuring the integral sums to 1 guarantees that all possible outcomes have been accounted for.

Exploring the Distribution Function

The distribution function provides a cumulative insight into the probability density function, specifying the probability that a random variable is less than or equal to a particular value. It bridges the gap between the abstract representation of the density function and practical probability calculations. For the given function, the distribution function, \(F(x)\), is defined by integrating the density function up to \(x\). Let's break it down:

• When \(x < 0\), \(F(x) = 0\) because the probability of \(x\) being less than 0 in this range is zero.
• For \(0 \leq x < 2\), \(F(x) = \frac{1}{2}x\). This linear relation rises steadily as \(x\) increases within this range.
• If \(x \geq 2\), \(F(x) = 1\), implying that the total probability up to and exceeding this point is complete.

Thus, the distribution function neatly reflects the cumulative probability, showing how probabilities build up as \(x\) progresses. It can be crucial for calculating probabilities within specific intervals or below certain thresholds.