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 valid density function and the distribution function is \( F(x) = \begin{cases} 0 & \text{if } x < 0, \\ \frac{1}{2}x & \text{if } 0 \leq x < 2, \\ 1 & \text{if } x \geq 2. \end{cases} \)

Step 1: Verify Non-Negativity

For a function to be a density function, it must be non-negative for all real numbers. Here, \( f(x) = \frac{1}{2} \) for \( 0 < x < 2 \) and \( f(x) = 0 \) otherwise. Since both these values are non-negative, the function fulfills this criterion.

Step 2: Check that Total Area Integrates to 1

The integral of the density function over its entire range should equal 1. Therefore, we need to evaluate \[ \int_{-\infty}^{\infty} f(x) \, dx = \int_{0}^{2} \frac{1}{2} \, dx. \] Evaluate the integral: \( \int_{0}^{2} \frac{1}{2} \, dx = \left[ \frac{1}{2}x \right]_0^2 = \frac{1}{2} \times 2 - \frac{1}{2} \times 0 = 1. \) Thus, the total area under the curve \( f(x) \) equals 1, confirming it as a density function.

Step 3: Define the Distribution Function

The distribution function \( F(x) \) is defined as the integral of \( f(x) \) from \(-\infty\) to \( x \). We calculate \( F(x) \) in parts: 1. **For** \( x < 0 \), \( F(x) = 0 \).2. **For** \( 0 \leq x < 2 \), evaluate \( F(x) = \int_{0}^{x} \frac{1}{2} \, dt = \frac{1}{2}x. \)3. **For** \( x \geq 2 \), \( F(x) = 1 \). Hence, \( F(x) \) 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

Distribution Function

In the realm of probability, the distribution function is crucial for understanding how probabilities distribute over outcomes. For a density function like our example, the distribution function, denoted as \( F(x) \), is established by integrating the density from \(-\infty\) to \( x \). This process accumulates the probabilities up to \( x \).

To articulate \( F(x) \) in different scenarios, we express it in a piecewise function:

• \( F(x) = 0 \) when \( x < 0 \). Before reaching the starting point (0), there's no accumulated probability.
• For \( 0 \leq x < 2 \), the integral calculates as \( F(x) = \frac{1}{2}x \), representing a linear increase in probability as \( x \) grows from 0 to just below 2.
• When \( x \geq 2 \), \( F(x) = 1 \), showing that the entire probability is accumulated beyond 2, as our density does not extend past this bound.

The shape of \( F(x) \) generally rises from 0 to 1, aligning with the total probability of the sample space.

Non-Negativity

A fundamental property of a probability density function (pdf) is its non-negativity over all possible values. This ensures that probabilities are never less than zero. After all, a probability can't be negative. For the given function, \( f(x) = \frac{1}{2} \) when \( 0 < x < 2 \) and \( f(x) = 0 \) otherwise. Both values conform to non-negativity as they are either zero or positive.

When verifying non-negativity:

• Check that for any \( x \), \( f(x) \ge 0 \).
• Within the defined range, maintain positivity or zero values.

This stipulation is essential for a valid probability density function, ensuring that the model aligns with real-world interpretations of probability.

Integration over Range

Integration over the function's range is pivotal to ensure it sums to 1. This condition signifies that all probability is accounted for, indicating no missing or excess probability.

We compute

• \( \int_{-\infty}^{\infty} f(x) \, dx = \int_{0}^{2} \frac{1}{2} \, dx \).
• This integral evaluates to \( \left[ \frac{1}{2}x \right]_0^2 = 1 \), confirming the total probability is complete and equals 1.

A valid density must integrate to 1, defining the total probability across its range. Calculating the integration not only confirms completeness but also validates that the pdf is properly normalized over its domain.

Calculus

Calculus plays a vital role in understanding and working with probability functions, particularly through the use of integration and differentiation. The integral helps in accumulating probabilities as seen in distribution functions, while derivatives might be used in more advanced topics like finding expectations or other statistical measures.

In our context,

• Integration converts the density function into a distribution function, \( F(x) \), showing probabilities accrued up to \( x \).
• Calculus assures that transformations between density and distribution conserve properties like total area (or probability).

Mastering the calculus involved in probability concepts allows us to better model and predict outcomes, addressing complex problems with mathematical precision.