Problem 78 - An Introduction to Mathematical Statistics and Its Applications

Question

Recall Question \(3.4 .4\), where the length of time \(Y\) (in years) that a malaria patient spends in remission has pdf \(f_{Y}(y)=\frac{1}{9} y^{2}, 0 \leq y \leq 3\). What is the average length of time that such a patient spends in remission?

Step-by-Step Solution

Answer

The average length of time that a malaria patient spends in remission is 2.25 years.

Step 1: Understand the formula for the expected value

The formula for the expected value or mean (\(E[Y]\)) of a continuous random variable \(Y\) is given by \(E[Y]=\int_{-\infty}^{+\infty} y f_{Y}(y) dy\).

Step 2: Integrate to find the expected value

First recall that the pdf of \(Y\) is given by \(f_{Y}(y)=\frac{1}{9} y^{2}, 0 \leq y \leq 3\). Then we integrate the product of this pdf and \(y\) over its given range to find the expected value: \(E[Y] = \int_{0}^{3} y \cdot \frac{1}{9} y^{2} dy\).

Step 3: Simplify and solve the integral

Solve this integral. Notice that we can simplify \(y \cdot \frac{1}{9} y^{2}\) to \( \frac{1}{9} y^{3}\). Then \(E[Y] = \frac{1}{9} \int_{0}^{3} y^{3} dy = \frac{1}{9} \cdot [\frac{1}{4} y^{4}]_{0}^{3} = \frac{1}{9} \cdot [\frac{3^{4}}{4} - 0]\).

Step 4: Compute the expected value

Finally, compute the expected value: \(E[Y] = \frac{1}{9} \cdot \frac{81}{4} = \frac{81}{36} = 2.25\) years.

Key Concepts

Continuous Random Variable

When studying probability and statistics, a continuous random variable is a type of variable that takes an infinite number of possible values. Unlike discrete random variables that list out all possible outcomes, continuous variables can take any value within a given range.
A good example is measuring time, like in the original exercise. Here, the time a malaria patient spends in remission is measured in years and can be any value between 0 and 3.
This concept allows us to model and analyze situations where different outcomes are not countable but rather measured on a continuous scale. Some characteristics of continuous random variables include:

• They are real-valued, meaning they are associated with numerical values or measurements.
• The probability of the variable assuming a specific value is zero; instead, probabilities are assigned over intervals.
• They require a probability density function (pdf) to specify the distribution of probabilities over intervals on the variable's range.

Understanding continuous random variables is essential for comprehending topics like expected values and probability distributions.

Probability Density Function

A Probability Density Function (pdf) is crucial when dealing with continuous random variables. It's designed to provide a way to understand how probabilities are distributed over different values of the variable.

The pdf helps to determine the likelihood of the variable falling within a specific interval. For example, in the exercise, the pdf, given as \( f_{Y}(y) = \frac{1}{9} y^{2} \) for \( 0 \leq y \leq 3 \), describes how the probability is spread across the possible remission years.Some key points about pdfs include:

• The area under the pdf curve within an interval gives the probability that the random variable falls within that interval.
• The total area under the pdf curve is always equal to 1, representing the total probability.
• To find probabilities for specific intervals, you integrate the pdf over that interval.

Mastering the use of pdfs is vital for effectively working with continuous random variables and calculating expected values.

Integration in Probability Theory

Integration is a mathematical process that is often used in probability theory to find various statistical measures, such as expected values. It involves finding the area under a curve which, in the context of probability, can represent a pdf.
This is exactly what you do when calculating the expected value of a continuous random variable, as shown in the original exercise.The process includes:

• Identifying the variable's pdf—like \( f_{Y}(y) = \frac{1}{9} y^{2} \).
• Multiplying this pdf by the random variable to form an integrable expression.
• Performing the integration over the given range of the variable, which in the exercise was from 0 to 3.

For example, to find the expected value \( E[Y] \), you'd calculate \( \int_{0}^{3} y \cdot f_{Y}(y) \, dy \), which simplifies to evaluating an integral of the form \( \frac{1}{9} \int_{0}^{3} y^{3} \, dy \).
Through integration, you derive a numerical value that represents the average or mean outcome expected from the distribution of values across the random variable's possible range.