Problem 41 - Calculus for Biology and Medicine

Question

Suppose that \(X_{1}, X_{2}\), and \(X_{3}\) are independent and uniformly distributed over \((0,1) .\) Define $$ Y=\max \left(X_{1}, X_{2}, X_{3}\right) $$ Find \(E(Y) .\) [Hint: Compute \(P(Y \leq y)\), and use it to deduce the density of \(Y .]\)

Step-by-Step Solution

Answer

The expected value \(E(Y)\) is \(\frac{3}{4}\).

Step 1: Understand the Distribution and Define Y

The random variables \(X_1, X_2, X_3\) are independent and uniformly distributed on \((0, 1)\). This means that each \(X_i\) has a probability density function \(f(x) = 1\) for \(0 < x < 1\). We define \(Y = \max(X_1, X_2, X_3)\). Our goal is to find \(E(Y)\), the expected value of \(Y\).

Step 2: Calculate Cumulative Distribution Function (CDF)

To find the expected value of \(Y\), we first need to find its cumulative distribution function \(P(Y \leq y)\). For \(Y \leq y\) to hold, all \(X_i\) must be less than or equal to \(y\). Since \(X_i\) are independent, \(P(X_1 \leq y, X_2 \leq y, X_3 \leq y) = P(X_1 \leq y) \cdot P(X_2 \leq y) \cdot P(X_3 \leq y)\). For a uniform distribution, \(P(X_i \leq y) = y\) for \(0 \leq y \leq 1\). Therefore, \(P(Y \leq y) = y^3\).

Step 3: Derive the Probability Density Function (PDF)

The PDF of \(Y\), denoted by \(f_Y(y)\), is the derivative of the CDF, \(P(Y \leq y)\). Thus, we calculate \(f_Y(y) = \frac{d}{dy}[y^3] = 3y^2\), valid for \(0 < y < 1\).

Step 4: Compute the Expected Value

The expected value of \(Y\) is given by the integral \(E(Y) = \int_0^1 y \cdot f_Y(y) \, dy = \int_0^1 y \cdot 3y^2 \, dy\). Simplifying, the integral becomes \(E(Y) = 3 \int_0^1 y^3 \, dy\).

Step 5: Solve the Integral

Evaluate the integral: \(3 \int_0^1 y^3 \, dy = 3 \left[ \frac{y^4}{4} \right]_0^1 = 3 \left( \frac{1^4}{4} - \frac{0^4}{4} \right) = \frac{3}{4}\). Thus, the expected value, \(E(Y)\), is \(\frac{3}{4}\).

Key Concepts

Uniform Distribution

The uniform distribution is a type of probability distribution where every outcome has an equal chance of occurring. In the context of the exercise, the random variables \(X_1, X_2,\) and \(X_3\) are independent and uniformly distributed over the interval \((0,1)\). This interval representation means that every value between 0 and 1 is equally probable for each \(X_i\). The probability density function (PDF) for a uniform distribution across an interval \((a,b)\) is given by:

• \(f(x) = \frac{1}{b-a}\) if \(a < x < b\)
• \(f(x) = 0\) otherwise

For \(X_i\), each \(f(x) = 1\) since \(b-a = 1-0 = 1\). This simply tells us that the likelihood of \(X_i\) taking any particular value in \((0, 1)\) is uniform across the whole interval. Understanding this helps in finding the expected value of \(Y = \max(X_1, X_2, X_3)\), as we need to apply the uniform distribution to determine how probable it is for all these variables to be below a certain threshold to compute \(P(Y \leq y)\).

Probability Density Function

The probability density function (PDF) of a continuous random variable is a function that describes the likelihood of the variable taking on a particular value. For the uniformly distributed random variable \(Y = \max(X_1, X_2, X_3)\), the PDF is crucial because it helps us understand how likely different values of \(Y\) are, along the interval \((0,1)\). In this exercise, we first find the PDF of \(Y\) by differentiating its cumulative distribution function (CDF).
Given that the CDF of \(Y\) is \(P(Y \leq y) = y^3\), we find the PDF by taking the derivative:

• \(f_Y(y) = \frac{d}{dy}[y^3] = 3y^2\) for \(0 < y < 1\)

This tells us that the likelihood of \(Y\) taking a specific value \(y\) is proportional to \(3y^2\). Knowing the PDF allows us to calculate the expected value, \(E(Y)\), by integrating the product of \(y\) and \(f_Y(y)\) over the interval \((0,1)\).

Cumulative Distribution Function

The cumulative distribution function (CDF) of a random variable is a function that describes the probability that the variable is less than or equal to a certain value. For \(Y = \max(X_1, X_2, X_3)\), finding \(P(Y \leq y)\) helps us understand the behavior of \(Y\) by showing how likely it is for \(Y\) to be below a given threshold \(y\).
Since the \(X_i\)'s are independent and uniformly distributed, the CDF for \(Y\) is calculated using the product of the probabilities that each \(X_i\) is \(\leq y\):

• \(P(Y \leq y) = P(X_1 \leq y) \cdot P(X_2 \leq y) \cdot P(X_3 \leq y) = y \cdot y \cdot y = y^3\)

This calculation assumes all \(X_i\) values are independently less than \(y\), making the CDF an essential tool in obtaining the PDF and solving for the expected value of \(Y\). The CDF provides insight into the distribution shape of \(Y\), which is pivotal for further statistical analysis.