Problem 42 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The expected value \(E(Y)\) is \(\frac{1}{4}\).

Step 1: Understand the Problem

We are given three independent random variables, \(X_1, X_2, X_3\), each uniformly distributed over \((0,1)\). We need to find the expected value, \(E(Y)\), of the random variable \(Y\), where \(Y = \min(X_1, X_2, X_3)\).

Step 2: Define the Probability Expression

To find the cumulative distribution function (CDF) of \(Y\), we first calculate \(P(Y > y)\) for \(0 < y < 1\). Since \(Y = \min(X_1, X_2, X_3)\), this probability is equivalent to \(P(X_1 > y)P(X_2 > y)P(X_3 > y)\).

Step 3: Calculate \(P(X_i > y)\)

Each \(X_i\) is uniformly distributed over \((0,1)\), so \(P(X_i > y) = 1 - y\) for any \(i\). Thus, \(P(X_1 > y) = P(X_2 > y) = P(X_3 > y) = 1 - y\).

Step 4: Compute \(P(Y > y)\)

Using the independence of \(X_1, X_2, X_3\), we have \(P(Y > y) = (1-y)^3\). This is true for any \(0 < y < 1\).

Step 5: Find the CDF of \(Y\)

The cumulative distribution function of \(Y\) is \(F_Y(y) = 1 - P(Y > y) = 1 - (1-y)^3\).

Step 6: Derive the PDF of \(Y\)

To find the probability density function (PDF) of \(Y\), we differentiate its CDF with respect to \(y\). This yields \(f_Y(y) = \frac{d}{dy}[1 - (1-y)^3] = 3(1-y)^2\).

Step 7: Calculate the Expected Value \(E(Y)\)

The expected value \(E(Y)\) can be found by integrating \(y \cdot f_Y(y)\) over the interval \([0,1]\). Thus, \[E(Y) = \int_0^1 y \cdot 3(1-y)^2 \, dy.\]Computing this integral, we get \[E(Y) = \int_0^1 3y(1 - 2y + y^2) \, dy = \int_0^1 (3y - 6y^2 + 3y^3) \, dy.\]

Step 8: Evaluate the Integral

Evaluating the integral,\[E(Y) = \left[ \frac{3y^2}{2} - 2y^3 + \frac{3y^4}{4} \right]_0^1 = \frac{3}{2} - 2 + \frac{3}{4} = \frac{1}{4}.\]

Step 9: Conclude the Solution

Hence, the expected value of \(Y\), which is the minimum of three independent uniformly distributed random variables over \((0,1)\), is \(\frac{1}{4}\).

Key Concepts

Uniform Distribution

A uniform distribution is a type of probability distribution in which all outcomes are equally likely. When we say each random variable, such as \(X_1, X_2, \) and \(X_3\), is uniformly distributed over the interval \((0, 1)\), it implies that every number within this interval has an equal chance of being selected.
In mathematical terms, the probability density function (PDF) of a uniform distribution over the interval \([a, b]\) is given by: \[f(x) = \frac{1}{b-a} \text{ for } a \leq x \leq b.\] For our interval \((0, 1)\), this simplifies to: \[f(x) = 1 \text{ for } 0 \leq x \leq 1. \] This means that the probability is spread evenly across the interval and outcomes like \(X_1 > y\) are found easily by subtracting the lower interval bound from the upper interval bound.

Independent Random Variables

Independent random variables are variables where the occurrence of an event related to one variable does not affect the probability distribution of another variable. In our exercise, \(X_1, X_2,\) and \(X_3\) are such independent random variables. This is crucial for calculating probabilities because, for independent variables, the probability of combined events is just the product of their individual probabilities.
For instance, if we want to find the probability that each of \(X_1, X_2,\) and \(X_3\) is greater than some value \(y\), we calculate: \[P(Y > y) = P(X_1 > y) \times P(X_2 > y) \times P(X_3 > y).\] Thanks to their independence, this multiplication simplifies calculating the specific probability needed to solve problems like finding \(E(Y)\).

Probability Density Function

The probability density function (PDF) provides a way to specify the likelihood of a random variable to take on a specific value. For continuous random variables, the PDF describes the relative likelihood for this random variable to occur at a given point. To find the PDF from a cumulative distribution function, you differentiate the CDF with respect to \(y\). For the random variable \(Y\), defined as the minimum of \(X_1, X_2,\) and \(X_3\), the PDF was found by differentiating its CDF, resulting in: \[f_Y(y) = \frac{d}{dy}[1 - (1-y)^3] = 3(1-y)^2.\] This function defines how likely it is for the minimum value of the independent and uniformly distributed variables to take on a certain value \(y\). The PDF thus obtained is crucial for finding the expected value \(E(Y)\).

Cumulative Distribution Function

A cumulative distribution function (CDF) describes the probability that a random variable will be less than or equal to a certain value. For \(Y\), which is the minimum of several independent random variables, understanding the CDF is essential in identifying how likely it is for \(Y\) to be less than or equal to some threshold. The CDF is calculated from the probability of \(Y\) being greater than \(y\). If we have: \[P(Y > y) = (1-y)^3,\] the CDF then becomes: \[F_Y(y) = 1 - (1-y)^3.\] This expression tells us how the chances of \(Y\) taking on values up to \(y\) increase as \(y\) moves from the lower bound up to the upper bound of the interval. Differentiating this function with respect to \(y\) gave us the PDF, necessary for computing expected values.