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: Understanding Uniform Distribution

Each random variable \(X_1, X_2,\) and \(X_3\) is independent and uniformly distributed over \((0,1)\). This means that the probability density function (PDF) for each \(X_i\) is \(f(x) = 1\) for \(0 < x < 1\) and \(0\) otherwise.

Step 2: Derive the Probability \(P(Y > y)\)

Since \(Y = \min(X_1, X_2, X_3)\), the event \(Y > y\) implies that \(X_1, X_2,\) and \(X_3\) are all greater than \(y\). For a uniform distribution, \(P(X_i > y) = 1-y\) when \(0 < y < 1\). Since the variables are independent, \(P(Y > y) = P(X_1 > y)P(X_2 > y)P(X_3 > y) = (1-y)^3\).

Step 3: Find the Cumulative Distribution Function (CDF) of \(Y\)

The CDF \(F_Y(y)\) is given by \(F_Y(y) = P(Y \leq y) = 1 - P(Y > y)\). Thus, \(F_Y(y) = 1 - (1-y)^3\) for \(0 < y < 1\).

Step 4: Derive the Probability Density Function (PDF) of \(Y\)

The PDF \(f_Y(y)\) is the derivative of the CDF \(F_Y(y)\). Calculate this by differentiating \(F_Y(y) = 1 - (1-y)^3\) to get \(f_Y(y) = 3(1-y)^2\).

Step 5: Calculate the Expected Value \(E(Y)\)

The expected value \(E(Y)\) is obtained by integrating \(yf_Y(y)\) over \((0, 1)\). \[ E(Y) = \int_0^1 y \, 3(1-y)^2 \, dy. \] Use integration by parts or substitution to solve the integral: \[ E(Y) = 3 \left[y - 2y^2 + \frac{y^3}{3} \right]_0^1 = \frac{1}{4}. \]

Step 6: Conclusion

Thus, the expected value of \(Y\) is \(\frac{1}{4}\).

Key Concepts

Uniform Distribution

A uniform distribution is a type of probability distribution where all outcomes are equally likely within the specified range. In this context, the random variables \(X_1, X_2,\) and \(X_3\) are uniformly distributed over the interval \((0, 1)\). This means each point within this interval has the same probability of occurring.

• Probability Density Function (PDF): For a uniform distribution over \((0, 1)\), the PDF is given by \(f(x) = 1\) for \(0 < x < 1\) and \(f(x) = 0\) otherwise. This reflects the even spread of probability.
• Interpretation: If you imagine drawing a number from this interval at random, any number between 0 and 1 is equally likely.

Cumulative Distribution Function

The cumulative distribution function (CDF) is a tool that tells us the probability that a random variable will take a value less than or equal to a certain level. For the minimum of \(X_1, X_2,\) and \(X_3\) (denoted as \(Y\)), we determined that the CDF is: \[ F_Y(y) = 1 - (1-y)^3 \] This formula was derived by first finding the probability that \(Y\) is greater than \(y\) and then using the relationship \(F_Y(y) = 1 - P(Y > y)\).

• How it works: The CDF accumulates the probabilities from the lowest value up to \(y\), giving a bit more insight into how the distribution distributes its values over the possible range.
• Usage: The CDF is particularly useful when calculating probabilities for ranges or when deriving related concepts, such as the probability density function.

Probability Density Function

The probability density function (PDF) provides a convenient way to evaluate the likelihood of a random variable coming close to a specific value. For the random variable \(Y\), or the minimum of our set \(X_1, X_2, X_3\), the PDF is derived from its CDF:\[ f_Y(y) = 3(1-y)^2 \]This formula comes from differentiating the CDF of \(Y\). Understanding the PDF helps in accurately identifying how the probabilities are distributed around specific values in the distribution.

• Shape & Insight: The formula \(3(1-y)^2\) results in a PDF that is skewed toward values closer to 0, which makes sense because \(Y\) represents the minimum of these random variables.
• Relevance: The PDF is vital for finding expected values and variances. It essentially maps out how density—or probability—is spread out over different values of \(Y\).

Independent Random Variables

Independent random variables are those whose outcomes do not affect each other. For instance, knowing the value of one variable does not give any information about another.

• Explanation: In the given problem, \(X_1, X_2,\) and \(X_3\) are independent, uniformly distributed random variables. The assumption of independence simplifies the calculation of probabilities, such as \(P(Y > y)\) because we can multiply the individual probabilities due to independence.
• Example: If \(P(X_i > y) = 1 - y\), and all \(X_i\) are independent, then \(P(X_1 > y, X_2 > y, X_3 > y) = (1-y)^3\), leading to our expressions for the CDF and subsequent calculations.
• Significance: Understanding independence is crucial for correctly setting up and solving probability problems that involve multiple variables, as it affects how we can combine their respective properties.