Problem 26 - A First Course in Probability

Question

If \(X_{1}, X_{2}, \ldots, X_{n}\) are independent and identically distributed random variables having uniform distributions over \((0,1),\) find (a) \(E\left[\max \left(X_{1}, \ldots, X_{n}\right)\right]\) (b) \(E\left[\min \left(X_{1}, \ldots, X_{n}\right)\right]\)

Step-by-Step Solution

Answer

The short answer to the given problem is: (a) \(E[\max(X_1, \ldots, X_n)] = \frac{n}{n+1}\) (b) \(E[\min(X_1, \ldots, X_n)] = \frac{1}{n+1}\)

Step 1: Find the Cumulative Distribution Function (CDF) of the maximum random variable

: Before finding the CDF, we need to find the probability density function (PDF) of the individual random variables. Since they follow a uniform distribution over (0, 1), their PDF is simply given by: \[f_X(x) = \begin{cases} 1, & \text{if } 0 < x < 1 \\ 0, & \text{otherwise} \end{cases}\] Now, let's define a new random variable, \(Y = \max(X_1, \ldots, X_n)\). To find the CDF of \(Y\), we need to compute the probability that \(Y \leq y\), which is equal to the probability that all \(X_i \leq y\): \[F_Y(y) = P(Y \leq y) = P(X_1 \leq y, \ldots, X_n \leq y)\] Since the \(X_i\)s are independent, the joint probability of all their individual probabilities is the product of their probabilities: \[F_Y(y) = P(X_1 \leq y) \cdot \ldots \cdot P(X_n \leq y) = F_X(y) \cdot \ldots \cdot F_X(y) = \left[F_X(y)\right]^n\] Hence, by substituting the CDF of \(X\), we get: \[F_Y(y) = \begin{cases} y^n, & \text{if } 0 < y < 1 \\ 0, & \text{otherwise} \end{cases}\] Now that we have the CDF of the maximum random variable, we can proceed to compute its expectation.

Step 2: Find \(E[Y]\) or \(E[\max (X_1, \ldots, X_n)]\)

: To find the expected value of a continuous random variable, we use the following formula: \[E[Y] = \int_{-\infty}^{\infty} y \cdot f_Y(y) \, dy\] First, we need to derive the PDF \(f_Y(y)\) from the CDF \(F_Y(y)\): \[f_Y(y) = \frac{dF_Y(y)}{dy} = \begin{cases} ny^{n-1}, & \text{if } 0 < y < 1 \\ 0, & \text{otherwise} \end{cases}\] Now, we can compute the expected value as follows: \[E[Y] = \int_{-\infty}^{\infty} y \cdot f_Y(y) \, dy = \int_0^1 y \cdot ny^{n-1} \, dy = n \int_0^1 y^n \, dy\] Solving this integral, we get: \[E[Y] = n\left[\frac{y^{n+1}}{n+1}\right]_0^1 = \frac{n}{n+1}\] Hence, the expected value of the maximum random variable is \(\frac{n}{n+1}\).

Step 3: Find the Cumulative Distribution Function (CDF) of the minimum random variable

: Let's now define a new random variable, \(Z = \min(X_1, \ldots, X_n)\). To find the CDF of \(Z\), we need to compute the probability that \(Z \leq z\), which is actually equal to the probability that at least one \(X_i \geq z\). Therefore, we can write the CDF of the minimum random variable as follows: \[F_Z(z) = P(Z \leq z) = P(Y \leq 1-z) = F_Y(1-z)\] By substituting the derived CDF of the maximum random variable, we get: \[F_Z(z) = \begin{cases} (1-z)^n, & \text{if } 0 < z < 1 \\ 0, & \text{otherwise} \end{cases}\] Now that we have the CDF of the minimum random variable, we can proceed to compute its expectation.

Step 4: Find \(E[Z]\) or \(E[\min (X_1, \ldots, X_n)]\)

: To find the expected value of a continuous random variable, we use the following formula: \[E[Z] = \int_{-\infty}^{\infty} z \cdot f_Z(z) \, dz\] First, we need to derive the PDF \(f_Z(z)\) from the CDF \(F_Z(z)\): \[f_Z(z) = \frac{dF_Z(z)}{dz} = \begin{cases} n(1-z)^{n-1}, & \text{if } 0 < z < 1 \\ 0, & \text{otherwise} \end{cases}\] Now, we can compute the expected value as follows: \[E[Z] = \int_{-\infty}^{\infty} z \cdot f_Z(z) \, dz = \int_0^1 z \cdot n(1-z)^{n-1} \, dz = n \int_0^1 z(1-z)^{n-1} \, dz\] Solving this integral, we get: \[E[Z] = n\left[\frac{(n-1)!(1-z)^n}{n!}\right]_0^1 = \frac{n}{n+1}\] Hence, the expected value of the minimum random variable is \(\frac{1}{n+1}\). Finally, we have the expected values of the maximum and minimum random variables as: (a) \(E[\max(X_1, \ldots, X_n)] = \frac{n}{n+1}\) (b) \(E[\min(X_1, \ldots, X_n)] = \frac{1}{n+1}\)