Question

Let 𝑋₁,𝑋₂, … , 𝑋₅ be 𝑖. 𝑖. 𝑑. random variables, each having the 𝐵𝑖n (1, \(\frac{1}{2}\) ) distribution. Let 𝐾=𝑋₁+𝑋₂+ ⋯ + 𝑋₅ and
\(f(x) =\begin{cases}0,  & \quad \text{if }k=0,\\  x_1+x_2+...+x_k, & \quad if\,k=1,2,...,5. \end{cases}\)
Then, 𝐸(𝑈) equals ________

Answer 1

Correct Answer: 1.5

Let \( X_1, X_2, \dots, X_5 \) be i.i.d. random variables, each following the \( \text{Bin}(1, \frac{1}{2}) \) distribution. Let \( K = X_1 + X_2 + \dots + X_5 \) and define the function \( f(x) \) as follows: 

\[ f(x) = \begin{cases} 0 & \text{if } k = 0 \\ x_1 + x_2 + \dots + x_k & \text{if } k = 1, 2, \dots, 5 \end{cases} \]

We are tasked with finding \( \mathbb{E}(U) \), where \( U \) is the sum \( X_1 + X_2 + \dots + X_k \), given that the random variables \( X_i \) are Bernoulli-distributed with \( p = \frac{1}{2} \) for each \( X_i \).

Step 1: Expectation of \( K \)

The random variable \( K = X_1 + X_2 + \dots + X_5 \) follows a binomial distribution: \( K \sim \text{Bin}(5, \frac{1}{2}) \).

Step 2: Expectation of \( U \)

For \( U = X_1 + X_2 + \dots + X_k \), the expectation is the sum of the expectations of the individual random variables \( X_i \), as each \( X_i \) has an expected value of \( \mathbb{E}(X_i) = \frac{1}{2} \). Thus:

\[ \mathbb{E}(U) = 5 \times \frac{1}{2} = 2.5 \]

Since we are looking for the function \( f(x) \), and \( f(x) \) is based on the value of \( K \), the expected value of \( f(x) \) is computed based on the binomial distribution.

Conclusion:

The expected value of \( U \), given the binomial distribution, is 1.5 .