Let's break down the problem step by step.

Problem Overview:

• We are given independent random variables $X_1, X_2, \ldots, X_n$, all having the same probability distribution: $P(X_i = 0) = \frac{1}{4}, \quad P(X_i = 2) = \frac{1}{4}, \quad P(X_i = 1) = \frac{1}{2}.$

• The random variable $X$ is defined as the sum of these independent variables: $X = \sum_{i=1}^n X_i.$

• The goal is to find the following:

  1. $\mathbb{E}[X_i]$ (the expectation of $X_i$),
  2. $\mathbb{E}[X]$ (the expectation of $X$),
  3. $\text{Var}(X_i)$ (the variance of $X_i$),
  4. $\text{Var}(X)$ (the variance of $X$).

1. Expectation of $X_i$ $\mathbb{E}[X_i]$

The expectation of a discrete random variable is given by: $\mathbb{E}[X_i] = \sum_{\text{values}} x \cdot P(X_i = x).$ For $X_i$, we have the following values and their probabilities: $\mathbb{E}[X_i] = (0) \cdot \frac{1}{4} + (2) \cdot \frac{1}{4} + (1) \cdot \frac{1}{2}.$ Simplifying this: $\mathbb{E}[X_i] = 0 + \frac{2}{4} + \frac{1}{2} = \frac{2}{4} + \frac{2}{4} = 1.$

2. Expectation of $X$ $\mathbb{E}[X]$

Since $X = \sum_{i=1}^n X_i$ and the $X_i$ are independent and identically distributed (i.i.d.), we can use the property of the linearity of expectation: [ \mathbb{E}[X] = \mathbb{E}\left[\sum_{i=1}^n X_i\right] = \sum_{i=1}^n \mathbb{E}[X_i]. ] Since $\mathbb{E}[X_i] = 1$, we have: $\mathbb{E}[X] = n \cdot 1 = n.$

3. Variance of $X_i$ $\text{Var}(X_i)$

The variance of a discrete random variable is given by: [ \text{Var}(X_i) = \mathbb{E}[X_i^2] - \left( \mathbb{E}[X_i] \right)^2. ] We already know $\mathbb{E}[X_i] = 1$, so we need to compute $\mathbb{E}[X_i^2]$: $\mathbb{E}[X_i^2] = (0^2) \cdot \frac{1}{4} + (2^2) \cdot \frac{1}{4} + (1^2) \cdot \frac{1}{2}.$ This simplifies to: $\mathbb{E}[X_i^2] = 0 + \frac{4}{4} + \frac{1}{2} = 1 + \frac{1}{2} = \frac{3}{2}.$ Now, using the formula for variance: $\text{Var}(X_i) = \frac{3}{2} - (1)^2 = \frac{3}{2} - 1 = \frac{1}{2}.$

4. Variance of $X$ $\text{Var}(X)$

Again, using the fact that $X = \sum_{i=1}^n X_i$, and the $X_i$'s are independent, we can use the property that the variance of a sum of independent random variables is the sum of their variances: $\text{Var}(X) = \sum_{i=1}^n \text{Var}(X_i) = n \cdot \text{Var}(X_i).$ Since $\text{Var}(X_i) = \frac{1}{2}$, we get: $\text{Var}(X) = n \cdot \frac{1}{2} = \frac{n}{2}.$

Final Answers:

1. $\mathbb{E}[X_i] = 1$,
2. $\mathbb{E}[X] = n$,
3. $\text{Var}(X_i) = \frac{1}{2}$,
4. $\text{Var}(X) = \frac{n}{2}$.

Let me know if you'd like more details or have any questions.

Here are 5 additional related questions to expand your understanding:

1. How would the variance of $X_i$ change if the probabilities were different?
2. What happens to $\mathbb{E}[X]$ and $\text{Var}(X)$ as $n \to \infty$?
3. Can you explain why the variance of the sum of independent variables is simply the sum of individual variances?
4. How would you compute the skewness or kurtosis for $X_i$?
5. How does the Central Limit Theorem apply to the sum $X = \sum_{i=1}^n X_i$?

Tip: When dealing with i.i.d. random variables, the linearity of expectation is a powerful tool that simplifies problems involving sums of these variables.