https://file.aiquickdraw.com/mathbot/file/1728854353676hap6xrcq.jpg

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.

Let X1,...,Xn be independent random variables with the same distribution given by P(Xi = 0) = P(Xi = 2) = 1/4 and P(Xi = 1) = 1/2. Moreover, we introduce X = sum(Xi). Determine E[Xi], E[X], V[Xi], V[X].

Solve the problem involving independent random variables X1, X2, ..., Xn with given probabilities, finding the expectation and variance of both individual variables and their sum. This detailed solution covers core probability concepts suitable for undergraduate students.

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