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The problem asks to find a value of \(\alpha\) such that the expression \(\alpha X_1 + (1 - 2\alpha) X_2\) serves as an unbiased estimator of \(p\), where the given probability distribution for \(X_1\) (and \(X_2\), since they are independent and identically distributed) is:

\[
P(X_1 = k) = C_2^k p^k (1 - p)^{2 - k}, \quad k = 0, 1, 2
\]

This is a binomial distribution for \(n = 2\), where \(C_2^k\) is the binomial coefficient.

### Steps to solve:

1. **Expectation of \(X_1\)**:
   The random variable \(X_1\) follows a binomial distribution with parameters \(n = 2\) and \(p\). So, the expected value of \(X_1\) is:
   \[
   \mathbb{E}[X_1] = 2p
   \]
   This comes from the general expectation formula of a binomial distribution, \(\mathbb{E}[X] = n p\).

2. **Expectation of \(X_2\)**:
   Since \(X_1\) and \(X_2\) are identically distributed, the expectation of \(X_2\) is the same:
   \[
   \mathbb{E}[X_2] = 2p
   \]

3. **Expectation of the linear combination**:
   The expression given is \(\alpha X_1 + (1 - 2\alpha) X_2\). To be an unbiased estimator for \(p\), the expectation of this expression must equal \(p\). Thus:
   \[
   \mathbb{E}[\alpha X_1 + (1 - 2\alpha) X_2] = p
   \]

   Using the linearity of expectation:
   \[
   \mathbb{E}[\alpha X_1 + (1 - 2\alpha) X_2] = \alpha \mathbb{E}[X_1] + (1 - 2\alpha) \mathbb{E}[X_2]
   \]
   Substituting \(\mathbb{E}[X_1] = 2p\) and \(\mathbb{E}[X_2] = 2p\):
   \[
   \alpha \cdot 2p + (1 - 2\alpha) \cdot 2p = p
   \]
   Simplifying:
   \[
   2p (\alpha + 1 - 2\alpha) = p
   \]
   \[
   2p (1 - \alpha) = p
   \]
   Dividing both sides by \(p\) (assuming \(p \neq 0\)):
   \[
   2(1 - \alpha) = 1
   \]
   Solving for \(\alpha\):
   \[
   1 - \alpha = \frac{1}{2}
   \]
   \[
   \alpha = \frac{1}{2}
   \]

Thus, \(\alpha = \frac{1}{2}\) is the value that makes \(\alpha X_1 + (1 - 2\alpha) X_2\) an unbiased estimator of \(p\).

---

Would you like more details on any step?

Here are five related questions:
1. How do we derive the expectation formula for a binomial distribution?
2. What are the conditions for an estimator to be unbiased?
3. How does the linearity of expectation work in more complex distributions?
4. How would the solution change if the distribution of \(X_1\) were different?
5. How can we estimate \(p\) using maximum likelihood estimation?

**Tip:** In problems involving estimators, checking for bias is crucial to ensure the estimator provides accurate results on average.

Независимая выборка из распределения с формулой P(X_1 = k) = C_2^k p^k (1 - p)^{2 - k}, k = 0,1,2. Найти α, при котором αX_1 + (1 - 2α)X_2 является несмещенной оценкой для p.

This problem focuses on finding the value of α such that the linear combination αX_1 + (1 - 2α)X_2 is an unbiased estimator of p, given a binomial distribution. The solution involves using the expectation formula for a binomial distribution and the concept of unbiased estimation.

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