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This is a martingale problem related to a sequence of independent and identically distributed (i.i.d.) random variables. Let me walk you through how to show that the given process \( M_n = S_n^2 - n \) is a martingale with respect to the filtration \( \mathcal{F}_n \), where \( S_n = X_1 + X_2 + \cdots + X_n \) and \( \mathcal{F}_n \) is the filtration generated by the random variables \( X_1, X_2, \dots, X_n \).

### Step-by-Step Solution:

1. **Given Conditions:**
   - The random variables \( X_1, X_2, \dots \) are i.i.d. with:
     - Mean: \( \mathbb{E}[X_i] = 0 \)
     - Variance: \( \mathrm{Var}(X_i) = 1 \)
   - \( S_n = X_1 + X_2 + \dots + X_n \), so \( S_n \) is the sum of the first \( n \) variables.
   - \( M_n = S_n^2 - n \) is the process we are asked to show is a martingale.

2. **Definition of a Martingale:**
   A stochastic process \( \{M_n\}_{n \geq 0} \) is a martingale with respect to the filtration \( \{\mathcal{F}_n\} \) if:
   \[
   \mathbb{E}[M_{n+1} | \mathcal{F}_n] = M_n, \quad \forall n \geq 0
   \]
   That is, the conditional expectation of \( M_{n+1} \) given \( \mathcal{F}_n \) must equal \( M_n \).

3. **Computing \( M_{n+1} \):**
   We need to express \( M_{n+1} \):
   \[
   M_{n+1} = S_{n+1}^2 - (n+1)
   \]
   Since \( S_{n+1} = S_n + X_{n+1} \), we expand \( S_{n+1}^2 \):
   \[
   S_{n+1}^2 = (S_n + X_{n+1})^2 = S_n^2 + 2S_n X_{n+1} + X_{n+1}^2
   \]
   Therefore:
   \[
   M_{n+1} = S_n^2 + 2S_n X_{n+1} + X_{n+1}^2 - (n+1)
   \]
   Simplifying:
   \[
   M_{n+1} = (S_n^2 - n) + 2S_n X_{n+1} + (X_{n+1}^2 - 1)
   \]

4. **Taking the Conditional Expectation:**
   Now we compute the conditional expectation \( \mathbb{E}[M_{n+1} | \mathcal{F}_n] \):
   - Since \( S_n \) is \( \mathcal{F}_n \)-measurable, we have \( \mathbb{E}[S_n | \mathcal{F}_n] = S_n \).
   - Also, \( X_{n+1} \) is independent of \( \mathcal{F}_n \), and given \( \mathbb{E}[X_{n+1}] = 0 \), we have:
     \[
     \mathbb{E}[2S_n X_{n+1} | \mathcal{F}_n] = 2S_n \mathbb{E}[X_{n+1}] = 0
     \]
   - Moreover, \( \mathbb{E}[X_{n+1}^2] = 1 \), so:
     \[
     \mathbb{E}[X_{n+1}^2 - 1 | \mathcal{F}_n] = \mathbb{E}[X_{n+1}^2] - 1 = 0
     \]

   Thus:
   \[
   \mathbb{E}[M_{n+1} | \mathcal{F}_n] = S_n^2 - n = M_n
   \]

5. **Conclusion:**
   Since \( \mathbb{E}[M_{n+1} | \mathcal{F}_n] = M_n \), the process \( \{M_n\}_{n \geq 0} \) is indeed a martingale with respect to the filtration \( \{\mathcal{F}_n\} \).

Would you like a deeper explanation on any of these steps? Let me know if you'd like to go further.

### Related Questions:
1. What are the properties of a martingale in general?
2. How does the concept of a filtration relate to the definition of a martingale?
3. What are some typical applications of martingales in probability and finance?
4. How would the result change if the random variables \( X_i \) had non-zero mean?
5. Can we use this martingale to apply optional stopping theorems, and under what conditions?

**Tip:** Remember that martingales are often used in stopping time problems, especially in gambling and finance, due to their "fair game" property, where the future expectation equals the current value.

Let X_1, X_2,... be independent identically distributed random variables with mean E[X_i] = 0 and variance Var[X_i] = 1. Let S_0 = 0 and S_n = X_1 + X_2 + ... + X_n. Show that M_n = S_n^2 - n is a martingale with respect to the filtration F_n generated by X_1, X_2,...

This page provides a detailed proof showing that M_n = S_n^2 - n is a martingale with respect to the filtration F_n generated by independent and identically distributed random variables X_1, X_2, ... having zero mean and unit variance. It covers key probability theory concepts including martingales, filtration, and stochastic processes.

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