Math Problem Statement

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,...

Solution

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 Mn=Sn2nM_n = S_n^2 - n is a martingale with respect to the filtration Fn\mathcal{F}_n, where Sn=X1+X2++XnS_n = X_1 + X_2 + \cdots + X_n and Fn\mathcal{F}_n is the filtration generated by the random variables X1,X2,,XnX_1, X_2, \dots, X_n.

Step-by-Step Solution:

  1. Given Conditions:

    • The random variables X1,X2,X_1, X_2, \dots are i.i.d. with:
      • Mean: E[Xi]=0\mathbb{E}[X_i] = 0
      • Variance: Var(Xi)=1\mathrm{Var}(X_i) = 1
    • Sn=X1+X2++XnS_n = X_1 + X_2 + \dots + X_n, so SnS_n is the sum of the first nn variables.
    • Mn=Sn2nM_n = S_n^2 - n is the process we are asked to show is a martingale.
  2. Definition of a Martingale: A stochastic process {Mn}n0\{M_n\}_{n \geq 0} is a martingale with respect to the filtration {Fn}\{\mathcal{F}_n\} if: E[Mn+1Fn]=Mn,n0\mathbb{E}[M_{n+1} | \mathcal{F}_n] = M_n, \quad \forall n \geq 0 That is, the conditional expectation of Mn+1M_{n+1} given Fn\mathcal{F}_n must equal MnM_n.

  3. Computing Mn+1M_{n+1}: We need to express Mn+1M_{n+1}: Mn+1=Sn+12(n+1)M_{n+1} = S_{n+1}^2 - (n+1) Since Sn+1=Sn+Xn+1S_{n+1} = S_n + X_{n+1}, we expand Sn+12S_{n+1}^2: Sn+12=(Sn+Xn+1)2=Sn2+2SnXn+1+Xn+12S_{n+1}^2 = (S_n + X_{n+1})^2 = S_n^2 + 2S_n X_{n+1} + X_{n+1}^2 Therefore: Mn+1=Sn2+2SnXn+1+Xn+12(n+1)M_{n+1} = S_n^2 + 2S_n X_{n+1} + X_{n+1}^2 - (n+1) Simplifying: Mn+1=(Sn2n)+2SnXn+1+(Xn+121)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 E[Mn+1Fn]\mathbb{E}[M_{n+1} | \mathcal{F}_n]:

    • Since SnS_n is Fn\mathcal{F}_n-measurable, we have E[SnFn]=Sn\mathbb{E}[S_n | \mathcal{F}_n] = S_n.
    • Also, Xn+1X_{n+1} is independent of Fn\mathcal{F}_n, and given E[Xn+1]=0\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, E[Xn+12]=1\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: E[Mn+1Fn]=Sn2n=Mn\mathbb{E}[M_{n+1} | \mathcal{F}_n] = S_n^2 - n = M_n

  5. Conclusion: Since E[Mn+1Fn]=Mn\mathbb{E}[M_{n+1} | \mathcal{F}_n] = M_n, the process {Mn}n0\{M_n\}_{n \geq 0} is indeed a martingale with respect to the filtration {Fn}\{\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 XiX_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.

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Math Problem Analysis

Mathematical Concepts

Martingales
Filtration
Stochastic Processes
Probability Theory

Formulas

E[X_i] = 0
Var[X_i] = 1
S_n = X_1 + X_2 + ... + X_n
M_n = S_n^2 - n
E[M_{n+1} | F_n] = M_n

Theorems

Martingale Property
Conditional Expectation
Law of Total Expectation

Suitable Grade Level

University-Level Probability Theory