Math Problem Statement

If T1 is the random variable for the first time the process reaches height 1, use an appropriate law of total probability involving X1 to show that P (T0 = k) = P (T1 = k − 1), k = 2, 3, . . . .

Solution

To show that P(T0=k)=P(T1=k1)P(T_0 = k) = P(T_1 = k - 1) for k=2,3,k = 2, 3, \ldots, we can use an appropriate application of the law of total probability involving X1X_1, the first step of the process.


Explanation:

  1. Define the random variables:

    • T0T_0: The first time the process reaches height 00.
    • T1T_1: The first time the process reaches height 11.
    • X1X_1: The value of the process after the first step.

    The law of total probability allows us to break down the probabilities of T0T_0 and T1T_1 by conditioning on the first step X1X_1.

  2. Key property of the process: Suppose the process starts at height 11. Then:

    • To reach height 00 for the first time at step kk, the process must reach height 11 for the first time at step k1k - 1.

    Mathematically, this means: {T0=k}={T1=k1}\{T_0 = k\} = \{T_1 = k - 1\}

  3. Transitioning from height 1: When the process starts at height 11, we consider the contribution of all possible outcomes of X1X_1 (the value of the process after the first step).

    By conditioning on X1X_1, the law of total probability gives: P(T0=k)=xP(T0=kX1=x)P(X1=x)P(T_0 = k) = \sum_{x} P(T_0 = k \mid X_1 = x) \cdot P(X_1 = x)

  4. Connection to T1T_1: Similarly, the probability P(T1=k1)P(T_1 = k - 1) can be written as: P(T1=k1)=xP(T1=k1X1=x)P(X1=x)P(T_1 = k - 1) = \sum_{x} P(T_1 = k - 1 \mid X_1 = x) \cdot P(X_1 = x)

    Since {T0=k}\{T_0 = k\} and {T1=k1}\{T_1 = k - 1\} describe the same event under the key property above, their probabilities are equal: P(T0=k)=P(T1=k1)P(T_0 = k) = P(T_1 = k - 1)

  5. Range of kk: This relationship holds for k=2,3,k = 2, 3, \ldots because for k=1k = 1, T0T_0 does not align with T1+1T_1 + 1.


Final Justification:

The equality P(T0=k)=P(T1=k1)P(T_0 = k) = P(T_1 = k - 1) follows directly from the recursive structure of the process and the law of total probability applied to the outcomes of X1X_1, ensuring the first event of height 11 corresponds to the first event of height 00 shifted by one step.

Would you like more detailed steps, or an example to illustrate this?


Follow-Up Questions:

  1. Would you like an example of a stochastic process (e.g., a random walk) to solidify this concept?
  2. Do you want the derivation broken down further in terms of conditional probabilities?
  3. Would you like an explanation of the significance of the range k=2,3,k = 2, 3, \ldots?
  4. Should I illustrate how the law of total probability applies specifically to T0T_0 and T1T_1?
  5. Do you want to extend this result to other height levels or general stopping times?

Tip: Always start by identifying the key recursive properties of the random process—this makes it easier to apply probabilistic tools effectively!

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Math Problem Analysis

Mathematical Concepts

Probability Theory
Stochastic Processes
Random Variables

Formulas

Law of Total Probability: P(A) = ∑ P(A | B_i) * P(B_i)
Key Equality: P(T0 = k) = P(T1 = k - 1)

Theorems

Law of Total Probability
Recursive Structure of Stochastic Processes

Suitable Grade Level

Undergraduate (Probability and Statistics)