Understanding the Law of Iterated Expectation in Probability Derivations

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Discussion Overview

The discussion revolves around the application of the law of iterated expectation in probability derivations, specifically in the context of calculating the expected number of flips required to achieve a certain number of consecutive heads in a coin flipping scenario.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant presents a derivation involving the equation $$E(X_n | X_{n-1}) = X_{n-1} + f$$ and questions how this leads to $$E(X_n) = E(X_{n-1}) + f$$ using the law of iterated expectation.
  • Another participant requests clarification on the definitions of E, X_n, and X_n|X_{n-1} to better understand the context of the discussion.
  • A later reply reiterates the initial question about the application of the law of iterated expectation, suggesting that a link to a website might provide clearer explanations.
  • Two participants provide similar steps in their reasoning, indicating that taking the expected value of both sides leads to $$E \left [ E(X_n | X_{n-1}) \right ] = E \left [ X_n \right ]$$ as a consequence of the law of iterated expectation.
  • One participant expresses frustration and self-doubt regarding their understanding of the topic, attributing their confusion to exhaustion.

Areas of Agreement / Disagreement

The discussion does not reach a consensus, as participants are still seeking clarification and understanding of the application of the law of iterated expectation. There are multiple viewpoints regarding the derivation and its implications.

Contextual Notes

Participants have not fully defined the variables involved, and there is some ambiguity regarding the transition from the conditional expectation to the expected value. The discussion reflects varying levels of understanding among participants.

member 428835
I'm reading a website where they're doing a derivation. Within the derivation they write $$E(X_n | X_{n-1}) = X_{n-1} + f \implies E(X_n) = E(X_{n-1} ) + f$$. Evidently the implication stems from the law of iterated expectation, but I can't see how. If it helps, the question asked is "what is the expected number of flips for a coin to achieve ##n## consecutive heads.
 
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Could you tell me more detail on the setting ? What are E, X_n and X_n|Xn-1 ?
 
anuttarasammyak said:
Could you tell me more detail on the setting ? What are E, X_n and X_n|Xn-1 ?
Sorry, I realize I didn't explain this well. Rather than retype everything, and since the website is very clear, perhaps the link is easier? It's here. I'm wondering how they applied the law of iterated expectation to arrive from equation 3 to 4.
 
We are given ## E(X_n | X_{n-1}) = X_{n-1} + f ##.
Take the expected value of both sides: ## E \left [ E(X_n | X_{n-1}) \right ] = E \left [ X_{n-1} + f \right ] ##.
From the law of iterated expectation we have ## E \left [ E(X_n | X_{n-1}) \right ] = E \left [ X_n \right ] ##.
 
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pbuk said:
We are given ## E(X_n | X_{n-1}) = X_{n-1} + f ##.
Take the expected value of both sides: ## E \left [ E(X_n | X_{n-1}) \right ] = E \left [ X_{n-1} + f \right ] ##.
From the law of iterated expectation we have ## E \left [ E(X_n | X_{n-1}) \right ] = E \left [ X_n \right ] ##.
Wow, I feel like a moron. Can we just say I was exhausted and that's why I was confused? Sheesh...thanks though!
 

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