I Understanding the Law of Iterated Expectation in Probability Derivations

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The discussion centers on the application of the law of iterated expectation in a probability derivation involving expected values of random variables. The key equation discussed is E(X_n | X_{n-1}) = X_{n-1} + f, leading to E(X_n) = E(X_{n-1}) + f. Participants clarify that taking the expected value of both sides allows the use of the law of iterated expectation, resulting in E[E(X_n | X_{n-1})] = E[X_n]. The confusion expressed by the original poster highlights the complexity of understanding these concepts, particularly in the context of calculating the expected number of flips for achieving n consecutive heads. Overall, the thread emphasizes the importance of the law of iterated expectation in deriving expected values in probability.
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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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