Understanding the Law of Iterated Expectation in Probability Derivations

[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.

 

[anuttarasammyak]

Could you tell me more detail on the setting ? What are E, X_n and X_n|Xn-1 ?

 

[member 428835]

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.

 

[pbuk]

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 $E \left [ E(X_n | X_{n-1}) \right ] = E \left [ X_n \right ]$.

 

[member 428835]

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!