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We define $X_{n+1}=X_n(ap+(1-p)Y_{n+1})$, where $\{Y_n\}$ are IID. Is $\{\log(X_n)\}$ IID, and if so how do I show it? $\log(X_{n+1})=\log(X_n)+\log(ap+(1-p)Y_{n+1})$ and I'm not sure where to go from there.

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    Here is a question. Let $(Z_n)$ be i.i.d. and $S_n=Z_1+\cdots+Z_n$. Do you think the random variables $S_n$ are i.i.d.?2011-10-25
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    Nevermind, I worded it very poorly and I know what I need to do now. How do I delete the question?2011-10-25
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    Calling $Z_n = \log(X_n)$ , it's quite clear that they are not independent ($Z_{n+1}$ depends on $Z_n$). Are you sure you got it right?2011-10-25
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    To get some intuition, take an extreme case, $Y_k$ identically $0$ for all $k$. (Yes, they are independent.) Then $X_{n+1}=apX_n$. Take $X_0$ be $1$ or $2$, each with probability $1/2$. Are $X_0$ and $X_1$ independent? Are their logarithms?2011-10-25
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    Yes, I was asking the wrong question. I was meaning to ask if $\{log(ap+(1-p)Y_{n})\}$ was IID so I could apply SLLN to $log(X_n)/n$2011-10-25

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