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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.? – Did Oct 25 '11 at 22:08
Nevermind, I worded it very poorly and I know what I need to do now. How do I delete the question? – DumbQuestion Oct 25 '11 at 22:08
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? – leonbloy Oct 25 '11 at 22:10
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? – André Nicolas Oct 25 '11 at 22:13
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 $log(X_n)/n$ – DumbQuestion Oct 25 '11 at 22:15

1 Answer

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Let $Z_n=u(Y_n)$ with $u(y)=\log(ap+(1-p)y)$. Since $(Y_n)_{n\geqslant1}$ is i.i.d., $(Z_n)_{n\geqslant1}$ is i.i.d. Furthermore, the sequence $(X_n)_{n\geqslant0}$ is not independent in general, neither is the sequence $(\log X_n)_{n\geqslant0}$, but $\log X_n=\log(X_0)+\sum\limits_{k=1}^nZ_k$ for every $n\geqslant0$.

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