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.
Are these variables IID?
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probability
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0Yes, 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
1 Answers
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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 $n\geqslant0$.