It is given that $\{X_i\}_{i = 1}^{\infty}$ is a sequence of i.i.d. random variables distributed as $$ \Pr(X_i=2)=1/2, \Pr(X_i=1/2)=1/4, \Pr(X_i =1/4)=1/4$$ Also let $\displaystyle Z_n = z_0\prod_{i=1}^n X_i $. If I define $W_i =\log_{2}(X_i)$, then I have $\mathbb{E}[W_i] = -1/4 $ and $\displaystyle \log_2 \left(\frac{Z_n}{z_0} \right)=\sum_{i=1}^nW_i$. It follows from the SLLN that, $$\lim_{n\to\infty} \frac{1}{n}\log_2 \left(\frac{Z_n}{z_0} \right) = -\frac{1}{4}$$ with probability 1. Noting that $\displaystyle \frac{1}{n}\log_2 \left(\frac{Z_n}{z_0} \right) = \log_2 \left(\left(\frac{Z_n}{z_0} \right)^{\frac{1}{n}} \right)$, I finally concluded that $$\lim_{n \to \infty}\left(\frac{Z_n}{z_0}\right)^{\frac{1}{n}} = \frac{1}{2}^{\frac{1}{4}}$$ with probability 1. But I feel like I am making a mistake at some point. Does this seem fine? Thanks in advance!
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How did you get from $\log_2\left(\left(\frac{Z_n}{z_0}\right)^{\frac1n}\right)$ to $\lim_{n\to\infty}\left(\frac{Z_n}{z_0}\right)^{\frac1n} = \frac12^{\frac14}$? – Math1000 Dec 18 '19 at 21:11
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I don't see any mistake. I think your answer is correct. – Kavi Rama Murthy Dec 18 '19 at 23:25