Skip to main content
6 events
when toggle format what by license comment
Aug 18, 2021 at 11:32 vote accept DM-97
Aug 18, 2021 at 10:58 comment added DM-97 Thanks. Yeah, it seems surprising at first that $\log(\bar{X_n})$ and $\log(X^n)$ have the same variance, but I guess that actually makes a lot of sense, since $\log(\bar{X_n})=\log(\frac{1}{n}X^n)=\log(X^n)-\log(n)$.
Aug 16, 2021 at 23:48 answer added Geoffrey Johnson timeline score: 1
Aug 16, 2021 at 23:31 comment added Geoffrey Johnson Yes, this looks correct to me. Interestingly, I also get $\frac{\sigma^2}{n\mu^2}$ as the asymptotic variance for $\text{log}(\bar{X})$. I suppose the best way to double check all of this is through simulation. I'll give it a try...
Aug 16, 2021 at 22:01 history edited DM-97 CC BY-SA 4.0
fixed a mistake; changed notation to avoid confusion
Aug 16, 2021 at 21:37 history asked DM-97 CC BY-SA 4.0