Skip to main content
7 events
when toggle format what by license comment
Jan 31, 2019 at 11:18 vote accept Kagaratsch
Jan 31, 2019 at 11:17 comment added Kagaratsch I guess you mean that $I_n$ is the matrix $\delta_{i,j}$, while $1_{n\times n}$ is matrix of full dimension with all elements filled by the same number $1$. Honestly, without knowing what we want to represent $Y_i=X_i-\bar X_n=X_i-\frac{1}{n}\sum_i^nX_i$, I would not have guessed that from the notation... ^^
Jan 31, 2019 at 11:07 comment added Ben Those are $n \times n$ matrices, so they operate as linear functions $\mathbb{R}^n \rightarrow \mathbb{R}^n$.
Jan 31, 2019 at 11:06 comment added Kagaratsch I see, makes sense, it is only really zero for $n\to\infty$.
Jan 31, 2019 at 11:06 comment added Ben Zero mean of the distribution doesn't lead to zero sample mean. Instead you have $\bar{X}_n \sim \text{N}(0, \sigma^2/n)$.
Jan 31, 2019 at 11:03 comment added Kagaratsch Thank you, this looks very good! Especially knowing the variance and correlations is useful. I'm a bit confused with the notation though, could you perhaps clarify which vector spaces the $I_n$ and $1_{n\times n}$ act on? And shouldn't the sample mean $\bar X_n$ vanish due to zero mean of all initial distributions?
Jan 31, 2019 at 2:29 history answered Ben CC BY-SA 4.0