Nataly Best

2022-01-22

A quick question; is it possible to say in a way analogous to the single variable case that a multivariable function is "asymptotically equivalent" to a second multivariable function? For example, consider the function of ${n}_{1},{n}_{2}\in \mathbb{R}$ given by
$\text{Var}\left(\stackrel{^}{\mu }\right)=\frac{{\sigma }^{2}\left({n}_{1}+2{n}_{2}\right)}{{\left({n}_{1}+{n}_{2}\right)}^{2}}.$
where ${\sigma }^{2}$ is a constant.
Can we say that $\text{Var}\left(\stackrel{^}{\mu }\right)\approx \frac{1}{{n}_{1}+{n}_{2}}$ and then conclude that $\text{Var}\left(\stackrel{^}{\mu }\right)\to 0$ as ${n}_{1}\to \mathrm{\infty }$ and ${n}_{2}\to \mathrm{\infty }$?
$\underset{\left(x,y\right)\to \left(\mathrm{\infty },\mathrm{\infty }\right)}{lim}\frac{x+2y}{{\left(x+y\right)}^{2}}$
does not exist. Am I wrong to think of $\text{Var}\left(\stackrel{^}{\mu }\right)$ as a function of two variables?

utgyrnr0

The two previous answers are perfectly right, thus I just would like it emphasize the statistical perspective. Note that when you are talking about a variance of an estimator, n1 and n2 are sample sizes, hence they are (strictly) positive integers, i.e., ${n}_{1},{n}_{2}\in \mathbb{N}$. As such, when you take the limit, you cannot consider any possible route (as Wolfram does) in ${\mathbb{R}}^{2}$. The extreme cases in this variance are ${n}_{1}>>{n}_{2}$ or ${n}_{2}>>{n}_{1}$, that dont

Karli Kaiser

You can conclude $Var\left(\mu \right)\to 0$ as ${n}_{1}\to \mathrm{\infty }$ and ${n}_{2}\to \mathrm{\infty }$, but you can't say $Var\left(\mu \right)\approx \frac{1}{{n}_{1}+{n}_{2}}$ because when ${n}_{1}>>{n}_{2}$ it is approximatly $\frac{1}{{n}_{1}+{n}_{2}}$ but when ${n}_{1}<<{n}_{2}$ it is near $\frac{2}{{n}_{1}+{n}_{2}}$

Do you have a similar question?