# Testing that a multivariate mean approximately equals a vector of constants

Suppose we have a $$p$$-variate random vector that has a multivariate Normal distribution, $$\boldsymbol{X}\sim MVN(\boldsymbol{\mu},\Sigma)$$. My hypothesis is that the mean vector contains only zeros, $$\boldsymbol{\mu}=\boldsymbol{0}$$. I think I can test that using Hotelling's $$T^2$$.

Now, I do not really believe my model is exactly correct, but I need it to be a good enough approximation. And I have a clear measure of what is good enough. I consider three cases of such a measure:

1. $$\boldsymbol{\mu}^\top\boldsymbol{\mu},
2. $$\sum_{j=1}^p|\mu_j|,
3. $$\max_{j\in 1,\dots,p}|\mu_j|.

How would I test 1., 2. or 3.?

• Cool question! You would use a different $c$ in $(1)$ than $(2)$, right? // Particularly in $(1)$, you might consider something like (squared) Mahalanobis distance, rather than the usual notion of distance. Something similar could apply to $(2)$, but I do not know what. // I would be curious to know how robust proposed tests are to deviations for the multivariate normal assumption.
– Dave
Commented Feb 21, 2023 at 18:45
• @Dave, thanks! Yes, $c$ would be different in (1) vs. (2). I would only do one of the options, (1) or (2), not both, so I would not have to compare $c$ from one to $c$ from the other. Regarding distance, I am actually interested in (2), but I doubt it can be easily done, so I consider (1) as a simplification. I suppose it does not get simpler than that. Regarding the assumption of multivariate normality, that is a valid concern. But again, I am starting from the simplest possible case to probe what is possible. My actual problem is a little more complex than the one I have posted. Commented Feb 21, 2023 at 18:58
• I wonder if some kind of bootstrap confidence ellipsoid would be useful here. Bootstrap your data, calculate either distance metric, and calculate a confidence ellipsoid. If that ellipsoid is contained within an acceptable distance (defined by $c$), then you're within the tolerance. (This makes sense in the univariate case, right?)
– Dave
Commented Feb 21, 2023 at 19:09
• @Dave, could you elaborate just a bit? I do not quite follow. Commented Feb 21, 2023 at 19:19
• Maybe it's just a usual bootstrap. Resample your data, calculate $(1)$ or $(2)$, and repeat this over and over. Then you can use a bootstrap confidence interval of the many calculated metrics. If that confidence interval does not contain $c$, then you have statistical evidence that your vector is no more than $c$ away from the zero-vector.
– Dave
Commented Feb 21, 2023 at 19:22

My first thought is to use a bootstrap approach.

1. Take a bootstrap sample of your data.

2. Calculate and store quantity $$(1)$$, $$(2)$$, or $$(3)$$ of your bootstrap sample

3. Repeat, repeat, repeat...

Now you have a bunch of numbers. Use them to compute a confidence interval using one of the usual bootstrap methods (e.g., BCa).

If that entire confidence interval is lower than $$c$$, then you have statistical evidence that the population-level value of $$(1)$$, $$(2)$$, or $$(3)$$ is less than $$c$$. If you need a p-value, you might figure out the confidence level at which the upper limit of the confidence interval just touches $$c$$.

AS AN ASIDE, it might make more sense to take the square root of $$(1)$$. Then all three of your measures are $$L_p$$ norms and reasonably called distances between your mean vector and the zero vector. Then again, if you have reason to care about $$(1)$$ as it is written, go for it!

• I look forward to other ideas, particularly if there are analytical solutions to this and how robust this bootstrap approach is to deviations from multivariate normality.
– Dave
Commented Feb 21, 2023 at 19:49