# Minimum number of observation in multivariate regression

Given a multivariate regression, in a form bellow, what would be the minimum number of observations ($$n$$)? $$\mathbf{Y}=\mathbf{X}\mathbf{B}+\mathbf{E},$$ where $$\mathbf{Y}, \mathbf{X}$$ and $$\mathbf{E} \in R^{n\times m}$$ $$(n>m)$$ and $$\mathbf{B}\in R^{m \times m}$$. You can treat each column of $$\mathbf{Y}$$ and $$\mathbf{X}$$ as a time series.

Based on Frank Harrell's book, Regression Modeling Strategies, if you expect to be able to detect reasonable-size effects with reasonable power, you need 10-20 observations per parameter (covariate) estimated. I wonder if this would still hold in multivariate regression rather univariate? In the univariate the dependent variable $$Y$$ is one time series (vector format) and in multivariate one, it consists of more than one time series (matrix format).

I believe that it is better to think in terms of power to detect effects of interest, rather than rules of thumb. That said, if you want a quick-and-dirty baseline value, the standard rule of thumb is $$N=10$$ per parameter. Note that in a multivariate context, the number of parameters is greater than the number of covariates. Consider a multiple regression with one $$Y$$ and three $$X$$'s: that would be a minimum of $$N=30$$. However, if there were three $$Y$$'s, that would be $$N=90$$.
Note further that this ignores covariance parameters (people often care primarily about the means). I don't know of an analogous rule of thumb for covariances, but if you wanted reasonable estimates of the covariances, they will grow much more rapidly as the number of $$Y$$'s increases: with $$p$$ $$Y$$'s, the variance-covariance matrix will have $$^{p(p+1)}/_2$$ parameters to estimate for each $$X$$ variable. Moreover, since these depend on the means already being stable, you should think of those extra tens being added on top of the original requirement. That is, in our three $$Y$$, three $$X$$ example, you would need $$N = 60$$ (per $$X$$), times three $$X$$'s, is $$N = 180$$, plus the original ninety, is $$N=270$$. That certainly seems like a lot, but if you care about those parameters, there's a lot of them.