10
votes
multinomial covariance matrix is singular?
You should have expected this, since for the multinomial distribution $\sum_{i=1}^k X_i = n$, so have variance zero. But let us also show this explicitly, with linear algebra.
Let $\Sigma_0 = \frac1n \...
- 82.8k
6
votes
Accepted
multinomial covariance matrix is singular?
There is a general answer that requires no special calculation and provides insight into how random variables, their distributions, and covariances are inter-related.
Let's begin with some definitions ...
- 334k
2
votes
Accepted
Covariance matrix in terms of $X^TX$
In this notation with an expectation $$\text{Cov}(X) = \mathbb{E}[(X-\mu_X)(X-\mu_X)^T]$$ and $$\text{Cov}(X) = \mathbb{E}[XX^T]$$ is seems like $X$ is a vector instead of a matrix.
$\mathbf{X}$ is a ...
- 86.4k
2
votes
If $n\operatorname{var}( \sum_{ij}M_{ij}v_{i}v_{j}) = (\sum_{i}v_{i}^{2})^{2} - \sum_{i}v_{i}^{4}$ for any $v_i$, what can we say about $M_{ij}$?
We start from the equality
$$\operatorname{var}\left( \sum_{ij}M_{ij}v_{i}v_{j} \right)
= \frac{1}{n} \left[ \left( \sum_{i}v_{i}^{2} \right)^{2} - \sum_{i}v_{i}^{4} \right]$$
This is an equality of ...
- 4,552
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