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Let $D(\mu, \Sigma)$ represent an arbitrary distribution with mean $\mu$ and covariance $\Sigma$.

I am given that $\beta_k$ is a vector of length $M_k$: $$\beta_k \sim D\left(0, \frac{\sigma^2_k}{M_k} I_{M_k} \right)$$ where I am given the value $\sigma^2_k$.

What exactly does this mean? Are we saying that the covariance of the $M_k$ elements in $\beta_k$ is is given by $\frac{\sigma^2_k}{M_k}$? In that case, where would the matrix $I_{M_k}$ come into play? This doesn't really seem like a distribution then?

Thanks in advance for any help!

mathz2003
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  • Statement is confusing. If D is one dimensional, $\Sigma$ is variance. What are $\beta_k$? Terms need to be defined. – herb steinberg Jul 09 '22 at 22:13
  • I am pretty sure $D$ is a multivariate distribution. The $0$ is the mean vector and the $\frac{\sigma^2_k}{M_k} I_{M_k}$ is the variance matrix. Here, $\sigma^2_k$ is a number ranging from $0$ to $1$ and $M_k$ is some whole number. So, $\beta_k$ is a vector of length $M_k$. What I'm not understanding is what exactly an arbitrary distribution $D$ looks like and how to generate a $\beta_k$ from such an arbitrary distribution. – mathz2003 Jul 09 '22 at 22:34
  • Context is needed to clarify. – herb steinberg Jul 10 '22 at 17:43

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