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The Bayesian Multivariate Linear Regression wikipedia article states that posterior parameters of a multivariate linear regression model with a Inverse-Wishart prior can be obtained from the following equations:

$$ \mathbf{V_n}=\mathbf{V_0}+(\mathbf{Y}-\mathbf{XB_n})^{\rm T}(\mathbf{Y}-\mathbf{XB_n})+(\mathbf{B_n}-\mathbf{B_0})^{\rm T}\boldsymbol\Lambda_0(\mathbf{B_n}-\mathbf{B_0}) $$ $$ \boldsymbol\nu_n=\boldsymbol\nu_0+n $$ $$ {B_n}=(\mathbf{X}^{\rm T}\mathbf{X}+\boldsymbol\Lambda_0)^{-1}(\mathbf{X}^{\rm T}\mathbf{Y}+\boldsymbol\Lambda_0\mathbf{B_0}) $$ $$ \boldsymbol\Lambda_n=\mathbf{X}^{\rm T}\mathbf{X}+\boldsymbol\Lambda_0 $$

However, from this equations we get that the dimensions of the parameter $V_n$ would be 1x1, meaning that the scale matrix of the Inverse-Wishart distribution, and therefore the sampled covariance matrix is a scalar.

Is this correct? And if it is, what is the difference of using the Inverse-Wishart prior instead of the Inverse-Gamma distribution?

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In the article you link to, the dimensions of $\mathbf{Y}$ are given as $n \times m$ (where $m$ is the number of outcome variables). From this (and consistent with the other matrix dimensions listed in the article), it follows that the $(\mathbf{Y}-\mathbf{XB_n})^{\rm T}(\mathbf{Y}-\mathbf{XB_n})$ term must have dimensions $m\times m$, and so those must also be the dimensions of $\mathbf{V_n}$.

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