1
$\begingroup$

I'm trying to understand the definition of the wishart distribution. In wiki,

$X_{(i)}{=}(x_i^1,\dots,x_i^p)\sim N_p(0,V).$ What do they mean by this? Each component is drawn from a univariate $N(0,V)$? I was reading it as a multivariate, but then some lines later they write when $p=V=1$ we have a chi-squared distribution.

Any help would be appreciated.

$\endgroup$
0
$\begingroup$

I think that it is a multivariate, $V$ is the variance-covariance matrix for the multivariate normal distribution ($0$ describes a vector of $0$ I think). If $p = 1$ then it is not a multivariate case anymore, hence you can get to a univariate chi-squared distribution.

$\endgroup$
2
$\begingroup$

Yes, those are $p$-dimensional vectors and $V$ is a $p \times p$-dimensional positive definite matrix.

The article's insistence on explaining the Wishart in terms of a correspondence to the chi-squared distribution is sort of strange to me, since the chi-squared is just a special case of the gamma, and the Wishart is more properly a multivariate generalization of the gamma. You can see the correspondence pretty cleanly when you disregard normalizing constants and use a 'rate' parameterization rather than scale.

Consider $w \sim \text{gamma}(\alpha, \beta)$, which has the following density:

\begin{equation} f(w) \propto w^{\alpha - 1} e^{-\beta w} \end{equation}

Compare it to the corresponding density for $\mathbf{W} \sim \text{Wishart}(\alpha, \mathbf{B})$, where $\mathbf{W} \in \mathbb{R}^{p \times p}$, and $\mathbf{B} \in \mathbb{R}^{p \times p}$ is positive-definite:

\begin{equation*} f(\mathbf{W}) \propto \det(\mathbf{W})^{\alpha - \frac{(p + 1)}{2}} e^{-\text{tr}(\mathbf{BW})} \end{equation*}

If $p = 1$ then both $\mathbf{W}$ and $\mathbf{B}$ are scalars, so $\det(\mathbf{W}) = \mathbf{W}$ and $\text{tr}(\mathbf{BW}) = \mathbf{BW}$. The result is

\begin{equation*} f(\mathbf{W}) \propto \mathbf{W}^{\alpha - 1} e^{-\mathbf{BW}} \end{equation*}

which is equivalent to the gamma density given previously.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.