2
$\begingroup$

Suppose $S$ is distributed as a Wishart matrix with $n$ degrees of freedom and scale matrix $\Sigma$, and let $\vec{a}$ be a fixed vector. It is well known that $\vec{a}^{\top}S\vec{a}$ is equal to $\vec{a}^{\top}\Sigma\vec{a}$ times a Chi-square random variable with $n$ degrees of freedom.

I am curious about the distribution of $\vec{a}^{\top}S^{-1}\vec{a}$. My guess was that it would be disributed as an inverse gamma, with scale $\vec{a}^{\top}\Sigma^{-1}\vec{a} / 2$, and shape $n/2$. My simulations (which are somewhat questionable) indicate this is not the case. Does this follow a well-known distribution?

$\endgroup$
  • 1
    $\begingroup$ You might be interested in Chapter 8 from M. L. Eaton (2007), Multivariate statistics: A vector-space approach, Inst. Math. Stat. Lecture Notes, Monograph series. See, in particular, Proposition 8.9 on pages 312ff. $\endgroup$ – cardinal Oct 23 '11 at 21:51
2
$\begingroup$

It does have an inverse gamma distribution, but not that one: it is actually an inverse gamma with parameters $\vec{a}^\top \Sigma^{-1} \vec{a} /2$ and $(n-p+1)/2$, where $p$ is the dimension of $S$.

The confusion is due to the marginal distribution of submatrices of the inverse Wishart. For example, if you choose $\vec{a} = (1,0,\ldots,0)^\top$, then: $$ \vec{a}^\top S^{-1} \vec{a} = [S^{-1}]_{11} $$ using the parameterisation of the Wikipedia page, this has a $W^{-1}\big( [\Sigma^{-1}]_{11},n-(p-1) \big)$ distribution.

Note that some papers, such as this one, use a different parameterisation of the inverse Wishart, such that if $U \sim IW(\delta; \Phi)$, then $U_{11} \sim IW(\delta; \Phi_{11})$, so you should check which one an author is using.

$\endgroup$
1
$\begingroup$

Here is the simulation confirming @Simon Byrne's answer:

#demonstrate a'S a is inverse gamma when S is inverse Wishart W^-1(n,Sg^-1),
#for vector a, with scale a'Sg^-1 a / 2...
rwish <- function(n,Sg) {
  #generate variates from W(n,Sg) wishart distribution, the slow way
  #rwishart from bayesm library is probably faster
  C <- chol(Sg)
  p <- dim(Sg)[1]
  X <- matrix(rnorm(n*p),nrow=n) %*% C
  Z <- t(X) %*% X
}
rinvwishprod <- function(n,a,Sg=diag(x=1,nrow=length(a),ncol=length(a))) {
  #a' X a, where X is inv-wishart with parameters n and Sg^-1
  X <- rwish(n,Sg)
  t(a) %*% solve(X,a)
}
set.seed(1729)
n <- 256
p <- 16
ntrial <- 2048
#generate a PSD matrix Sg
X <- matrix(rnorm(2*p*p),nrow=2*p,ncol=p)
Sg <- t(X) %*% X
#generate vector a
a <- matrix(rnorm(p),nrow=p)
#draw many variates from this process
many.rv <- replicate(ntrial,rinvwishprod(n,a,Sg))
#try to transform to uniform RV using the inverse gamma law
alpha <- (n - p + 1) / 2
beta <- (t(a) %*% solve(Sg,a)) / 2
is.it.uniform <- pgamma(1 / many.rv,shape=alpha,rate=beta)
#plot the results
plot(ecdf(is.it.uniform))

Basically I generate, for fixed $\Sigma, n, \vec{a}$, variates of the form $v = \vec{a}^{\top}S^{-1}\vec{a}$, where $S \sim \mbox{W}\left(n,\Sigma\right)$, then compute the CDF under the inverse gamma relationship (shape parameter $\alpha = (n - p + 1)/2$ and scale parameter $\beta = \vec{a}^{\top}\Sigma^{-1}\vec{a}/2$). The resultant 'p-value' should be uniform, which is confirmed visually: p-value under inverse wishart to inverse gamma

$\endgroup$

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.