2
$\begingroup$

it's known the conjugate priors for multivariate normal distribution are the normal & inverse-whishart distributions. but i'm interest in very specific case where the correlation matrix is $R$ multiply by random variable $\kappa^2$ with inverse-gamma prior summed with $\sigma^2 I$ . I'm trying to get the closed-form posterior from an inverse gamma prior and a likelihood based on a multivariate normal distribution expecting to get inverse gamma posterior but I don't have any success . I appreciate it if you could help me get the posterior. the model:

$$ Y | \kappa \sim \mathcal{N}_j(0,( \kappa^2 R + \sigma^2 I))$$ $$ \kappa \sim \text{InverseGamma} (A_0, B_0)$$

Where $A_0$, $B_0$, and $\sigma^2$ are known constants and $R$ is $j$ by $j$ known matrix we can even say it is rank(1) matrix . $I$ is a $j$ by $j$ identity matrix.

$\endgroup$
1
  • $\begingroup$ I wouldn't be surprised if there is no closed form posterior. $\endgroup$ Apr 3, 2017 at 13:45

1 Answer 1

4
$\begingroup$

[This answer refers to a previous version of the question with the covariance matrix being the inverse of what it is now]

Since the distribution of $Y$ given $\kappa$ is an exponential family, as a sample $(y_1,\ldots,y_n)$ from $\mathcal{N}_j(0,( \kappa^2 R + \sigma^2 I)^{-1})$ has density $$|\kappa^2 R + \sigma^2 I|^{n/2}\exp-\frac{1}{2}\text{tr}\left\{ (\kappa^2 R + \sigma^2 I)\sum_{i=1}^n Y_iY_i^t\right\}$$ i.e., $$|\kappa^2 R + \sigma^2 I|^{n/2}\exp-\frac{1}{2}\left\{ \kappa^2 \text{tr}\left(R\sum_{i=1}^n Y_iY_i^t\right) + \sigma^2 \text{tr}\sum_{i=1}^n Y_iY_i^t\right\}$$ there exists a conjugate family on $\kappa$ but not of the inverse Gamma variety.

[This answer refers to the new version of the question]

Since the distribution of $Y$ given $\kappa$ is a curved exponential family, as a sample $(y_1,\ldots,y_n)$ from $\mathcal{N}_j(0,( \kappa^2 R + \sigma^2 I)^{-1})$ has density $$|\kappa^2 R + \sigma^2 I|^{-n/2}\exp-\frac{1}{2}\text{tr}\left\{ (\kappa^2 R + \sigma^2 I)^{-1}\sum_{i=1}^n Y_iY_i^t\right\}$$ which factorises through the sufficient statistic $\sum_{i=1}^n Y_iY_i^t$, there exists a conjugate family on $\kappa$ but not of the inverse Gamma variety, since this family is actually $j(j+1)/2$ dimensional.

$\endgroup$
2
  • $\begingroup$ actually the density suppose to by like that $$|\kappa^2 R + \sigma^2 I|^{-n/2}\exp-\frac{1}{2}\text{tr}\left\{ (\kappa^2 R + \sigma^2 I)^{-1}\sum_{i=1}^n Y_iY_i^t\right\}$$ it is possible to use woodbary decomposition but still what is the family of $\kappa$? thanks $\endgroup$ Apr 3, 2017 at 18:54
  • $\begingroup$ yes you right i edit the question. $\endgroup$ Apr 4, 2017 at 8:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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