I am trying to calculate a posterior predictive distribution for the magnitude (Euclidean norm) of a 3D displacement vector. Displacement in each dimension is independent and normally distributed (but each may have different means and variances).

$$ X_i \sim N(\mu_i,\sigma_i^2); (i = 1:3) $$

$$ D = \sqrt{\sum\limits_{i=1}^3 X_i^2} $$

There is plenty of literature on how to deal with normally distributed data with both frequentist and bayesian approaches.

However the magnitude isn't normally distributed, and I believe it should follow a noncentral chi distribution. This is defined by number of degrees of freedom k = 3, and the lambda parameter, which combines the ratio of the mean and standard deviation for each dimension (en.wikipedia.org/wiki/Noncentral_chi_distribution).

$$ \lambda = \sqrt{\sum\limits_{i=1}^k \left(\frac{\mu_i}{\sigma_i}\right)^2} $$

I don't know any population parameters - but I have informative priors on the parameters governing displacement in each dimension, and would like to use them:

$$ \mu_i \sim N(m_i,a_i^2) $$ $$ \sigma_i^2 \sim N(s_i^2,b_i^2) $$ Where $m_i$ and $s_i^2$ are my best guesses for the mean and variances of data in each displacement dimension, and $a_i^2$ and $b_i^2$ are hyperparameters quantifying my uncertainty in these beliefs.

How can I convert prior information on the gaussian displacement parameters into a prior on the lambda parameter of the non-central chi distribution?

$$ \lambda \sim f(m_i,a_i^2,s_i^2,b_i^2)? $$

Next I will combine this prior with observed data (either in each dimension, or the magnitude directly) to obtain a posterior distribution for lambda. I will then integrate over all lambda to get my posterior predictive distribution for the magnitude. I think this is a sensible approach, but I need a decent prior first.

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    $\begingroup$ @Xi'an Edited to add equations - I've done my best (I'm not very familiar with LaTeX), hopefully that makes the question clearer. $\endgroup$
    – Arun
    Commented Jan 6, 2017 at 10:27

1 Answer 1


Note that

  1. the distribution of $D^2$ does not depend directly on $\lambda$ but rather on $||\mathbf{\mu}||$ and $(\sigma_1,\ldots,\sigma_k)$. For instance,$$\mathbb{E}[D^2]=||\mathbf{\mu}||^2+\sum_{i=1}^k \sigma^2_i$$
  2. this distribution is definitely not a non-central $\chi^2$ distribution (*) as explained in this SE answer

Since $\lambda$ is a transform of $(\mu_1,\sigma_1,\ldots,\sigma_3)$, $$\lambda=T(\mu_1,\sigma_1,\ldots,\sigma_3)$$ the prior density of $\lambda$ is the (mathematical) projection of the joint density of $(\mu_1,\sigma_1,\ldots,\sigma_3)$, which is given by the Jacobian formula. If this proves impossible to derive in a closed-form formula, a simulation-based representation can be achieved by simulating from the joint prior and constructing the resulting $\lambda$'s.

If one observes the $X_i$'s (rather than $D$), the posterior of $(\mu_1,\sigma_1,\ldots,\sigma_3)$ given the $X_i$'s is available in closed form. And can again be simulated to produce a projected posterior on $\lambda$.

If one only observes the norm $D$, the posterior of $(\mu_1,\sigma_1,\ldots,\sigma_3)$ given $D$ is harder to manage if only because the distribution of $D$ is not standard and I would suggest treating the $X_i$'s as missing data in a simulation that would then condition on those $X_i$'s [in a Gibbs sampler] to simulate $(\mu_1,\sigma_1,\ldots,\sigma_3)$.

(*) The reason why $D^2$ is not distributed as a non-central $\chi^2$ variate is that it can be expressed as $$D^2=\sum_{i=1}^3 X_i^2=\sum_{i=1}^3 (\mu_i+\sigma_i\epsilon_i)^2\qquad\epsilon_i\sim\text{N}(0,1)$$thus that the different scales $\sigma_i$ prevent a non-central $\chi^2$ representation.

To achieve a non-central $\chi^2$ distribution, as defined on the Wikipedia page, one would need to divide the $X_i^2$'s by the \sigma_i^2$'s, which is impossible since those are unknown.

  • $\begingroup$ Thank you for your response. My primary goal is to create a posterior predictive distribution for $D$. In the case of perfect parameter knowledge, (all $\mu_i$ and $\sigma_i^2$ known), is $D$ chi-distributed as defined by $\chi$ and $k$? $\endgroup$
    – Arun
    Commented Jan 9, 2017 at 14:02
  • $\begingroup$ I'm afraid I am confused - in the case of known sigma's, why is the equation for lambda not valid? $\endgroup$
    – Arun
    Commented Jan 9, 2017 at 17:12
  • $\begingroup$ @Xi'an would you say that a normal prior for $\sigma^2$ could be considered dangerous/inconvenient/ temerarious? $\endgroup$
    – mugen
    Commented Jan 9, 2017 at 23:18
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    $\begingroup$ @mugen: a normal prior on $\sigma^2$ needs to be truncated at zero. Other than that, it is a proper prior, hence leads to a valid Bayesian analysis of the problem. I do not see how qualifications like dangerous and others apply to a prior: a prior is a prior (is a prior). $\endgroup$
    – Xi'an
    Commented Jan 10, 2017 at 8:51
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    $\begingroup$ @Xi'an That makes much more sense, thank you. I think my original question is probably now tangential to what I want to do - calculate a posterior predictive interval for future $D$, given either acquired measurements of $D$, or measurements of $X_i$ with priors on the parameters underlying the generation of $X_i$. Time to try some simulations I think. $\endgroup$
    – Arun
    Commented Jan 13, 2017 at 16:53

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