I have data $x_1,...,x_n$, $y_1,...,y_m$ and $z_1,...,z_p$ where $$x_1,...,x_n\sim N(\mu_x,\sigma^2_x)$$ and $$y_1,...,y_m\sim N(\mu_y,\sigma^2_y)$$ and $$z_1,...,z_p\sim N(\mu_z,\sigma^2_z)$$

Now let's assume I want to take a Bayesian approach and place the following priors: $p(\mu_x,\sigma^2_x)\propto (\sigma^2_x)^{-1}$, $p(\mu_y,\sigma^2_y)\propto (\sigma^2_y)^{-1}$, and $p(\mu_z,\sigma^2_z)\propto (\sigma^2_z)^{-1}$. Given these priors, I know what the posterior distribution is, but more importantly, I know that the conditional marginal distributions are

$$\mu_x|\sigma^2_x,x_1,...,x_n\sim N(\bar{x},\sigma^2_x/n)$$ and $$\mu_y|\sigma^2_y, y_1,...,y_m\sim N(\bar{y},\sigma^2_y/m)$$ and $$\mu_z|\sigma^2_z,z_1,...,z_n\sim N(\bar{z},\sigma^2_z/p)$$

where $\bar x$ is the average of the $x$'s. Similarly, for the case of the $y$'s and $z$'s.

I am interested in deriving the joint distribution of $\mu_x-\mu_y$, $\mu_y-\mu_z$, $\mu_y-\mu_z$ . Does the following approach make sense?

We know that the conditional posterior distribution of $\mu_x$ is $$p(\mu_x|\sigma^2_x,x_1,...,x_n) = N\left(\bar x, \frac{\sigma^2_x}{n}\right)$$

and similarly for $p(\mu_y|\sigma^2_y,y_1,...,y_m)$ and $p(\mu_z|\sigma^2_z,z_1,...,z_p)$.

Now, define \begin{align} \Delta := \begin{bmatrix} \delta_{xy}\\\\ \delta_{xz}\\\\ \delta_{yz} \end{bmatrix} = \begin{bmatrix} \mu_x-\mu_y\\\\ \mu_x - \mu_z\\\\ \mu_y - \mu_z \end{bmatrix} = \begin{bmatrix} 1 & -1 & 0\\\\ 1 & 0 & -1\\\\ 0 & 1 & -1 \end{bmatrix}\begin{bmatrix} \mu_x\\\\ \mu_y\\\\ \mu_z \end{bmatrix} =: A\mu, \end{align} then we have that $$\Delta|\sigma^2_x,\sigma^2_y,\sigma^2_z, x_1,...,x_n, y_1,...,y_m, z_1,...,z_p \sim N_3\left(A \mathbb{E}(\mu), A\mathbb{C}\text{ov}(\mu)A^T\right)$$ where \begin{align*} \mathbb{E}(\mu) = \begin{bmatrix} \bar{x}\\\\ \bar{y}\\\\ \bar{z} \end{bmatrix} \text{ and } \mathbb{C}\text{ov}(\mu) = \begin{bmatrix} \sigma^2_x/n & 0 & 0 \\\\ 0 & \sigma^2_y/m & 0 \\\\ 0 & 0 & \sigma^2_z/p \end{bmatrix}. \end{align*}

So my ultimate questions are the following:

  1. Would the distribution above for $\Delta$ be the correct joint conditional posterior distribution of the differences?

  2. And if so, would the appropriate strategy to obtain posterior samples from $\Delta$ be to first samples $\sigma^2_x, \sigma^2_y,$ and $\sigma^2_z$ from their joint posterior distribution?

  3. And if so for number 2) does the joint posterior distribution of $\sigma^2_x, \sigma^2_y,$ and $\sigma^2_z$ have a closed form?


1 Answer 1

  1. The conditional analysis is correct (assuming posterior conditional independence between $\mu_x,\mu_y,\mu_z$), as already indicated in the answer to your earlier question!

  2. Simulating $σ_x$, $σ_y$, and $σ_z$ from their joint (marginal) posterior distribution, then $\mu_x,\mu_y,\mu_z$ from the corresponding conditional posterior is correct. However, if simulating from the (marginal) posterior proves complex, a Gibbs sampling approach may be preferable. But...

  3. ...since $$\int_{-\infty}^{+\infty} \exp\left\{-\left[n(\bar x_n-\mu_x)^2+\sum_{i=1}^n(x_i-\bar x_n)^2\right]\big/2\sigma^2_x\right\}\,\frac{\text d\mu_x}{\sigma_x^{n+2}}\\\propto\exp\left\{-\sum_{i=1}^n(x_i-\bar x_n)^2\big/2\sigma^2_x\right\}\sigma_x^{-n-1}$$ and the equivalent closed forms for $σ^2_y$, and $σ^2_z$, the marginal posterior on $(\sigma^2_x,\sigma^2_y,\sigma^2_z)$ is available.

  • $\begingroup$ Thanks for confirming 1) and 2). For 3) I understand that the marginal posteriors for the $\sigma^2$s are available in closed form (inverse-gammas), but are you implying that that joint (marginal) of the $\sigma^2$'s is not available in closed form (I was hoping that it reduced some how to, say, and inverse wishart)? Also, is there any reason or probability rule I am breaking by sampling from the marginals of each $\sigma^2$ and then plugging those into the joint distribution of $\Delta$ to obtain a posterior sample of $\Delta$ from the corresponding Normal distribution? $\endgroup$
    – John Smith
    Commented Nov 19, 2022 at 14:55
  • $\begingroup$ Sorry for the multiple questions, but if I interpret your answer correctly, that is to say I have $p(\Delta, \sigma^2_x, \sigma^2_y,\sigma^2_z|data) = p(\Delta| \sigma^2_x, \sigma^2_y,\sigma^2_z,data) p( \sigma^2_x, \sigma^2_y,\sigma^2_z|data)$ and here $ p(\sigma^2_x, \sigma^2_y,\sigma^2_z|data) = p( \sigma^2_x|data)p( \sigma^2_y|data)p( \sigma^2_z|data)$ by independence? $\endgroup$
    – John Smith
    Commented Nov 19, 2022 at 17:10
  • $\begingroup$ Thank you so much! That was extremely to walk through that. $\endgroup$
    – John Smith
    Commented Nov 19, 2022 at 17:22
  • $\begingroup$ Sorry, one more question. As I was going to code this up, I realize I don't recognize the kernel of the last expression in your 3). At first I had assumed it was inverse-gamma, however inside the exponential you have $\sigma^2_x$ and outside the exponential you have $\sigma_x^{-n-1}$. In order to get it to match upI assume a bit of reworking is needed. I want to confirm that the marginal of $\sigma^2_x$ is Inverse-gamma($n/2 - 1/2, \Sigma(x_i-\bar{x})^2/2$) $\endgroup$
    – John Smith
    Commented Nov 20, 2022 at 16:35
  • 1
    $\begingroup$ I get it now! I was forgetting about the original posterior distribution of $(\mu,\sigma^2)$ and was only thinking in terms of $\Delta$. $\endgroup$
    – John Smith
    Commented Nov 23, 2022 at 14:40

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