Let $X_1,...,X_{n_1}$ be an i.i.d. sample from $N_p(\mu_1,\Sigma)$ and let $Y_1,...,Y_{n_2}$ be an independent sample from $N_p(\mu_2,\Sigma)$, for some $\mu_1,\mu_2 \in \mathbb{R}^p$ and some invertible, $p\times p$ positive definite matrix $\Sigma$. I found that:

\begin{align} \hat{\mu_1} &= \frac{1}{n_1}\sum^{n_1}_{i=1}x_i, \\ \hat{\mu_2} &= \sum^{n_2}_{i=1}y_i, \\ \hat{\Sigma} &= \frac{1}{n_1+n_2}\biggl(\sum^{n_2}_{i=1}(x_i-\hat{\mu_1})(x_i-\hat{\mu_1})^T+\sum^{n_2}_{i=1}(y_i-\hat{\mu_2})(y_i-\hat{\mu_2})^T\biggr) \end{align}

I'd like to find the joint distribution $(\hat{\mu_1},\hat{\mu_2},\hat{\Sigma})$.

I know that $\hat{\mu_1}\sim N_p(\mu_1,\frac{1}{n_1}\Sigma),\hat{\mu_2}\sim N_p(\mu_2,\frac{1}{n_2}\Sigma)$

For $\hat{\Sigma}$, I'm not quite sure actually. So, I know that for example, if I set $S_n=\frac{1}{n-1}\sum^n_{i=1}(x_i-\overline{x})(x_i-\overline{x})^T$ then $(n-1)S\sim \mathcal{W}_p(\Sigma,n-1)$. Also, if $S_m=\frac{1}{m-1}\sum^{m}_{i=1}(x_i-\overline{x})(x_i-\overline{x})^T$ is independent from $S_n$, then $S_n+S_m\sim \mathcal{W}_p(\Sigma,n+m-2)$

I think I can then set $(n_1+n_2)S_x=\frac{n_1+n_2}{n_1+n_2}\bigl(\sum^{n_1}_{i=1}(x_i-\hat{\mu_1})(x_i-\hat{\mu_1})^T\bigr) \sim \mathcal{W}_p(\Sigma,n_1+n_2)$ and $S_y=\frac{1}{n_1+n_2}\bigl(\sum^{n_2}_{i=1}(y_i-\hat{\mu_2})(y_i-\hat{\mu_2})^T\bigr)$, I can then say that $\hat{\Sigma}=S_x+S_y$. So by the fact that they are independent, (and since $S_n+S_m\sim \mathcal{W}_p(\Sigma,n+m-2)$) I can say that $(n_1+n_2)\hat{\Sigma}\sim \mathcal{W}_p(\Sigma,n_1+n_2)$.

Would this be correct?

I also know that $\hat{\mu_1},\hat{\mu_2},\hat{\Sigma}$ are all independent. So what exactly would be their joint distribution $(\hat{\mu_1},\hat{\mu_2},\hat{\Sigma})$?

  • $\begingroup$ I think you're missing a $\frac{1}{n_2}$. $\endgroup$ Commented Jan 21, 2021 at 7:47
  • $\begingroup$ Is your problem with finding the (independent) distributions $\hat\mu_1, \hat\mu_2, \hat\Sigma$ or with how to combine them to a joint distribution? $\endgroup$ Commented Jan 22, 2021 at 17:02
  • $\begingroup$ @SextusEmpiricus yes, how to combine them to a joint distribution $\endgroup$
    – user255658
    Commented Jan 23, 2021 at 8:02

1 Answer 1


If you have two independent multivariate normal distributions then their joint distribution is another multivariate normal distribution with a block scheme covariance matrix

$$\begin{bmatrix} \Sigma_1&0\\0&\Sigma_2 \end{bmatrix}$$

When you also consider the joint distribution with a Wishart distribution then you get a Normal-Wishart distribution

  • $\begingroup$ Thank you! I don't quite understand the notation on wikipedia, could you show how it would look like on my example? $\endgroup$
    – user255658
    Commented Jan 23, 2021 at 8:03

Your Answer

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