How to merge two Gaussians

Ok I solved it :)

Since covariance matrix is diagonal we can assume having multiple univariates. And then variance combination is as

$$\hat{\mu} = \frac{n_1\mu_1 + n_2\mu_2}{n_1+n_2}$$

$$\hat{\sigma}^2 = \frac{(\sigma_1^2 + \mu_1^2)n_1 + (\sigma_2^2 + \mu_2^2)n_2}{ (n_1+n_2)} - \hat{\mu}^2$$

Here, I used $\sigma^2 = E[x^2] - E[x]^2$

thanks again

I might be wrong or misinterpreted the question, but trying to reproduce the result in the accepted and upvoted answer, I get a different result:

Let $x \sim N(\mu, \sigma^2)$. From the definition of the variance follows $$ \sigma^2 = E[x^2] - E[x]^2 = E[x^2] - \mu^2 $$$$ or \quad E[x^2] = \sigma^2 + \mu^2 $$

Now let $x$ be the random variable defined as the weighted average $x = \frac{n_1x_1 + n_2x_2}{n_1 + n_2}$, where $x_1 \sim N(\mu_1, \sigma_1^2)$ and $x_2 \sim N(\mu_2, \sigma_2^2)$ are independent.

We have easily $$ E[x] = \frac{n_1\mu_1 + n_2\mu_2}{n_1 + n_2} := \mu $$

By the above formula for $E[x_1^2]$ and $E[x_2^2]$, and since $x_1$ and $x_2$ are independent ($E[x_1x_2] = E[x_1]E[x_2]$) we have

\begin{align} E[x^2] &= E[(\frac{n_1x_1 + n_2x_2}{n_1 + n_2})^2] \\ &= \frac{1}{(n_1 + n_2)^2} E[n_1^2 x_1^2 + n_2^2 x_2^2 + 2n_1n_2x_1x_2] \\ &= \frac{1}{(n_1 + n_2)^2} (n_1^2 E[x_1^2] + n_2^2 E[x_2^2] + 2n_1n_2E[x_1]E[x_2]) \\ &= \frac{(\sigma_1^2 + \mu_1^2)n_1^2 + (\sigma_2^2 + \mu_2^2)n_2^2 + 2n_1n_2\mu_1\mu_2}{(n_1+n_2)^2} \\ &= \frac{\sigma_1^2 n_1^2 + \sigma_2^2 n_2^2 + (n_1\mu_1 + n_2\mu_2)^2}{(n_1+n_2)^2} \end{align}

We can use this to calculate the pooled variance: $$ \sigma^2 = E[x^2] - E[x]^2 = \frac{\sigma_1^2 n_1^2 + \sigma_2^2 n_2^2 + (n_1\mu_1 + n_2\mu_2)^2}{(n_1+n_2)^2} - \mu^2 $$

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In the multivariate case, if the covariance matrix is diagonal, we can apply this formula on each dimension separately. Otherwise, let $X \sim N(\mu, \Sigma)$, where $\mu \in \mathbb{R}^n$ and $\Sigma \in \mathbb{R}^{n \times n}$.

Again start with the general definition of the covariance matrix that gives

$$ \Sigma = E[(X-\mu)(X-\mu)^T] = E[XX^T] - \mu \mu^T $$$$ or \quad E[XX^T] = \Sigma + \mu\mu^T $$

Let $X$ be the random variable defined as the weighted average $X = \frac{n_1X_1 + n_2X_2}{n_1 + n_2}$, where $X_1 \sim N(\mu_1, \Sigma_1)$ and $X_2 \sim N(\mu_2, \Sigma_2)$ are independent.

To simplify, let $D = \frac{1}{(n_1 + n_2)^2}$ . We have now

\begin{align*} E[XX^T] &= D \cdot E[(n_1X_1 + n_2X_2)(n_1X_1 + n_2X_2)^T] \\ &= D \cdot E[n_1^2 X_1X_1^T + n_2^2 X_2X_2^T + 2n_1n_2X_1X_2^T] \\ &= D \cdot (n_1^2 E[X_1X_1^T] + n_2^2 E[X_2X_2^T] + 2n_1n_2E[X_1]E[X_2^T]) \\ &= D \cdot (n_1^2 (\Sigma_1 + \mu_1\mu_1^T) + n_2^2 (\Sigma_2 + \mu_2\mu_2^T) + 2n_1n_2\mu_1\mu_2^T) \\ &= D \cdot (n_1^2 \Sigma_1 + n_2^2 \Sigma_2 + (n_1\mu_1 + n_2\mu_2)(n_1\mu_1 + n_2\mu_2)^T) \end{align*}

Finally the pooled covariance matrix is: $$ \Sigma = E[XX^T] - E[X]E[X]^T = \frac{n_1^2 \Sigma_1 + n_2^2 \Sigma_2 + (n_1\mu_1 + n_2\mu_2)(n_1\mu_1 + n_2\mu_2)^T}{(n_1 + n_2)^2} - \mu\mu^T $$