EM algorithm - What happens with the standard deviation?

Question

What have you tried?

So I watched this video. According to the video, we've to calculate the variance $\sigma^2$ as follows: $$ \sigma_{k}^{2} = \frac { p_{1} \left(x_{1} - \mu_{k_1}\right)^{2} + \ldots + p_{n} \left(x_{n} - \mu_{k_n}\right)^{2} } { p_{1} + p_{2} + \ldots + p_{n} } $$

Where

• $k$ is the index of the current Gaussian class.
• $p$ represents the possibility of one point of the set
• $n$ is the amount of points in the set

So what's your question?

I don't understand, why we have to use $\mu_{1}$, $\mu_{2}$, $\ldots$, $\mu_{n}$. I expected to calculate it follows: $$ \sigma_{k}^{2} = \frac { p_{1} \left(x_{1} - \mu_k\right)^{2} + \ldots + p_{n} \left(x_{n} - \mu_k\right)^{2} } { p_{1} + p_{2} + \ldots + p_{n} } $$

Because why do I have to use the "current-calculated" mean of the Gaussian and not the "final" one?

Answer 1

Here's the answer of @plop:

At the very time point in the video that your link takes you the speaker mentions that there is a typo.