## Intuition

It's an easy question and @Henry basically answered it. However, I think it would be nice to *add some intuition* on the second equation:

$$\mu_k=\mu_{k-1}+\frac{x_k-\mu_{k-1}}{k}$$

The idea is to represent the new value $x_k$ value by a part that is equal to the previous mean $\mu_{k-1}$ plus the remaining part $x_k-\mu_{k-1}$. The new mean $\mu_k$ we then contains $k$ times the previous mean and one time the remaining part of the new value, divided by the total count. Okay, so how to get there?

## Derivation

Let's say you already have the mean $\mu_{k-1}$ for the elements $x_0,...,x_{k-1}$. Of course, we can easily incorporate an additional value $x_k$ by undoing the division, adding the new value, and redoing the scaling (but the new count):

$$\mu_k=\frac{(k-1)*\mu_{k-1}+x_k}{k}$$

Now, let's represent the new value by the previous mean and the difference to the previous mean:

$$x_k=\mu_{k-1}+(x_k-\mu_{k-1})$$

Putting that into our first equation:

$$\mu_k=\frac{(k-1)*\mu_{k-1}+\mu_{k-1}+(x_k-\mu_{k-1})}{k}$$

Look how we now got $k$ times the previous mean:

$$\mu_k=\frac{k*\mu_{k-1}+(x_k-\mu_{k-1})}{k}$$

The nice thing is that we don't have to undo and redo the division on the previous mean anymore:

$$\mu_k=\mu_{k-1}+\frac{x_k-\mu_{k-1}}{k}$$

As you can see, we still have to do the division on the difference of the new value to the previous mean. Since the new value comes in "sum-space" already, we don't have to undo anything on it, just divde by the total count.

## Application

An application of this form of incremental mean can be found in the field of *reinforcement learning* where a value function is averaged over many experienced rewards. In this setting, you can think of $x_k-\mu_{k-1}$ as a temporal-difference error term. Since we usually don't know the total number of experiences here, we multiply by a small learning rate rather than dividing by $k$.