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 $x_k-\mu_{k-1}$. For the new mean $\mu_k$ we then have $k$ times the previous mean plus one $k$-th of the remaining part of the new value. 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 it 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 $k$.

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 $k$.

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 $x_k-\mu_{k-1}$. For the new mean $\mu_k$ we then have $k$ times the previous mean plus one $k$-th of the remaining part of the new value. Okay, so how to get there?

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 it 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.

A practical application of the formula can be found in the field of reinforcement learning where a value function is approximated over many experienced rewards. Since we usually don't know the total number of experiences here, we multiply by a learning rate instead of dividing by $k$.

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 $x_k-\mu_{k-1}$. For the new mean $\mu_k$ we then have $k$ times the previous mean plus one $k$-th of the remaining part of the new value. 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 it 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 $k$.