Average, exponential moving average, identities/splitting input parts I would like to know if certain identities for averages (mean) also hold for the exponential moving average (EMA). I can verify the mean case, but not the exponential case. Can somebody tell me if the same basic identities hold?
$$\begin{align*}
\mathrm{mean}(A_i + B_i) &= \mathrm{mean}(A_i ) + \mathrm{mean}(B_i)\\
\mathrm{mean}(R * A_i) &= R* \mathrm{mean}(A_i)
\end{align*}$$
Now, for an EMA we have the basic formula:
$$\mathrm{EMA} = S_i = S_{i-1} + \alpha * (A_i - S_{i-1})$$
Where $\alpha$ is a weighting factor $(0 < \alpha < 1)$. 
So the question is, "are the following also true?":
$$\begin{align*}
\mathrm{EMA}(A_i + B_i) &= \mathrm{EMA}(A_i) + \mathrm{EMA}(B_i)\\
\mathrm{EMA}(R *A_i) &= R * \mathrm{EMA}(A_i)
\end{align*}$$
For the second case it seems to hold based on a quick analysis (as I can completely factor out $R$). But for the first case I'm not so good with factoring the recursive relationship, so I don't know.
 A: Yes, you start with $ema(A_1+B_1)=ema(A_1)+ema(B_1)$.  Then consider what happens when you add another term.  Nothing disturbs the relation, so it continues.  This can be formalized by induction. 
Added:  to show the induction, assume it is true for $n$ terms.  I will use $ema(A_n)$ to mean the exponential moving average of the first $n$ terms of $A$.  So $ema((A+B)_n)=ema(A_n)+ema(B_n).$
$ema((A+B)_{n+1})=ema((A+B)_n)+\alpha((A_{n+1}+B_{n+1})-ema((A+B)_n))$
$=ema(A_n)+ema(B_n)+\alpha (A_{n+1}-ema(A_n))+\alpha (B_{n+1}-ema(B_n))=ema(A_{n+1})+ema(B_{n+1})$
where the first equality is the definition of $ema$, the second uses the induction hypothesis and the distributive law, and the third reuses the definition of $ema$.
A: To add to the post answer (nailed by Ross), notice that the exponential moving average (also called recursive average, AR(1) filter, one-pole smoothing, etc) can also be expressed as 
$\displaystyle s_i = \beta \; s_{i-1} + \alpha \; a_i \;\;\; $ 
  with $\beta = 1 - \alpha $
which is computationally equivalent (except perhaps for numerical stability), but conceptually more enlightening: it shows that the output is a weighted average of the current input and the  previous output.
