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.