The Pearson mean is a weighted mean defined as
$$m_t = \frac {\sum_{i=0}^t w_ix_i} {\sum_{i=0}^t w_i}$$
The exponential mean weight average (EMWA) is the following recurrence where $0 < \alpha <1$.
$$m_t = (1-\alpha)m_{t-1} + \alpha x_t = m_{t-1} + \alpha(x_t - m_{t-1})$$
Is this recurrence a Pearson mean with an exponentially decaying weight ?
If we develop the EMWA, we have
$$m_t = \sum_{i=0}^t \alpha (1-\alpha)^i x_{t-i}$$
But it seam that the denominator is missing.
Yes, it is. The easiest way to think of this is that in the limit you have $w_i = \alpha (1-\alpha)^i$. The denominator then is $$\sum_{i=0}^\infty w_i = \sum_{i=0}^\infty \alpha (1-\alpha)^I$$ , and summing the geometric series gives $$\alpha \times {1 \over 1 - (1-\alpha)} = 1$$ so the "missing" denominator is in fact 1.
But of course you don't have infinite data in practice. Let $m_0 = x_0$ - that is, your initial value for the EWMA is just the initial value of the time series. Then you have
$$m_1 = m_0 + \alpha (x_1 - m_0) = \alpha x_1 + (1-\alpha) m_0 = \alpha x_1 + (1 - \alpha) x_0$$
$$m_2 = m_1 + \alpha (x_2 - m_1) = \alpha x_2 + (1-\alpha) m_1 = \alpha x_2 + (1-\alpha) (\alpha x_1 + (1-\alpha) x_0) = \alpha x_2 + \alpha (1-\alpha) x_1 + (1-\alpha)^2 x_0$$ and so on for larger values of $t$.
If you continue to develop this you find that
$$m_t = \left( \sum_{i=0}^{t-1} \alpha (1-\alpha)^i x_{t-i} \right) + (1-\alpha)^t x_0$$
and again you can sum the geometric series to see that the weights sum to 1.
