I'm working with exponential smoothing for a series, using the formula $z_i=\lambda \bar x_i + (1- \lambda) z_{i-1}$, as explained in this wiki article.
Using this method, should $z_{1}$ be defined to equal $\bar x_{1}$, or $\lambda\bar x_{1}$? I think of the second option as defining a $z_0=0$
I think a main problem is that I don't understand how the effect of the lambda is compounded for less recent observations. Isn't this the basic idea: $x_1\to\lambda^1x_1,\,x_2\to\lambda^2x_2,\,\, x_3\to\lambda^3x_3...?$
The value of $z_{\,1}$ is basically arbitrarial: its definition is left to your “guess” on how the data series starts, which means how you guess might be its values before the recording period (time <= $0$). According to the nature of the process, you may hypothize that it was a) almost flat zero or b) almost steady equal to the starting value, i.e. $ \bar x_{\,1}$, or other assumptions. So the two expressions you indicate are both admissible.
If you roll down futher the recursion you will arrive at:
$$ z_{\,n} = \lambda \,\sum\limits_{0\, \le \,k\, \le \,n - 1} {\left( {1 - \lambda } \right)^k \bar x_{\,n - k} } + \left( {1 - \lambda } \right)^{n - 1} z_{\,1} $$ which clearly shows how $z_{\,n} $ depends on previous $\bar x$ values.