Lag in financial trading vs Group Delay in signal processing
i'm gonna work on this using the EE DSP notation i think is most conventional in the EE DSP field.  your exponentially-decaying 1-pole filter has this simple difference equation:

$$ y[n] = (1-p) \, x[n] \ + \ p \, y[n-1] $$

turns out (as we shall see below) that $p$ is the pole.  applying the Z transform gets you

$$ Y(z) = (1-p) \, X(z) \ + \ p \, z^{-1}Y(z) $$

from which we can solve for the transfer function:

$$ H(z) \triangleq \frac{Y(z)}{X(z)} = \frac{1-p}{1-p\,z^{-1}} = (1-p)\frac{z}{z-p} $$

so there is a zero at $z=0$ and a (real) pole at $z=p$.  the zero would go away if we took our input from $x[n-1]$ instead of $x[n]$ which would create a pole at $z=0$ and cancel the zero, but all that does is introduce another sample period of delay.  i don't think you wanna do that.

the (Kronecker) impulse response is

$$ \begin{align}
h[n] &amp; = (1-p) p^n u[n] \\
&amp; = (1-p) e^{\log(p) \, n} u[n] \\
&amp; = (1-p) e^{n / \tau} u[n] \\
\end{align} $$

where u[n] is the Heaviside unit step function

$$ u[n] \triangleq
\begin{cases}
1, &amp; \text{if }n \ge 0 \\
0, &amp; \text{if }n &lt; 0
\end{cases} $$

and $\tau$ is the "time constant" of the decaying exponential.  in order for the filter to be stable, the exponential must be decaying which means that $\tau &gt; 0$, $\log(p) &lt; 0$, and $0 &lt; p &lt; 1$

okay, so now the complex frequency response of $H(z)$ is

$$ \begin{align}
H(z)\bigg|_{z=e^{j\omega}} = H(e^{j\omega}) &amp; = \frac{1-p}{1-p \, e^{-j\omega}} \\
&amp; = \frac{1-p}{1-p \, \left(\cos(\omega) + j\sin(\omega) \right)} \\
\end{align} $$

the magnitude response is

$$ \begin{align}
|H(e^{j\omega})| &amp; = \left|\frac{1-p}{1-p \, e^{-j\omega}} \right| \\
&amp; = \left| \frac{1-p}{1-p \, \left(\cos(\omega) + j\sin(\omega) \right)} \right| \\
&amp; = \frac{1-p}{\sqrt{ (1-p\cos(\omega))^2 + (p\sin(\omega))^2 }} \\
&amp; = \frac{1-p}{\sqrt{ 1 + p^2 - 2 p \cos(\omega) }} \\
\end{align} $$

this is pretty familiar.  less oft observed is the phase response

$$ \begin{align}
\phi(\omega) \triangleq \arg \{ H(e^{j\omega}) \} &amp; = \arg \left\{ \frac{1-p}{1-p \, e^{-j\omega}} \right\} \\
&amp; = \arg \{ 1-p \} - \arg \{1-p \, e^{-j\omega} \} \\
&amp; = 0 - \arg \{1-p (\cos(\omega) + j \sin(\omega)) \} \\
&amp; = - \arctan \left( \frac{p \sin(\omega)}{1-p \cos(\omega)} \right) \\
\end{align} $$

the phase delay is

$$ \tau_\phi(\omega) \triangleq -\frac{\phi(\omega)}{\omega} $$

and the group delay is

$$ \tau_g(\omega) \triangleq -\frac{d\phi(\omega)}{d\omega} $$

gotta go.  i'll post this now and finish it later.  perhaps someone else wants to finish it.