What is the effective alpha of 2 cascaded EWMA IIR filters?
It is important to define in what sense two cascaded EWMA filters should be "equivalent" to a single EWMA filter. One possibility would be to define an "effective memory" or "effective averaging window size" and say that the two filters are equivalent if they have that same effective memory.

The second question to be answered is how to define the effective memory length of such a filter. Again we have to make a reasonable choice, and one possibility is to look at the filter's impulse response $h[n]$ and choose the length $N$ such that the sum over the first $N$ coefficients equals a certain percentage of the sum over all coefficients:

$$\frac{\sum_{n=0}^{N-1}h[n]}{\sum_{n=0}^{\infty}h[n]}=r,\quad 0\ll r&lt;1\tag{1}$$

That is, we consider the coefficients with indices greater than $N$ negligible. The impulse response coefficients $h[n]$ are the weights to compute the output signal from the input signal:

$$y[n]=\sum_{k=0}^{\infty}h[k]x[n-k]\tag{2}$$

For the filter given in your question it can be shown that the impulse response is given by

$$h[n]=\alpha(1-\alpha)^n,\quad n\ge 0\tag{3}$$

Furthermore,

$$\sum_{n=0}^{\infty}h[n]=1\tag{4}$$

and

$$\sum_{n=0}^{N-1}h[n]=1-(1-\alpha)^N\tag{5}$$

With $(4)$ and $(5)$, Eq. $(1)$ reduces to

$$1-(1-\alpha)^N=r\tag{6}$$

which, for a given percentage $r$, can be solved for either $\alpha$ or $N$:

$$\begin{align}\alpha&amp;=1-(1-r)^{1/N}\\
N&amp;=\frac{\log(1-r)}{\log(1-\alpha)}\end{align}\tag{7}$$

So, e.g., with $r=0.9$ and $\alpha=0.1$, the effective averaging length is $N\approx 22$. Alternatively, for a desired length $N=100$, $\alpha$ must be chosen as $\alpha=0.023$.

The same calculations can be done for the cascade of two such filters, the only problem being that it is not possible to give analytical expressions analogous to Eq. $(7)$. However, the finite sum given in $(5)$ can be easily computed using the filter itself. If you cascade two EWMA filters and use a constant input $x[n]=1$, $n\ge 0$, the output will be the sum $(5)$. The first output sample is the sum for $N=1$, the next one for $N=2$, etc. So by this computation one can figure out the value of $N$ for a given percentage $r$. This value is given by the number of output samples necessary for the output to exceed the given value $r$.

It's time for an example. Let's again choose $r=0.9$, i.e. the effective memory size is define by the number of coefficients which add up to $90\%$ of the total sum of all coefficients. Let the two $\alpha$-values of the cascaded filters be $\alpha_1=0.1$ and $\alpha_2=0.2$. To find out its effective averaging length (at least according to our definition), we feed the filter with a constant input $x[n]=1$ until the output exceeds the value of $r=0.9$. The output is shown in the figure below. The output values are blue, and $r$ is shown by the red line.


The effective averaging length $N$ turns out to be $N=27$. In order to find out which first order averaging filter has the same effective length, we need to compute $\alpha$ from Eq. $(7)$ for $N=27$ and $r=0.9$, which gives $\alpha=0.082$. Hence, in the above defined sense, we can say that two cascaded EWMA filters with $\alpha_1=0.1$ and $\alpha_2=0.2$ are equivalent to a single EWMA filter with $\alpha=0.082$.