The exponential moving average (EMA) operator is defined as:
$$y_t(x, \lambda) = (1-\lambda) \sum_{i=0}^\infty \lambda^i x_{t-i}$$ where $1-\lambda$ is the normalization factor, and the operator is a convolution filter with unbounded support and strictly $\lambda<1$
A double EMA is application of this operator twice on the input process, $x_t$:
$$y_t(x,\lambda_1, \lambda_2) = (1-\lambda_1)(1-\lambda_2) \sum_{i=1}^\infty \sum_{j=0}^\infty \lambda_1^i \lambda_2^j ~x_{t-i-j}$$
Is it possible to express this double summation as a combination of single summations instead, either as a closed form solution or asymptotically under certain conditions?
$$\sum_{i=0}^\infty\sum_{j=0}^\infty\lambda_1^i\lambda_2^j a_{i+j}=\sum_{k=0}^\infty a_k\sum_{j=0}^k\lambda_1^{k-j}\lambda_2^j=\sum_{k=0}^\infty c_k(\lambda_1,\lambda_2)a_k,\\c_k(\lambda_1,\lambda_2)=\begin{cases}\frac{\lambda_1^{k+1}-\lambda_2^{k+1}}{\lambda_1-\lambda_2},&\lambda_1\neq\lambda_2\\\hfill(k+1)\lambda^k,\hfill&\lambda_1=\lambda_2=\lambda\end{cases}$$
(thus for $\lambda_1\neq\lambda_2$ we reduce to two single convolutions; the case $\lambda_1=\lambda_2$ is also clear).