Assume that a signal $ y $ is a noisy perturbation of time-shifted copies of a given waveform $ f(t) $ defined on K time bins $ \{ 0, \cdots, K-1 \} $: \begin{equation} \forall t \in \{1, \cdots, T\}, \quad y(t) = \sum_{i=1}^{T} x_i f(t - i) + \eta(t) \end{equation} where $ \eta(t) $ is drawn from a Gaussian distribution of variance $ \sigma^2 $ and $ x = \{ x_i \}_{i \in \{ 1, \cdots, T \}} $ is a discrete process such that $ x_i = a_i b_i $ with $ a_i $ draw from a density $ g $ (such that $ g(a) = 0 $ when $ a \leq 0 $) and $ \{ b_i \}_{i \in \{1, \cdots, T\}} $ is a bernoulli process with parameter $ p $.
The Maximum A Posteriori (MAP) estimator is given by: \begin{equation} \theta_{MAP} = \underset{x}{\arg\max} \; f_{Y \mid X}(y \mid x) \times f_{X}(x) \end{equation} where: \begin{equation} f_{Y \mid X}(y \mid x) = \frac{1}{\sigma \sqrt{2 \pi}} \exp \left( - \frac{1}{2} \frac{\left\| y(t) - \sum_{i=1}^n x_i f(t - i) \right\|_2^2}{\sigma^2} \right) \end{equation} and: \begin{equation} f_{X}(x) = \prod_{i=1}^T \left[ (1 - p) \delta(x_i) + p g(x) \right] \end{equation} Taking the negative log of the prior $ f_{X} $ gives: \begin{align} - \log(f_{X}(x)) &= - \sum_{i=1}^T \log((1-p) \delta(x_i) + p g(x)) \\ &= - \sum_{i=1,x_i=0}^T \log(1 - p) - \sum_{i=1,x_i\neq0}^T \log(pg(x)) \\ &= - \sum_{i=1,x_i=0}^T \log(1 - p) - \log(p) \| x \|_0 - \sum_{i=1,x_i\neq0}^T \log(g(x)) \end{align}
What is the proper (and standard) approximation to get rid of the first term $ \sum_{i=1,x_i=0}^T \log(1 - p) $ ?
It seems that I have not pushed the calculation to the end. In fact: \begin{align} \sum_{i=1,x_i=0}^T \log(1 - p) &= \log(1 - p) \left( T - \| x \|_0 \right) \\ &= \log(1 - p) T - \log(1 - p) \| x \|_0 \\ &= \log(1 - p) T + \left[ p + o(p) \right] \| x \|_0 \quad \text{for } p \rightarrow 0 \end{align} The first term doesn't depend on $ x $ and the second one tends to $ 0 $ when $ p $ tends to $ 0 $.