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Let $x_i \sim N(m, 1)$ for $i \in \lbrace 1, \dots, N\rbrace$ and define $y_i \equiv x_i$ if $x_i > 0$ and $y_i \equiv 0$ if $x_i \leq 0$. The data in this case is the sample $Y = \lbrace y_1, \dots, y_N \rbrace$. So any likelihood based method (Bayes, MLE, etc) starts with the likelihood for $Y$. m$(or sampling model for$Y$depending on how you want to look at it). Letting$n$denote the number of exactly zero observations, this likelihood can be simply expressed as $$\Phi(-m)^n \times \prod_{j \;\mid \; y_j \;\neq \;0} N^+(y_j \;;\; m, 1),$$ where$\Phi(\cdot)$is the standard Normal CDF and$N^+(\ \cdot \mid \;mean, variance)$denotes a positive truncated Normal pdf. By explicitly writing the normalizing constant, we can rewrite this as $$\frac{\Phi(-m)^n }{\left(1-\Phi(-m)\right)^{N-n}} \times \prod_{j \;\mid \; y_j \;\neq \;0} N(y_j \;;\; m, 1).$$ This expression makes clear why the zeros$do$matter, because they are informative about$m$via the ratio of zeros to non-zeros observed. 1 Others have hit on this, but I thought I'd contribute how I'd write the problem down (briefly): Let$x_i \sim N(m, 1)$for$i \in \lbrace 1, \dots, N\rbrace$and define$y_i \equiv x_i$if$x_i > 0$and$y_i \equiv 0$if$x_i \leq 0$. The data in this case is the sample$Y = \lbrace y_1, \dots, y_N \rbrace$. So any likelihood based method (Bayes, MLE, etc) starts with the likelihood for$Y$. Letting$n$denote the number of exactly zero observations, this likelihood can be simply expressed as $$\Phi(-m)^n \times \prod_{j \;\mid \; y_j \;\neq \;0} N^+(y_j \;;\; m, 1),$$ where$\Phi(\cdot)$is the standard Normal CDF and$N^+(\ \cdot \mid \;mean, variance)$denotes a positive truncated Normal pdf. By explicitly writing the normalizing constant, we can rewrite this as $$\frac{\Phi(-m)^n }{\left(1-\Phi(-m)\right)^{N-n}} \times \prod_{j \;\mid \; y_j \;\neq \;0} N(y_j \;;\; m, 1).$$ This expression makes clear why the zeros$do$matter, because they are informative about$m\$ via the ratio of zeros to non-zeros observed.