Suppose $X_i$ are independently distributed random variables from a discrete distribution given by $P[X_i=0]=\frac{1}{2}, P[X_i=1]=\frac{\theta}{4}, P[X_i=2]=\frac{1-\theta}{2}, P[X_i=3]=\frac{\theta}{4}$ where $\theta \in (0,1)$ is an unknown parameter. For each $i$ we observe $Y_i$, which equals $X_i$ in case $X_i>0$, but equals $Y_i=1$ in case $X_i=0$.
I need to find the distribution of $Y_i$ and also the observed likelihood.
Approach
See that $P[Y_i=1]=P[Y_i=1|X_i>0]P[X_i>0]+P[Y_i=1|X_i=0]P[X_i=0]=P[X_i=1]P[X_i >0]+1.P[X_i=0]=\frac{\theta+4}{8}$.
Similarly, $P[Y_i=2]=P[X_i=2]P[X_i>0]=\frac{1-\theta}{4}$ and $P[Y_i=3]=P[X_i=3]P[X_i>0] =\frac{\theta}{8}$.
But they do not sum to $1$ somehow! Probably I am missing any case? Can anyone help out?
I understand that you use $\mathbb P(Y_1=1\vert X_1>0)=\mathbb P(X_1=1)$ because $Y_1=X_1$ when $X_1>0$. This is wrong. You should write $$ \mathbb P(Y_1=1\vert X_1>0)=\mathbb P(X_1=1\vert X_1>0)=\frac{\mathbb P(X_1=1,X_1>0)}{\mathbb P(X_1>0)}=\frac{\mathbb P(X_1=1)}{\mathbb P(X_1>0)}=\frac\theta2\cdot $$
You have the same error for the computation of $\mathbb P(Y_2=2)$ and $\mathbb P(Y_3=3)$.