may I have some key words about how to due with a hypothesis test with restricted domain?
Here is an sample question to my problem. Alice have a dice with $1/3$ chances to get $1$, $2/3$ to get $0$. Bob have another dice with $1/4$ and $3/4$ chances instead.
Given $n$ dices and with $r$ $1$, assuming the number of dices from Alice is $k$, and $n-k$ from Bob. The likelihood of the value of $k$ is
$$C^n_r (k * 1/3)^k ((n-k)* 1/4)^{n-k}$$
in this case, $k \in [0, n]$. However the distribution are still continuing beyond the domain.
What is the proper way to handle this scenario? I couldn't search some relevant results up to this moment, may I have some keywords about some topic discussion this kind of question?
Too long for a comment.
Let us assume that we have two Bernoulli distributions $\mathsf{Ber}(\frac{1}{3})$ and $\mathsf{Ber}(\frac{1}{4}).$ If $k$ independent and identically distributed experiments of $\mathsf{Ber}(p)$ are performed and $X$ is number of successes then $X \sim \mathsf{Bin}(k, p).$ In your context, you want to estimate $k$ knowing that there is a total of $n$ trials with $r$ sucesses (so $n$ and $r$ are fixed while $k$ is unknown). So, you have two random variables $X \sim \mathsf{Bin}(k, \frac{1}{3})$ and $Y \sim \mathsf{Bin}(n-k, \frac{1}{4}).$ Under these circumstances, we have $X + Y = r.$ You need to write $$ \{X+Y = r\} = \bigcup_{s = 0}^{\min(r, k)} \{X = s\} \cap \{Y = r-s\}. $$ Then, $$ P(X + Y = r) = \sum_{s = 0}^{\min(r, k)} \dbinom{k}{s} \left( \frac{1}{3} \right)^s \left( \frac{2}{3} \right)^{k-s} \dbinom{n-k}{r-s} \left( \frac{1}{4} \right)^{r-s} \left( \frac{3}{4} \right)^{n-k-r+s}. $$ Even if you assumed $r \leq k,$ this sum is not easy. It also seems that you will need to do computer estimates directly.