Sufficient statistic for function of exponential random variable

Question

Let $X_1, \dots, X_{20} \overset{\text{iid}}{\sim} \text{Exp}(\lambda)$; that is, $$f_{X_1}(x) = \lambda e^{-\lambda x} \cdot I(x > 0)\text{.}$$ However, only the $U_i$ are observed, with $$U_i = \begin{cases} 1, & X_i \leq 1 \\ X_i, & 1 < X_i < 5 \\ 5, & X_i \geq 5\end{cases}$$ for each $i = 1, \dots, 20$.

I wish to find a three-dimensional sufficient statistic for $\lambda$, dependent on $U_1, \dots, U_{20}$.

Denote $a$ as the number of $U_i$ that are $1$, and $b$ as the number of $U_i$ which are $5$.

Then the likelihood function is, according to the solution I have, $$\begin{align*} L(\lambda) &=\left(\underbrace{1-e^{-\lambda}}_{\mathbb{P}(U_i = 1)}\right)^a\left(\underbrace{e^{-5\lambda}}_{\mathbb{P}(U_i = 5)}\right)^{b}\lambda^{20-a-b}\prod_{\{i: 1 < X_i < 5\}}e^{-\lambda u_i} \\ &=\left(1-e^{-\lambda}\right)^a\left(e^{-5\lambda}\right)^{b}\lambda^{20-a-b}\exp\left(-\lambda \sum_{\{i: 1 < X_i < 5\}}u_i\right)\text{.} \end{align*}$$

Two questions:

1. Why are we using the PDF for when $1 < X_i < 5$ rather than, say, the probability that $1 < X_i < 5$ for when $U_i = X_i$ in $L$?
2. Given the above, the solution I have says that $\left(a, b, \sum_{\{i: 1 < X_i < 5\}}U_i\right)$ is a sufficient statistic. By the factorization theorem, to identify a sufficient statistic, we have to be able to factor $L$ into two functions: one depending on the statistic and $\lambda$, and one only dependent on the $u_i$. So, then, would this mean that the function dependent on the $u_i$ is just equal to $1$?

Answer 1

1. Note that $U_i $ has support $[1,5]$. In particular, $P(U_i=1) = 1-e^{-\lambda} $ and $P(U_i=5) = e^{-5 \lambda}$. But for $x \in (1,5)$, the distribution of $U_i$ is continuous; i.e. $P( U_i \in (x,x+dx)) = f_X(x) dx$ and so we have to use the pdf of $X$ to describe the likelihood here.
2. Yes, that is fine.