2
$\begingroup$

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 $1$?
$\endgroup$
2
$\begingroup$
  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.
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.