State the likelihood function Im learning something about likelihood function and there is a task in my book. But I'm not sure if I did it correctly could you help with some tips?
Following task:
Clients are lining up in a post office. We record the time $t_1, \ldots, t_N$ in minutes required to serve the $N$ consecutive
clients. We distinguish between two types of clients, those that are coming to send a packet, and those that are
coming to send a letter (and whose service is typically twice faster). Service times for all clients are independent,
and drawn from an exponential distributions with rate dependent on whether the client sends a packet or a letter:
$$p(t_i\mid\theta) = \theta \exp(-\theta t_i) \tag{packet}$$
$$q(t_i\mid\theta) = 2\theta \exp(-2\theta t_i) \tag{letter}$$
and where $\theta$ is a parameter between $0$ and $\infty$ to be learned.
Consider six clients, the first two wanted to send a packet, and stayed at the post office for $2$ and $5$ minutes
respectively. The last four clients wanted to send a letter and were served in $1$ minute each.
State the likelihood function measuring the joint probability of observing all these events.
My problem here is, that I have the Data $D$ and the information about six Clients. How do I combine this informations?
My solution(s):
1:
$$P(t_1, \ldots, t_ND\mid\theta) = \prod_{i=1}^N (\theta \exp(-\theta t_i))^2 \cdot (2\theta \exp(-2\theta t_i))^4$$
2: Just consider the six given clients.
$$P(2,5,1,1,1,1\mid\theta) = \theta \exp(-2\theta ) \cdot \theta \exp(-5\theta) \cdot (2\theta \exp(-2\theta))^4  $$
 A: Two points to emphasize.

*

*The likelihood function is a function of $\theta$ with $t_1,\ldots,t_N$ fixed. The notation can make this explicit, i.e. you can say $L(\theta) = \theta \exp(-2\theta ) \cdot \theta \exp(-5\theta) \cdot (2\theta \exp(-2\theta))^4$ or you can say the function is $\theta \mapsto \theta \exp(-2\theta ) \cdot \theta \exp(-5\theta) \cdot (2\theta \exp(-2\theta))^4,$ where the notation $\theta\mapsto\cdots$ means that when $\theta$ is the input then the expression after that arrow is the output. (This arrow $\text{“} {\mapsto} \text{”}$ does not mean the same thing as $\text{“} {\to} \text{”}.$)


*I don't know how your instructor does things but I would accept the following (plus some subsequent simplification) as a correct answer: $$ L(\theta) \propto \theta\exp(-2\theta)\cdot\theta\exp(-5\theta)\cdot \theta^4\exp(-2\theta), $$ where the symbol $$\text{“} {\propto} \text{”}$$ means "is proportional to," and here it matters that this means proportional as a function of $\theta$, not proportional as a function of $t_1,\ldots,t_N.$ I know of no context in which anything matters about the likelihood function except which proportionality class it belongs to. This can save work in some situations. For example, if $\lambda$ is the expected value of a sequence of i.i.d. Poisson-distributed random variables, then you would have $$ L(\lambda) = \frac{\lambda^{x_1} e^{-\lambda}}{x_1!} \cdots \frac{\lambda^{x_n} e^{-\lambda}}{x_n!} \propto \lambda^{x_1+\cdots+x_n} e^{-n\lambda}. $$ Those factorials in the denominators don't matter in this context. They can distract from what information can be gleaned from a likelihood function. For another example, suppose you have the gamma distribution $$ \frac 1{\Gamma(6)} (\sigma x)^{6-1} e^{-\sigma x} (\sigma\,dx) $$ and you take an i.i.d. sample $x_1,\ldots,x_n. $ Then $$ L(\sigma) \propto \sigma^{6n} \exp(-\sigma(x_1+\cdots+x_n)). $$ You can discard the factor $(x_1\cdots x_n)^{6-1}$ and the factor $\big( 1/\Gamma(6)\big)^n$ because they don't depend on $\sigma.$
Finally, notice the simplification:
\begin{align}
L(\theta) \propto {} & \theta \exp(-2\theta ) \cdot \theta \exp(-5\theta) \cdot (\theta \exp(-2\theta))^4 \\[8pt]
= {} & \theta^6 \exp(-(2+5+2\cdot4)\theta) \\[8pt]
= {} & \theta^6\exp(-15\theta).
\end{align}
