Is likelihood function always unique? I am reading the following part of a book. 



What I do not understand in Sample 3 is that why $F_M(m)$ is not the likelihood function. As in sample 2, we produced the probability that $0.9< X < 4$. Similarly, $F_M(m)$ (can be expressed as $F_M(m,\theta)$ can provide the probability of $X_i\le3.5,i = 1,...,5$ for $m = 3.5.$ 
In other words, if I consider all the individual samples independent from each other, I can get likelihood of each sample as follows:
$G_i(\theta) = \Phi(3.5-\theta)$ as $-\infty\le X_i\le 3.5$
After this, I can extract the likelihood as follows:
$L_3(\theta) = \prod\limits_{i = 1}^5 {{G_i(\theta)}} = (\Phi(3.5-\theta))^5$.
What do I miss? Why is my result different than the book?
Should the likelihood $L_3(\theta)$ be unique or my likelihood in above equation is also correct?
 A: Is likelihood function always unique?
The likelihood function is as unique as the probability distribution function. Different variables may have different distribution functions, but there is only one distribution function for each variable.


why $F_M(m)$ is not the likelihood function

$F_M(m)$ is the cumulative distribution/density function (CDF). 
$$\begin{array}\\
F_M(m) &=& \int_{-\infty}^{m} f(t) \text{d}t\\ &=& Pr(M \leq m)
\end{array}$$
The likelihood function is related to the probability density distribution (PDF), not to the cumulative density/distribution function. 
$$\mathcal{L}(\theta \vert m) = f_M(m \vert \theta) \neq F_M(m \vert \theta)$$

As in sample 2, we produced the probability that $0.9< X < 4$. Similarly, $F_M(m)$ (can be expressed as $F_M(m,\theta)$ can provide the probability of $X_i\le3.5,i = 1,...,5$ for $m = 3.5.$

The event that $M=m$ does not equal the event that each of the five $X_i\leq m$. (so you are coming up with the likelihood function for a different observation set) 
The difference is that at least one of the five inequalities $X_i\leq m$ must be an equality. 
$M=m$ implies $X_i\leq m$, but the other way around is not true.
So you get  $5\phi(m-\theta)\Phi(m-\theta)^4$ instead of $\Phi(m-\theta)^5$. 
The term $5\phi(m-\theta)$ refers to the equality relation, and the second part $\Phi(m-\theta)^4$ relates to the inequality relations.
A: The likelihood is unique for each of the three experiments, the difference is between the three experiments in that one gets three different summaries of the 5 Normal random variables:


*

*$x^\text{obs}=x_1=2.3$

*$x^\text{obs}=\mathbb{I}_{(.9,4)}(x_1)=1$

*$x^\text{obs}=\max\{x_1,\ldots,x_5\}=3.5$


The three summaries have different distributions hence the likelihood differs across the three cases. 
In the last one, observing that $$X_{(5)}\stackrel{\text{def}}{=}\max\{x_1,\ldots,x_5\}=3.5$$which has density$$5\mathbb{P}_\theta(X_1\le 3.5)^4f_\theta(3.5)$$ is not equivalent to observing that all five Normal realisations are less than $3.5$, which has probability$$\mathbb{P}_\theta(X_1\le 3.5)^5$$ since in the former setting one of these realisations took the value $3.5$.
