I'm learning a statistics tutorial which considers
$$
X_i \sim \mathit{Bernoulli}\left(\frac{\theta}{3}\right)
$$
to describe an experiment where one draws a ball 4 times with replacement from a bag that contains 3 balls.
And then the tutorial gives another experiment where
$$
X_i \sim \mathit{Binomial}(3, \theta)
$$
and $(x_1,x_2,x_3,x_4)=(1,3,2,2)$.
I understand
$X_i \sim \mathit{Binomial}(3, \theta)$
when considering 3 the number of experiments where one experiment/trial flips one coin one time and $\theta$ denotes the success possibility for each trial. but I don't understand the definition of one experiment in that tutorial.
Could someone give me a hint?
Possible setting 1
Does 3 in $X_i \sim \mathit{Binomial}(3, \theta)$ mean there are 3 balls in the bag?
Note that in the second experiment, 3 blue balls are observed which means there already are 3 blue balls in the bag, which means, all 3 balls are blue, if the 3 in $X_i \sim \mathit{Binomial}(3, \theta)$ means there are 3 balls in the bag.
Alternatively, the experiments are using different bags, but that doesn't seem to be compatible with the definition of binomial distribution.
Possible setting 2
Does it make sense to define one experiment as follows?
"draw a ball, record if it's blue, and put it back in the bag, repeat 4 times"
In other words, one experiment draws 4 times, so the possible outcomes are ${0, 1, 2, 3, 4}$. And then, $X_i \sim \mathit{Binomial}(3,\theta)$ means doing the experiment 3 times.
In this setting, $\theta$ doesn't mean the possibility that one blue ball is observed, so what does $\theta$ mean?
Possible setting 3
Another possible definition of one experiment could be
"draw 3 balls from the bag, record how many blue balls observed, and then put the balls back in the bag"
So the possible outcome of one experiment could be ${0, 1, 2, 3}$. See the difference between this setting and setting_1?
The 3 in $X_i \sim \mathit{Binomial}(3, \theta)$ means doing the experiment 3 times.
Again, what does $\theta$ mean in this setting?
Best Answer
No. If you want to apply Example 8.8 to the 'balls in the bag' example that the tutorial gives earlier, the 3 means that you do the following action three times: draw a ball, record if its blue, and put it back in the bag (this last part is very important). The total number of blue balls you record is $X_i$ and is a binomial random variable. It will take on values of 0, 1, 2, or 3. You could do this if there were just one ball in the bag or if there were 100 balls in the bag.
In example 8.8, it is telling you that the entire process in the paragraph above was repeated 4 times, and hence you have a set of 4 numbers, each of which could be 0, 1, 2, or 3.
It was perhaps unwise of the tutorial author to use $\theta$ for both the $Bernoulli(\theta/3)$ example (where $\theta$ is the number of blue balls in the bag) and the $Binomial(3,\theta)$ example, where $\theta$ represents the probability of success (i.e. drawing one ball and getting a blue ball).
=====Addition in response to OP's edits=====
A Binomial with $n$ trials is defined as the sum of $n$ independent Bernoulli random variables having a common probability of success, and the sum of independent Binomial random variables each having a common probability of success is itself a Binomial random variable with number of trials equal to the sum of the constituent number of trials. Thus, if $\mu$ is a probability of success (I'm changing the parameter from $\theta$ to $\mu$ so that it is clear I'm not using the tutorial's parametrization), all of the following experiments yield equivalent statistical information (and have the same likelihood up to a multiplicative constant) about $\mu$:
(i) 20 independent draws from $Bernoulli(\mu)\equiv Binomial(1,\mu)$
(ii) 10 independent draws from $Binomial(2,\mu)$
(iii) 5 independent draws from $Binomial(4,\mu)$
(iv) 1 draw from $Binomial(20,\mu)$
The parameter $\mu$ has the same interpretation in all cases: the probability of success for a single Bernoulli random variable.