Suppose $U_1,...,U_n$ are independent Bernoulli random variables taking values $0$ or $1$ with $Pr(U_i=1) = \theta$ and $Pr(U_i=0)=1-\theta$
Let $S = U_1+...+U_n$ be the number of $U_i$ equal to one.
Show $S \sim Binomial(n, \theta)$
I'm unclear as to what I need to show in order to achieve this. Would it be enough to show:
$(1)$ That the sample of space of $S$ is the same as a Binomial variable with such parameters.
$(2)$ The probability of each element of the sample space and hence the probability distribution of $S$ is also the same as a Binomial variable with such parameters.
Or is there something specific way to show that $S$ is a Binomial random variable?
The easiest proof is to use Moment Generating Function and its properties
$$ \bbox[5px,border:2px solid black] { MGF_{U_i}=(1-\theta)+\theta e^t \qquad (1) } $$
$$ \bbox[5px,border:2px solid black] { MGF_S=\Pi_i MGF_{U_i}=[(1-\theta)+\theta e^t]^n \qquad (2) } $$
We immediately recognize in (2) the MGF of a binomial $B(n;\theta)$
To realize that, take 2 iid Bernulli rv's.
Observe that $S=U_1+U_2$ can take values in $\{0;1;2\}$ with probablities
$P(S=0)=P(U_1=0;U_2=0)=(1-\theta)^2$
$P(S=1)=P(U_1=0;U_2=1)+P(U_1=1;U_2=0)=2\theta(1-\theta)$
$P(S=2)=P(U_1=1;U_2=1)=\theta^2$
Using a compact notation you get
$$P(S=s)=\binom{2}{s}\theta^s(1-\theta)^{2-s}$$
$s=0;1;2$
that is a binomial $B(2;\theta)$