Suppose I have a finite population of size $N>1$ with units $U_1,U_2,...,U_n$. Now, I have this sampling plan:
(1)either the sample consists of only one unit $U_j$ with probability $\frac{1}{N+1}$ for any $j=1,2,...,N $
(2) or, it consists of the whole population with probability $\frac{1}{N+1}$
Find the expected sample size.
Now, I assumed $X$ be the number of units taken.
See that $P(X=N)=\frac{1}{N+1}$ and also, $P(X=1)=\frac{1}{N+1}$ .But $X$ is supposed to follow Bernoulli distribution but infact here it is not a pmf. I have a doubt whether $P(X=1)=\frac{N}{N+1}$? Help!
Okay first, we define $X$ as the number of units selected. We know that according to the sampling plan, we can either select exactly $1$ unit OR every unit, i.e. $N$ units. From the $(2)$, $$P(X=N)= \frac 1{N+1}$$ Since by definition of a probability distribution, $$P(X=N)+P(X=1)=1$$ Hence $$P(X=1)=1-P(X=N)=1-\frac 1{N+1}=\frac {N+1}{N+1}-\frac 1{N+1}=\frac N{N+1}$$
Now by definition of an expectation, $$E(X)=\sum_x xP(X=x)=1\cdot P(X=1)+N\cdot P(X=N)$$ $$=1\cdot \frac N{N+1}+N\cdot \frac 1{N+1}=\frac {2N}{N+1}$$
Lets also consider what happens when $N$ is large, i.e. asymptotics: $$\lim_{n\to \infty}E(X)=\lim_{n\to \infty} \frac {2N}{N+1}=2$$
Does that make sense?