Let me start by defining the random variable.
I have $n$ units. I run a (Bernoulli) experiment where a unit is assigned to group $U$ with probability $q$ or to group $U'$ with probability $1-q$.
Let $K=|U|$ after this experiment ends. I think of this as flipping $n$ coins with bias $q$ and counting the number of heads. Then, $K \sim \text{Binomial}(n, q)$.
Now, suppose I modify the procedure a little:
If by random chance, after I have flipped all $n$ coins, I have $|U| = \emptyset$, i.e. $K = 0$, I "reject" my results and redo the whole experiment. By this, I mean I reflip all $n$ coins and count the number of heads and that will be my new $K$. If this new $K$ is greater than $0$, I keep it. However, if by random chance I keep getting $K=0$, I start over again. Essentially, I can just never accept $K=0$ but I'm insisting on doing this coin flip procedure to determine $K$. It is clear that $K$ is no longer a binomial random variable… but this is where I get stuck. I don't know how to reason about the distribution of $K$ in general.
Specifically, I need to calculate
$$\mathbb{E}\big[\tfrac{1}{K}\big] = \sum_{i=1}^n \frac{1}{i}\cdot \text{Prob}(K=i),$$
but I don't know how to determine what those probabilities are or how to even begin reasoning about the distribution of $K$. How can I determine those probabilities?
Best Answer
As explained in the comment by Henry, we have a "truncated" or "chopped" Binomial distribution. The probabilities $\text{Prob}(K=i)$ need to be rescaled such that $\sum_{i=1}^n \text{Prob}(K=i) = 1$.
Let $X \sim \text{Binomial}(n,q)$ and let $K$ be distributed as described in the question. We can reason as follows. First, notice that $$\sum_{i=1}^n {n \choose i} q^i(1-q)^{n-i} = 1 - \text{Prob}(X=0) = 1 - (1-q)^n.$$ Dividing both sides of the equation by $1 - (1-q)^n$, we have $$ \frac{\sum_{i=1}^n {n \choose i} q^i(1-q)^{n-i}}{1 - (1-q)^n} = \frac{1 - (1-q)^n}{1 - (1-q)^n} = 1. $$
Thus, if we define $$\text{Prob}(K=i)= \frac{{n \choose i} q^i(1-q)^{n-i}}{1 - (1-q)^n}$$ for $i=1,2,\ldots,n$, we have a valid probability distribution for $K$.
Note that an alternative (but possibly equivalent?) solution would be to note that $$(1-q)^n + \sum_{i=1}^n {n \choose i} q^i(1-q)^{n-i} = 1,$$ and define $\text{Prob}(K=i)= {n \choose i} q^i(1-q)^{n-i} + \frac{(1-q)^n}{n}$.