I am looking at the following problem:
Let $C$ and $C'$ be two circles centered at $(0,0)$ such that $C'$ is inside $C$. Denote by "$+$" samples drawn from inside $C$ and "$-$" samples drawn from outside $C$. The probability of "$+$" samples falling in the gap between $C$ and $C'$ is $\delta$ (the distribution is unknown). We sample a total of $m$ samples from inside $C$ (i.e., supposed to be all "$+$").
If one "+" sample is mistakenly flipped into a "-" sample (it is unknown which sample, but there's definitely one), then (according to my calculations):
$$\Pr[\text{no “$+$” samples fall in } \delta\text{-gap}] \leq (1-\delta)^{m-1}$$
However, someone else insists it should be:
$$\Pr[\text{no “$+$” samples fall in } \delta\text{-gap}] = m(1-\delta)^{m-1} \delta $$
Is it the latter? If it is, I cannot see why (and the other person was unable to explain their understanding besides insisting it is binomial).
Any insight would be much appreciated!
It might help to understand my reasoning -- it does not matter where the false "$-$" falls since it is not a "$+$", and I am only interested in where the "$+$" (as I perceive them, error included) fall.
Here is how I derived mine:
$$\Pr[\text{no “$+$” samples fall in } \delta\text{-gap } | \text{one mistake}] = \Pr[\cup_{i=1}^m \{i \text{ is a mistake }\cap \text{ rest not in } \delta\text{-gap }\} \leq \sum_{i=1}^m\Pr[\text{ rest not in } \delta\text{-gap }\ | i \text{ is a mistake }]\Pr[i \text{ is a mistake }] $$
where the last transition is by the union-bound.
Now, since the samples are iid we obtain:
$$\sum_{i=1}^m\Pr[\text{ rest not in } \delta\text{-gap }\ | i \text{ is a mistake }]\Pr[i \text{ is a mistake }] = m(1-\delta)^{m-1}\frac{1}{m} = (1-\delta)^{m-1}$$
So I see now mine is an upper bound due to the union-bound (note to readers of previous version - I had a "=" and a multiplication by $m$ before, which is now fixed).
Still - could it be possible to obtain the other person't equality using the Binomial distribution?