Let $\{X_{n}\}_{n\in \mathbb{N}}$ be a sequence of i.i.d. random variables with density $f$. Also, define
$$\eta = \# \left\{k \leq n \mid X_{k} \geq \min_{i \leq n} X_{i} + 2\right\}.$$
What is the value of $\mathbb{E}\eta$?
I am really not so sure how to solve this question. The notation I am using $\#\{\cdot\}$ is used to denote the cardinality of a set. I am given the following hint:
Hint: Write $\eta$ as the sum of indicator functions and use symmetry arguments.
Even with this hint, I am not sure how to solve the problem. I guess that we should define $I_{k}$ to be $1$ if $k \leq n$ and $X_{k} \geq \min_{i\leq n} X_{i} + 2$ so that we can write $\eta = \sum_{k=0}^{n} I_{k}$? Then I need $\mathbb{E}[I_{k}] = P(k \leq n \text{ and } X_{k} \geq \min_{i \leq n} X_{i} + 2)$.
I am not even sure if this is correct and even if it is I don't know how to proceed.
I will appreciate your assistance in solving this question. I guess (but not sure) that the answer should be in terms of the density $f$. Otherwise, there is no reason for the problem to introduce this notation.
Actually, you can simply define $I_k$ for $1\leqslant k\leqslant n$. Then indeed, $\eta=\sum_{k=0}^nI_k$ and $$\mathbb E\left[I_k\right]=\mathbb P\left(X_k\geqslant \min_{0\leqslant i\leqslant n}X_i+2\right).$$ Observing that the vectors $\left(X_k,\min_{0\leqslant i\leqslant n}X_i\right)$ and $\left(X_0,\min_{0\leqslant i\leqslant n}X_i\right)$ have the same distribution (this is were the symmetry comes into play), it suffices to compute $\mathbb E\left[I_0\right]=\mathbb P\left(X_0\geqslant \min_{0\leqslant i\leqslant n}X_i+2\right)$. To this aim, we first notice that $$ \left\{X_0\geqslant \min_{0\leqslant i\leqslant n}X_i+2\right\}=\left\{X_0\geqslant \min_{1\leqslant i\leqslant n}X_i+2\right\}. $$ The probability of the complement is maybe easier to compute: by Fubini's theorem, $$ \mathbb P\left\{X_0\lt \min_{1\leqslant i\leqslant n}X_i+2\right\}=\int_{\mathbb R}f(x)\mathbb P\left\{x \lt \min_{1\leqslant i\leqslant n}X_i+2\right\}dx. $$ Then use independence to simplify this.