Q18
Discrete Random Variables
Expectation and Variance via Combinatorial Counting
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We consider a sequence of random variables $(X_n : \Omega \longrightarrow \{-1,1\})_{n \in \mathbf{N}}$ defined on the same probability space $(\Omega, \mathscr{A}, P)$, taking values in $\{-1,1\}$, mutually independent and centered. For every $n \in \mathbf{N}^*$, we denote $S_n = \sum_{k=1}^n X_k$. We fix the integer $n \geqslant 1$. The random variable $N_n : \Omega \longrightarrow \mathbf{N}$ counts, for every $\omega \in \Omega$, the number of equality indices of the path $(X_1(\omega), \cdots, X_{2n}(\omega))$. For every integer $i$ between $1$ and $n$, the event $A_i$ is defined by: $$A_i = \left\{\omega,\; 2i \text{ is an equality index of } (X_1(\omega), \cdots, X_{2n}(\omega))\right\}.$$ Show that the random variable $N_n$ has finite expectation and that its expectation $\mathbb{E}(N_n)$ equals: $$\mathbb{E}(N_n) = \sum_{i=1}^n \frac{\binom{2i}{i}}{4^i}.$$ [Hint: one may express the variable $N_n$ using indicator functions associated with the events $A_i$.]