We have sequence of random variables $X_0, X_1, X_2, \ldots$. Let $\eta=\inf\{n:X_n > X_0\}$. Find distribution $\eta$ and $E\eta$.

Question

We have sequence of random variables $X_0, X_1, X_2, \ldots$ independent identity distributed with continuous cdf. Let $\eta=inf\{n:X_n > X_0\}$. Find distribution $\eta$ and $E\eta$.

Is my approach correct? $$ \begin{split} P(\eta = n) &= \int_{-\infty}^{\infty}P(X_n > t \wedge X_0 \le t) dt \\ &= \int_{-\infty}^{\infty}P(X_n > t)P( X_0 \le t)dt \\ &= \int_{-\infty}^{\infty}F_{X_0}(t)(1-F_{X_n}(t))dt \end{split} $$

Answer 1

$$ \begin{split} P(\eta = n ) &= \int_{-\infty}^{\infty}P( X_1 \leq t \wedge X_2 \leq t ....\wedge ... X_{n-1} \leq t \wedge X_n > t | X_0 = t ) \cdot f_{X_0}(t) dt \\ &= \int_{-\infty}^{\infty}P( X_1 \leq t ) \cdot P( X_2 \leq t ) ... P( X_{n-1} \leq t ) \cdot P( X_n > t ) \cdot f_{X_0}(t) dt \\ &= \int_{-\infty}^{\infty} (F_X(t))^{n-1} \cdot ( 1 - F_X(t) ) \cdot f_X(t) dt \\ \end{split} $$

Using substitution $$ \begin{split} u = F_X(t) \end{split} $$ We get $$ \begin{split} P(\eta=n) = 1/(n(n+1)) \\ E[\eta] = \sum_{n=1}^{\infty} n \cdot P(\eta=n) = \sum_{n=1}^{\infty} 1/(n+1) = \infty \end{split} $$

Answer 2

Hint. Notice that it should hold $P[X_1 > X_0] = P[X_0 > X_1] = 1/2$. (Since the random variables are continuous, $P[X_i = X_j] = 0$ for $i \neq j$). This means $P[\eta = 1] = \frac12$. Can you generalize this reasoning?