Compute $E[\alpha]$ where $\alpha = \inf\{i > 0 : X_i > X_0\}$ and $\{X_i\}_{i=0}^{\infty}$ are IID. If $\{X_i\}_0^{\infty}$ are IID discrete, then I'm quite sure $E[\alpha] = \infty$ because
$$
E[\alpha] = \sum_{\{x_j \}_{j=0}^{N}} E[\alpha | X_0 = x_j] P(X_0 = x_j) = \infty = E[\alpha | X_0 = \sup\{x_j \}_{j=0}^{N}]\\
$$
In this set of values, you will eventually condition on $X_0$ being the maximum value, and as long as the probability of obtaining that maximum value is non-zero, then $E[\alpha | X_0 = \sup\{x_j \}_{j=0}^{N}]$ is necessarily $\infty$ because you will never find another $X_i$ such that $i > 0$ and $X_i > X_0$.
Now what if $\{X_i\}_0^{\infty}$ are IID continuous. I feel the same logic should apply, but for continuous distributions, there should be a probability of zero that $X_0$ realizes the maximum value (or any value), which would seem to mean $E[\alpha]$ is finite.
Is this where the "almost surely" comes in? Which would mean we could assume that there is a non-zero probability that $X_0$ realizes the maximum value of the continuous distribution, in which case the logic applied in the discrete case will still apply here.
 A: Note that, conditional on $X_0$, the events $\{X_i > X_0\}_i$ are independent and all have conditional probability $1-F(X_0)$. Thus, conditional on $X_0$, one has that $\alpha$ is a geometric random variable with chance of success $p(X_0) = 1-F(X_0)$.
$$
E[\alpha] = E[E[\alpha | X_0]] = E\left[ \frac{1}{p(X_0)} \right] = E\left[\frac{1}{1-F(X_0)}\right].
$$
Indeed, if $X_0$ is discrete with finite support or otherwise has positive probability of achieving its supremum, then there is a positive chance that $F(X_0) = 1$, so that the expectation is infinite.
In the continuous case we can easily see the expectation is infinite. We can simplify the expectation further if the distribution is continuous, by the Stieltjes change of variables formula for continuous finite variation processes,
$$
\begin{align}
&= \int_\mathbb{R} \frac{1}{1-F(x)}dF(x) \\ 
 &= (-\log(1-F(\infty))) - (-\log(1-F(-\infty)))\\
&= \infty-0.
\end{align}
$$
Thus in the continuous case the result still holds.
In the general case, we can consider arbitrary $y_n$ increasing and we find using that $1/(1-x)$ is increasing on $[0,1]$ that
$$
\begin{align}
E[\alpha] &\geq \sum_n E\left[\frac{1_{X_0 \in (y_n, y_{n+1}]}}{1-F(X_0)}\right]\\
& \geq \sum_n \frac{1}{1-F(y_n)}P(y_n< X_0 \leq y_{n+1})\\
& = \sum_n \frac{F(y_{n+1})-F(y_n)}{1-F(y_n)}
\end{align}
$$
Assuming that $X_0$ does not achieve its essential supremum with positive probability (that case was already covered), we can choose the $y_n$ cleverly to make the sum infinite.
Choose any $y_0$ such that $0 < F(y_0) < 1$ and recursively define $y_{n+1}$ to be such that $$1> F(y_{n+1}) - F(y_{n}) > \frac{1}{2}(1-F(y_n)).$$ We can always do this since the sup is never attained and $F(y)-F(y_n) \to 1-F(y_n)$ as $y \to \infty$.
Then we have
$$
E[\alpha] \geq \sum_n \frac{1}{2} = \infty.
$$
In summary, $E[\alpha] = \infty$ for all possible distributions of $X_0$ on $\mathbb{R}$.
A: Assume $X_i$ are IID. Then
$$P(\alpha \ge k) =P \bigl(X_0=\max\{X_0,X_1, \ldots,X_{k-1}\} \bigr) \ge 1/k$$ by symmetry. (If the variables are continuous, then the rightmost inequality is an equality.)
Therefore,$$E(\alpha)=E \left[\sum_{k=1}^\infty {\mathbf 1}_{\{k \le \alpha\}} \right]=
\sum_{k=1}^\infty  P(\alpha \ge k) \ge \sum_{k=1}^\infty 1/k =\infty \,.$$
