Expected number of updates of minimum
The trick is to use linearity of expectation. Let $E_k$ be the event that the $k$th input is a left-to-right minimum (i.e., requires an update), and let $X_k$ be an indicator variable for $E_k$, that is, $X_k$ is $1$ if $E_k$ happens and $0$ otherwise. Let $U = X_1 + \cdots + X_n$ be the number of updates. The expected number of updates is
$$ \mathbb{E}[U] = \sum_{k=1}^n \mathbb{E}[X_k] = \sum_{k=1}^n \Pr[E_k]. $$
It remains to compute $\Pr[E_k]$. We can construct a random permutation $\pi$ of $[n] = \{1,\ldots,n\}$ in the following way: take a random permutation of $[n]$, and randomly permute the first $k$ elements. This shows that the probability that $\pi(k) = \min(\pi(1),\ldots,\pi(k))$ is exactly $1/k$, and so $\Pr[E_k] = 1/k$. All in all, we get
$$ \mathbb{E}[U] = \sum_{k=1}^n \Pr[E_k] = \sum_{k=1}^n \frac{1}{k} = H_n, $$
the $n$th Harmonic number. It is well-known that $H_n = \ln n + \gamma + O(1/n)$ (Wikipedia contains the entire asymptotic expansion).

We can also compute the variance in this way:
$$
\begin{align*}
\mathbb{E}[U^2] &amp;= \sum_{k=1}^n \mathbb{E}[X_k^2] + 2\sum_{k=1}^{n-1} \sum_{\ell=k+1}^n \mathbb{E}[X_k X_\ell] \\ &amp;=
\sum_{k=1}^n \Pr[E_k] + 2\sum_{k=1}^{n-1} \sum_{\ell=k+1}^n \Pr[E_k \land E_\ell],
\end{align*}
$$
where $\land$ is logical AND. We already know that $\Pr[E_k] = 1/k$. In order to compute $\Pr[E_k \land E_\ell]$ (where $k &lt; \ell$), we follow the same route as before. With probability $1/\ell$, $\pi(\ell)$ is a left-to-right minimum. Given that, the probability that $\pi(k)$ is a left-to-right minimum is $1/k$. Therefore $\Pr[E_k \land E_\ell] = 1/(k\ell)$, and so
$$
\begin{align*}
2\sum_{k=1}^{n-1} \sum_{\ell=k+1}^n \Pr[E_k \land E_\ell] &amp;=
2\sum_{k=1}^{n-1} \sum_{\ell=k+1}^n \frac{1}{k\ell} \\ &amp;=
\left(\sum_{k=1}^n \frac{1}{k}\right)^2 - \sum_{k=1}^n \frac{1}{k^2} \\ &amp;=
H_n^2 - \sum_{k=1}^n \frac{1}{k^2}.
\end{align*}
$$
Therefore
$$
\begin{align*}
\mathbb{E}[U^2] &amp;= H_n + H_n^2 - \sum_{k=1}^n \frac{1}{k^2}, \\
\mathbb{V}[U] &amp;= H_n - \sum_{k=1}^n \frac{1}{k^2} = \ln n + \gamma - \frac{\pi^2}{6} + O\left(\frac{1}{n}\right).
\end{align*}
$$
We can compute all other moments in a similar way using (essentially) the inclusion-exclusion principle and the formula
$$
\mathbb{E}[U^d] = \sum_{i_1,\ldots,i_d=1}^n \prod_{i \in \{i_1,\ldots,i_d\}} \frac{1}{i}.
$$
If we are careful enough then we can probably establish the asymptotic normality of $U$.