Recently I met an interesting problem:

[Problem Definition] Given $n$ independent random variables $x_1, x_2, \ldots, x_n \sim \mathcal{U}(0,1)$, what is the expected value of the minimum among them? What is the expected value of the second smallest?

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Tranditional Approach: Probability Distribution

Denote $m_1 = \min_{i\in [n]}{x_i}$, we can find the distribution of $m_1$ as follows. For $0\le m\le 1$:
$$
\begin{aligned}
\Pr(m_1 \le m) =& 1-\Pr(m_1 > m) \\
=& 1-\Pr(\min_{i\in [n]}{x_i} > m) \\
=& 1-\Pr(x_1 > m \cap x_2 > m \cap \cdots \cap x_n > m)
\end{aligned}
$$

Note that $x_1, x_2,\ldots, x_n$ are independent, hence
$$
\Pr(x_1 > m \cap x_2 > m \cap x_n > m) = \prod_{i=1}^{n} \Pr(x_i > m) = (1-m)^n
$$

So the probability distribution function of $m_1$ is $F_{m_1}(m) = 1-(1-m)^n$ for $0\le m\le 1$.

Given this, we can find the expected value of $m_1$ using integral:
$$
\begin{aligned}
\mathbb{E}[m_1] =& \int_{0}^{1} m \text{d}F_{m_1}(m) \\
=& \int_{0}^{1} nm (1-m)^{n-1}\text{d}m \\
=& \left.\left(\frac{n}{n+1}(1-m)^{n+1}-(1-m)^n\right)\right|_{0}^{1} \\
=& \frac{1}{n+1}
\end{aligned}
$$

[Second Smallest Case] Denote $m_2$ is the second smallest value of $x_1,\ldots,x_n$. Similarly, we can try to find the probability distribution function of $m_2$ as follows:

\begin{aligned}
\Pr(m_2 > m) =& \Pr(m_1 > m, m_2 > m) + \Pr(m_1 < m, m_2 > m) \\
=& \Pr(m_1 > m) + \Pr(m_1 < m, m_2 > m) \\
=& \Pr(m_1 > m) + \sum_{i=1}^n\Pr(x_i < m\text{ and }x_{j} > m \text{ for all }j\ne i) \\
=& (1-m)^n + nm(1-m)^{n-1}
\end{aligned}

Hence, $F_{m_2}(m) = 1-((1-m)^n + nm(1-m)^{n-1})$. By integration we have $\mathbb{E}[m_2] = \frac{2}{n+1}$.

[$k$-th Smallest Case] Similarly, if we denote $m_k$ as the $k$-th smallest values among $x_1,x_2,\ldots,x_n$, we have $F_{m_k}(m) = 1-\sum_{i=0}^{k-1}\binom{n}{i}m^{i}(1-m)^{n-i}$, and $\mathbb{E}[m_k] = \frac{k}{n+1}$.

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Easy Approach: Property of Expected Value

The problem can be solved in an much easier way using property of expected value.

[Key Aspects] Note the following observations:
Observation 1: $\mathbb{E}[x_1 | x_1 > m] = \frac{m+1}{2}$
Observation 2: $\mathbb{E}[\mathbb{E}[x_1 | m_1 = m]] = \mathbb{E}[x_1]$

We focus on $x_1$. Note that $\mathbb{E}[x_1] = \frac{1}{2}$ as the property of uniform random variable.

In the following we conditional on $m_1 = m$ for certain $0\le m \le 1$. Let’s consider two events: $A_{=}$ be the event that $x_1$ is the smallest, and $A_{>}$ be the event that $x_1$ is not the smallest.

If $A_{=}$ is the smallest (happens with probability $\frac{1}{n}$), then we have $\mathbb{E}[x_1 | A_{=}, m_1=m] = m$. Otherwise, we have $\mathbb{E}[x_1 | A_{>}, m_1=m] = \frac{m+1}{2}$ (as $x_1$ can only have values between $[m,1]$ and the density is evenly distributed among the corresponding segment. Hence,

$$
\begin{aligned}
\mathbb{E}[x_1| m_1 = m] =& \mathbb{E}[x_1| A_{=}, m_1 = m]\Pr(A_{=}, m_1=m) \\
& + \mathbb{E}[x_1| A_{>}, m_1 = m]\Pr(A_{>}, m_1=m) \\
=& m\cdot \frac{1}{n} + \frac{m+1}{2}\cdot \frac{n-1}{n}
\end{aligned}
$$

As $\frac{1}{2} = \mathbb{E}[x_1] = \mathbb{E}[\mathbb{E}[x_1| m_1 = m]] = \frac{n-1}{2n} + \frac{n+1}{2n}\mathbb{E}[m_1]$, hence we have $\mathbb{E}[m_1]=\frac{1}{n+1}$.

[$k$-th Smallest Case] Similarly, conditional on $m_k = m$, we consider the following events: $A_{<}$ be the event that $x_1$ is less than $k$-th smallest, $A_{=}$ be the event that $x_1$ is $k$-th smallest, $A_{>}$ be the event that $x_1$ is larger than $k$-th smallest. These event happens with probability $\frac{k-1}{n}$, $\frac{1}{n}$, $\frac{n-k}{n}$ respectively. And we have $\mathbb{E}[x_1|A_{<},m_k=m]=\frac{m}{2}$, $\mathbb{E}[x_1|A_{=},m_k=m]=m$, $\mathbb{E}[x_1|A_{>},m_k=m]=\frac{m+1}{2}$. Hence,

$$
\mathbb{E}[x_1| m_k = m] = \frac{m}{2}\cdot\frac{k-1}{n} + m\cdot\frac{1}{n} + \frac{m+1}{2}\cdot\frac{n-k}{n}
$$

and $\frac{1}{2} = \mathbb{E}[x_1] = \mathbb{E}[\mathbb{E}[x_1| m_k = m]] = \frac{n-k}{2n} + \frac{n+1}{2n}\mathbb{E}[m_k]$, which gives $\mathbb{E}[m_k] = \frac{k}{n+1}$.