Question 197, Randomly select a permutation $\sigma=\left(\mathrm{x}_{1}, \mathrm{x}_{2}, \ldots, \mathrm{x}_{\mathrm{n}}\right)$ of $1, 2, \ldots, \mathrm{n}$. If $\mathrm{x}_{\mathrm{i}}=\mathrm{i}$, then $\mathrm{i}$ is called a fixed point of $\sigma$. The number of fixed points in the permutation is denoted as the random variable $\mathrm{X}_{\mathrm{n}}$. Try to find the expectation $\mathrm{E}\left(\mathrm{X}_{\mathrm{n}}\right)$ of $\mathrm{X}_{\mathrm{n}}$.
For each $\mathrm{j} \in\{1,2, \ldots, \mathrm{n}\}$, define the indicator random variable $\mathrm{I}_{\mathrm{l}}=$ $\left\{\begin{array}{c}1, \text { if } \mathrm{j} \text { is a fixed point } \\ 0, \text { if } \mathrm{j} \text { is not a fixed point }\end{array}\right.$ . Then $x_{n}=\sum_{j=1}^{n} \mathrm{I}_{\mathrm{j}}$. Therefore, we have
$$
E\left(X_{n}\right)=E\left(\sum_{j=1}^{n} 1_{j}\right)=\sum_{j=1}^{n} E\left(I_{i}\right)
$$

Notice that, the probability of $\mathrm{j}$ being a fixed point (i.e., $\mathrm{j}$ is exactly in the $j$-th position) is $\frac{1}{n}$, so $E\left(I_{1}\right)=1 \times \frac{1}{n}+0 \times$ $\frac{n-1}{n}=\frac{1}{n}$. Hence, $E\left(X_{n}\right)=\sum_{j=1}^{n} E\left(I_{j}\right)=\sum_{j=1}^{n} \frac{1}{n}=1$.