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For a uniform random permutation $\pi$ of $n\ge 1$ objects, let $M_n$ count its fixed points. Write $M_n={\sum }_{i=1}^n{1}_{\{\pi (i)=i\}}$. Each indicator random variable has expected value $1/n$, so linearity of expectation gives $\mathbb{E}[M_n]=1$ for every $n\ge 1$. This average does not require independence: for $n\ge 2$, fixing two specified objects has probability $1/[n(n-1)]$, greater than the product $1/n^2$ of their individual probabilities.

For four objects, exact enumeration gives counts $9,8,6,0,1$ for $0,1,2,3,4$ fixed points among the $24$ equally likely permutations. Their weighted arithmetic mean is $(0\cdot 9+1\cdot 8+2\cdot 6+4\cdot 1)/24=1$. Exactly three fixed points are impossible: the last object must then also be fixed.

Choose the $m$ fixed objects and derange the remaining $n-m$. Their count is $\left(\genfrac{}{}{0ex}{}{n}{m}\right)D_{n-m}$ among $n!$ equally likely permutations. The inclusion-exclusion principle for $D_{n-m}$ proves the displayed law, valid for $0\le m\le n$ and zero otherwise. With $D_0=1$, it includes the all-fixed case and the impossibility of exactly $n-1$ fixed objects. For fixed $m$, the law tends to $e^{-1}/m!$, the Poisson distribution of mean one.

For a uniform random permutation $a_1,\dots ,a_n$, define $R_i$ as the rank of $a_i$ among $a_1,\dots ,a_i$, with rank one assigned to the smallest value. Each sequence $r_i\in \{1,\dots ,i\}$ corresponds to exactly one permutation: insert index $i$ in position $r_i$ of the current increasing-value list, then assign the final list the values $1,\dots ,n$. The $n!$ equally likely permutations therefore correspond to the ${\prod }_{i=1}^{n}i=n!$ equally likely rank sequences. Consequently the $R_i$ are independent random variables, each with the displayed uniform distribution.

For integrable random variables $X_1,\dots ,X_n$ on the same probability space and fixed real numbers $a_1,\dots ,a_n$, linearity of expectation statesThe random variables can be dependent random variables. The formula follows by the linearity of integration; a finite linear combination of integrable random variables remains integrable. For example, if $X$ takes $-1,2,8$ with probabilities $1/2,3/10,1/5$, then $X$ and $2X$ are dependent random variables, yet $\mathbb{E}[X+2X]=3\mathbb{E}[X]=5.1$.

An especially useful application counts how many events occur. Put $I_i=1_{A_i}$, an indicator random variable, and $N={\sum }_i{I}_i$. Since $\mathbb{E}[I_i]=\mathbb{P}(A_i)$, we get $\mathbb{E}[N]={\sum }_i\mathbb{P}(A_i)$. There is no need to compute the whole probability distribution of $N$, or to assume independence. For a uniform random permutation of $n\ge 1$ objects, each object stays fixed with probability $1/n$, so the fixed point count of a uniform random permutation has expected value one for every $n$. ^[ref].

Linearity of expectation concerns addition and multiplication by fixed scalars. Products instead satisfy $\mathbb{E}[XY]=\mathbb{E}[X]\mathbb{E}[Y]$ under the hypotheses of expectation of a product of independent random variables. For the example above, $\mathbb{E}[X^2]=14.5$, whereas $(\mathbb{E}[X]{)}^2=2.89$. Applying a nonlinear function before averaging can therefore change the answer substantially; use the law of the unconscious statistician to compute that average.

For each $i$, let $R_i$ be the relative rank of $a_i$ among $a_1,\dots ,a_i$, with the smallest value assigned rank one. Then $Y_i=1_{\{R_i=1\}}$. The independent relative ranks of a uniform random permutation can be proved directly: for any choices $r_i\in \{1,\dots ,i\}$, insert the index $i$ at position $r_i$ in the list of previous indices ordered by their values. After all insertions, assign the values $1,\dots ,n$ along that list. This is a unique permutation with those relative ranks, and conversely every permutation yields one such insertion sequence.

There are ${\prod }_{i=1}^{n}i=n!$ possible insertion sequences, all equally likely because the uniform random permutation is uniform. ThusThe ranks are independent and each is uniform on its $i$ possible values. It follows that the independent record indicators haveThe first indicator is the constant one, which is also independent of the others. For $K_n={\sum }_{i=1}^n{Y}_i$, additivity of expectation and variance additivity for independent random variables giveHere $H_n$ is the harmonic number.

Let $T$ be the year index of the second record. For $2\le i\le n$, independence givesThe empty product for $i=2$ is one. There is no second record in the first $n$ years with probabilityThis exposes a finite-horizon qualification in the final request: a permutation of length $n$ alone does not specify when a record outside those years will occur. Under the usual continuation by an infinite sequence of independent, identically distributed observations with a continuous distribution, the same relative-rank argument holds for every finite prefix. Then $P(T=i)=1/[i(i-1)]$ for every $i\ge 2$, and $P(T>m)=1/m$. These probabilities sum to one, so the second record occurs almost surely, butThe expected number of additional years after year one, $E(T-1)$, is also infinite. This is the first-sample case of the waiting time for a new record after a fixed sample.

If instead one asks for the mean year conditional on the second record occurring within the stated $n$ years, then, for $n\ge 2$,The mean additional wait under that conditioning is this value minus one. These are distinct questions: assigning $T=\infty$ whenever no second record occurs in the finite horizon also gives an infinite unconditional mean, whereas a censored observation $\min (T,n)$ has finite expectation $1+H_{n-1}$ by the tail-sum formula for expectation.