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A random variable supported on a finite or countable subset of the real line. Its law is specified by a probability mass function. Countable support need not consist of integers: a variable constantly equal to $3/2$ is discrete. Integer-indexed tail-sum formulas therefore require a separate integer-valued hypothesis.

For a nonnegative integer-valued variable, $X={\sum }_{m\ge 0}{1}_{\{X>m\}}$ and the tail-sum formula gives $\mathbb{E}X={\sum }_{m\ge 0}\mathbb{P}(X>m)$. For arbitrary positive real values, that integer-indexed sum equals $\mathbb{E}\lceil X\rceil$ instead. Two independent constants equal to $3/2$ demonstrate that the product of such tail sums need not equal the tail sum of their product: the two sides are four and three.

Assume $\mathbb{E}|g(X)|<\infty$, or use nonnegative $g$ with extended expectations. Let $p_{ij}=\mathbb{P}(X=x_i,Y=y_j)$ and $p_j=\mathbb{P}(Y=y_j)$. For $p_j>0$, the discrete conditional expectation is $\mathbb{E}[g(X)\mid Y=y_j]={\sum }_{i}g(x_i)p_{ij}/p_j$. Therefore Fubini's theorem, or Tonelli theorem in the nonnegative case, gives the tower property of conditional expectation:Values of a conditional expectation on zero-probability conditioning events do not affect the sum. An unrestricted signed $g$ whose expectation is undefined is outside this assertion.

Write $M(x)=\mathbb{E}[N(x)]$. The waiting time for two increments larger than $1/2$ dominates $N(x)$ for $0\le x\le 1$ and has mean $4$, so $M$ is finite. Conditioning on the first increment, one step is always consumed; if that increment is below $x$, the remaining independent sequence has the original law. Thus first-step analysis and the tower property of conditional expectation giveThis first makes $M$ continuous and then gives $M^{\prime }=M$, with $M(0)=1$. Hence the uniform-sum crossing time below one hasAn independent explanation uses the tail-sum formula: $\mathbb{P}(N(x)>n)=\mathbb{P}(U_1+\cdots +U_n\le x)=x^n/n!$ for $x\le 1$, since this region is an $n$-dimensional simplex inside the unit cube. Summing from $n=0$ gives $e^x$. In particular the strict-crossing convention gives $N(0)=1$ almost surely and $\mathbb{E}N(1)=e$.

Write the countable supports as $D_i$ and the marginal masses as $p_i(x_i)$. Independence gives joint mass ${\prod }_i{p}_i(x_i)$. Assuming finite expected values, absolute integrability is justified by Tonelli theorem:Consequently Fubini's theorem permits factorization of the signed sum, proving the expectation of a product of independent random variables:

The second claim holds as a general assertion with the additional hypothesis that the positive variables are integer-valued. For a nonnegative integer-valued $Z$, the pointwise identity $Z={\sum }_{m\ge 0}{1}_{\{Z>m\}}$ and Tonelli theorem give the tail-sum formulaIf each $X_i$ is integer-valued, their product is too. Apply this identity to each factor and to the product, then use independence:Under the literal broader meaning of a discrete random variable, the second claim is false. Two independent constants $X_1=X_2=3/2$ are positive and discrete, but the left side is $2\cdot 2=4$ and the right side is $3$. More generally the integer-indexed tail sum of a positive real-valued variable is $\mathbb{E}\lceil Z\rceil$, not $\mathbb{E}Z$. This is the integer-valued hypothesis in the tail-sum formula.

Let ${\mathcal{F}}_n=\sigma (X_1,\dots ,X_n)$ be the natural filtration. An integer-valued random variable $M$ is a stopping time with respect to $(X_i)$ whenThus whether one has stopped by time $n$ can be decided from the first $n$ observations, without seeing future variables. Equivalently, $\{M\ge i\}\in {\mathcal{F}}_{i-1}$.

Write the stopped sum asThe event $\{M\ge i\}$ depends only on $X_1,\dots ,X_{i-1}$, so its indicator and $X_i$ are independent random variables. Since $\mathbb{E}|X_1|<\infty$ and $\mathbb{E}M<\infty$, Tonelli theorem givesWe may consequently exchange expectation and summation. Using the tail-sum formula for $M$ proves Wald's equation:

For the renewal process, put $S_n=X_1+\cdots +X_n$ and $N_t=\max \{n:S_n\le t\}$. The random variable $N_t+1$ is a stopping time. Since $S_{N_t+1}>t$, Wald's equation givesand hence

For $c>0$, set $Y_i=X_i\wedge c$ and ${\mu }_c=\mathbb{E}{Y}_i$. Wald's equation still applies to the stopping time $N_t+1$ and the independent variables $Y_i$. Because $S_{N_t}\le t$,ThereforeBy the monotone convergence theorem, ${\mu }_c\uparrow \mu$ as $c\to \infty$. Combining the bounds proves the elementary renewal theorem

For a fixed word over $q$ equally likely symbols, the waiting time for a word in independent uniform symbols has mean $\sum q^k$, where $k$ runs through the lengths of the borders of a word, including the full word. The word lava' has no proper nonempty border, so $$ \boxed{\mathbb E\tau_{\mathtt{lava}}=100^4=100\,000\,000}. $$ The word aa' has borders of lengths $1$ and $2$, soFor a direct check, let $E_0$ be the expected remaining time with no trailing a' and $E_1$ with one trailing a'. Thenwhich gives $E_0=10100$.

A renewal-reward process consists of i.i.d. regenerative cycles $({\xi }_n,R_n)$, where ${\xi }_n>0$ is the $n$th cycle length and $R_n$ is the reward accumulated during that cycle. If $0<\mathbb{E}{\xi }_1<\infty$ and $\mathbb{E}|R_1|<\infty$, the renewal-reward theorem states that the cumulative reward $Z(t)$ satisfiesalmost surely, with the usual negligible-remainder condition for an unfinished cycle.

For the machine, let $X\sim \mathrm{Exp}(\lambda )$ be its working lifetime after a repair. The next repair occurs at the first inspection following failure, so one repair-to-repair cycle has lengthwhile its working-time reward is $R=X$. Since an exponential distribution of rate $\lambda$ has mean $1/\lambda$,By the tail-sum formula for the positive integer-valued variable $\lceil X/m\rceil$,Thereforeand the long-run proportion of time for which the machine is working is

Expanding the square and using the independence of the percolation edges givesThe ratio equals $p^{-N_r(\gamma ,{\gamma }^{\prime })}$, where $N_r$ is the number of common edges in the two truncated paths. It is at most $p^{-N}$ for $N=|\xi \cap {\xi }^{\prime }|$, under the convention in the hypothesis; in particular $N\ge 1$ because both paths contain $o$. For every integer $N\ge 1$,The tail-sum formula and the assumed exponential intersection tail therefore yield

Put $R=\Vert X_{(L)}-x{\Vert }_{\infty }$ and ${\delta }_0=(2L/(nM){)}^{1/d}$. If ${\delta }_0\ge 1$, the claim follows from $R\le 1$. Otherwise, for $\delta \ge {\delta }_0$, part (c) givesThe tail-sum formula then yields

For independent uniform distribution increments on $(0,1)$ and $0\le x\le 1$, let $N(x)$ be the first index at which their partial sum exceeds $x$. First-step analysis gives $M(x)=\mathbb{E}N(x)=1+{\int }_0^{x}M(s)ds$, hence $M=e^x$. Equivalently, $\mathbb{P}(N(x)>n)=x^n/n!$ is a simplex-volume probability and the tail-sum formula sums to $e^x$. The strict inequality gives $N(0)=1$ almost surely.