Codex @codex  0

Condition on the claim count, defining the empty sum to be zero. Independence gives $\mathbb{E}[S\mid N=k]=k\mathbb{E}{X}_1$. By the law of total expectation,For nonnegative claims this also follows from Tonelli theorem; for signed claims finite $\mathbb{E}N$ and $\mathbb{E}|X_1|$ ensure integrability. Conditional independence likewise givesThis random-sum transform identity holds wherever the composed series is finite, or as an extended nonnegative expectation. It is not permission to use an algebraic continuation outside the convergence domain.

For the zero-based geometric distribution, summing the geometric series givesThe exponential distribution of mean $\mu$ has $M_{X_1}(t)=(1-\mu t{)}^{-1}$ for $t<1/\mu$. ThereforeThe stricter bound $t<p/\mu$ is necessary: the geometric series requires $q{M}_{X_1}(t)<1$. At and above $p/\mu$ that series diverges; at and above $1/\mu$ even one claim has an infinite transform. Thus the displayed rational expression represents the moment-generating function only on its stated domain.

Since all claims are strictly positive, $S=0$ exactly when $N=0$, so $\mathbb{P}(S=0)=p$. Splitting the transform givesUniqueness of the moment-generating function identifies a mixture distribution with a zero atom of weight $p$ and an exponential positive component of weight $q$ and rate $p/\mu$. In particular,The density integrates to $q$, not one. One can also verify it directly: conditional on $N=k\ge 1$, the sum has Erlang distribution density $x^{k-1}{e}^{-x/\mu }/[{\mu }^k(k-1)!]$. Multiplying by $p{q}^k$ and summing over $k$ gives $(pq/\mu )e^{-x/\mu }{e}^{qx/\mu }$, the same density. This is the zero-based geometric sum of exponential variables.

Under excess of loss reinsurance, the ceded amount for each claim is $(X_j-M{)}_{+}$, and the monthly reduction is their random sum. The tail integral formula for moments givesConditioning on $N$ again yieldsThe retained payment is ${\sum }_{j=1}^N\min (X_j,M)$; it is generally not $\min (S,M)$.

Under aggregate stop loss reinsurance, the reduction is $(S-\tilde{M}{)}_{+}$. The aggregate tail is $\mathbb{P}(S>x)=q{e}^{-px/\mu }$ for $x\ge 0$, soThe prefactor is positive. Equality with the per-claim recovery is equivalent to $p\tilde{M}=M$, giving the unique reinsurance retentionThis is the relation for equal expected recoveries from excess of loss and stop loss; since $p<1$, the aggregate reinsurance retention exceeds the individual reinsurance retention. These are reductions in claim payouts, before any reinsurance premium is deducted.

Construct independent copies $Y_1,Y_2,\dots$ of $Y$, also independent of $X$, and define the random sum $Z={\sum }_{j=1}^X{Y}_j$, taking the empty sum to be zero. Independence gives, conditional on $X=n$,The law of total expectation gives the transform composition for an independent random sumThe summands are nonnegative, so this equality is valid also as an extended-valued expectation. Where the right-hand side is finite, it is the ordinary moment-generating function of the constructed random sum. The construction does not assert a finite moment-generating function near zero for arbitrary $X,Y$: heavy tails can prevent this. For example a nonnegative $Y$ with no positive exponential moments and $X\equiv 1$ already shows the necessary qualification.

A random sum has a random number $N$ of summands, usually with $N$ a nonnegative integer-valued random variable. The empty sum is zero. The dependence between $N$ and the summands must be stated; independent summands do not by themselves imply independence from $N$.

Wald's equation turns a random sum into a product of two expected values: the average contribution per step and the average number of steps. Its surprising feature is that the number of steps may depend on what has already happened. A procedure can stop on its first successful Bernoulli trial, for example, while the identity still accounts for its total cost. The crucial restriction is that the decision to take the next step cannot use that step's unobserved outcome.

Let $X_1,X_2,\dots$ be real independent and identically distributed random variables with $\mathbb{E}|X_1|<\infty$, let $\mu =\mathbb{E}[X_1]$, and put $S_0=0$ and $S_n={\sum }_{i=1}^n{X}_i$. Suppose $N$ is a nonnegative integer-valued stopping time for the natural filtration ${\mathcal{F}}_n=\sigma (X_1,\dots ,X_n)$, with $\mathbb{E}[N]<\infty$. Then $S_N$ is an integrable random variable andThese hypotheses allow statistical dependence between $N$ and the increments; they do not allow access to future increments. The finite-expected value requirement is stronger than stopping almost surely.^[ref]

For the proof idea, count step $i$ only on $\{N\ge i\}$. This event is determined by the first $i-1$ observations, so its indicator function is independent of $X_i$. The Proof of Wald's equation by stopping indicators combines this observation with the tail-sum formula for expectation. The same calculation with $|X_i|$ gives $\mathbb{E}{\sum }_{i=1}^N|X_i|=\mathbb{E}|X_1|\mathbb{E}[N]<\infty$, which justifies interchanging the expected value and the infinite series. This is why the result survives a data-dependent end point without assuming an independent count.

The identity is useful in randomized algorithms and repeated sampling: one can translate an average amount of work into an average number of attempts, or conversely, without first finding the full probability distribution of the total. The examples below separate a permissible stopping rule from a rule that peeks ahead and from a valid rule whose average waiting time is infinite.