In its simplest form, it relates the expectation of a sum of randomly many finite-mean, independent and identically distributed random variables to the expected number of terms in the sum and the random variables' common expectation under the condition that the number of terms in the sum is independent of the summands.
be a sequence of real-valued, independent and identically distributed random variables and let N ≥ 0 be an integer-valued random variable that is independent of the sequence (Xn)n∈
By Wald's equation, the resulting value on average is Let (Xn)n∈
Assume that: Then the random sums are integrable and If, in addition, then Remark: Usually, the name Wald's equation refers to this last equality.
Assumption (2) controls the amount of dependence allowed between the sequence (Xn)n∈
Note that assumption (2) is satisfied when N is a stopping time for a sequence of independent random variables (Xn)n∈
[citation needed] Assumption (3) is of more technical nature, implying absolute convergence and therefore allowing arbitrary rearrangement of an infinite series in the proof.
In particular, the conditions (4) and (8) are satisfied if Note that the random variables of the sequence (Xn)n∈
The interesting point is to admit some dependence between the random number N of terms and the sequence (Xn)n∈
A standard version is to assume (1), (5), (8) and the existence of a filtration (Fn)n∈
For convenience (see the proof below using the optional stopping theorem) and to specify the relation of the sequence (Xn)n∈
0, the following additional assumption is often imposed: Note that (11) and (12) together imply that the random variables (Xn)n∈
An application is in actuarial science when considering the total claim amount follows a compound Poisson process within a certain time period, say one year, arising from a random number N of individual insurance claims, whose sizes are described by the random variables (Xn)n∈
Note that, when Z is not almost surely equal to the zero random variable, then (11) and (12) cannot hold simultaneously for any filtration (Fn)n∈
Intuitively, one might expect to have E[SN] > 0 in this example, because the summation stops right after a one, thereby apparently creating a positive bias.
be a sequence of independent, symmetric random variables, where Xn takes each of the values 2n and –2n with probability 1/2.
Since N is a stopping time with respect to the filtration generated by (Xn)n∈
Using assumption (1), define the sequence of random variables Assumption (11) implies that the conditional expectation of Xn given Fn–1 equals E[Xn] almost surely for every n ∈
Hence we can add the expectation of TN to both sides of Equation (13) and obtain by linearity Remark: Note that this proof does not cover the above example with dependent terms.
This proof uses only Lebesgue's monotone and dominated convergence theorems.
0 and since S0 = 0, it follows that The Lebesgue monotone convergence theorem implies that By the triangle inequality, Using this upper estimate and changing the order of summation (which is permitted because all terms are non-negative), we obtain where the second inequality follows using the monotone convergence theorem.
By assumption (3), the infinite sequence on the right-hand side of (15) converges, hence SN is integrable.
0 and since T0 = 0, it follows that As in step 1, the Lebesgue monotone convergence theorem implies that By the triangle inequality, Using this upper estimate and changing the order of summation (which is permitted because all terms are non-negative), we obtain By assumption (2), Substituting this into (17) yields which is finite by assumption (3), hence TN is integrable.
To prove Wald's equation, we essentially go through the same steps again without the absolute value, making use of the integrability of the random sums SN and TN in order to show that they have the same expectation.
Due to assumption (2) and the σ-additivity of the probability measure, Substituting this result into the previous equation, rearranging the summation (which is permitted due to absolute convergence, see (15) above), using linearity of expectation and the definition of the partial sum Ti of expectations given in (16), By using dominated convergence again with dominating random variable |TN|, If assumptions (4) and (5) are satisfied, then by linearity of expectation, This completes the proof.