23.3.1 Stopping times

We will motivate the concept discussed here by looking at the gambling situation. Some gamblers claim that they can overcome unfavourable odds by a clever ‘system’. This often takes the form of planning to stop at a certain point. For example, they will continue gambling until they have won $100 and then quit. That way, they claim, they are always a winner. Or, they will continue to bet on black until the wheel comes up black four times in a row, and then quit. They are using what is called a stopping time. Intuitively, a stopping time is a rule that tells you when to stop, and it must be such that you know about it when the time occurs. In other words, it depends only on the past and not the future. Stopping after four blacks in a row is a legitimate stopping time. A rule which says that whenever there are four blacks in a row, then you stop after the third one, is not a stopping time since you clearly will not be aware of that time when it occurs. (The concept is similar to that we encountered in Section 20.5 when discussing trading strategies).

Here is a more formal definition. A stopping time for a discrete-time stochastic process is a rule that assigns to each realization (x_n) of the process an integer k, the stopping time, in such a way that if we assign k to a realization (x_n), and (y_n) is a realization such that x_n = y_n for n = 1, 2, …, k, then we must assign the stopping time k to (y_n) as well.

To illustrate this definition, take the coin flipping example, starting with an initial surplus of 3, winning 1 for a head, and consider a rule which tells you to stop after the first head whenever you get two consecutive heads. This should not be a stopping time, and we can see that it does not satisfy the definition. A realization of the form (3, 4, 5, …) would be assigned 1, but a realization of the form (3, 4, 3, …), which agrees with the first one up to time 1, would not be assigned 1.

A stopping time will often be denoted by a letter such as S. A major example that we have already encountered is when S equals the time of ruin, which certainly satisfies the requirement for a stopping time. A particularly simple example of a stopping time is S = k, for some fixed k. That is, one stops at time k regardless of what has happened.

A fundamental fact about the fixed stopping time is that for any martingale (X_k), as defined in Section 18.3,

(23.5)

This is intuitively clear. In the case of a Markov chain, it is derived easily from

and by induction we derive (23.23).

Example 23.3 Consider again the game of flipping a fair coin, winning 1 for heads, starting with an initial surplus of 3. The gambler decides to stop at time 4 or whenever two consecutive heads come up, if earlier. What is the expected surplus at the end of the game?

Solution. This is complicated by the fact that we no longer have a Markov chain. There is, however, a useful general technique that allows us to recover the Markov property by adding states. In this example, instead of having a single state for each integer w, we insert a state wu to signify that there was a win on the previous play, following a loss on the play before; and we insert a state wd to signify a loss on the previous play. See Figure 23.1, where the shaded boxes indicate points where play stops. By counting paths, we calculate the expected surplus at stopping as

The stopping rule has not helped to raise the expectation above the initial stake.

23.3.2 The optional stopping theorem and its consequences

This motivates the question of whether it is always true in the case of a martingale that E(X_S) = E(X₀)? That is, does (23.5) hold when the fixed time k is replaced by an arbitrary stopping time? The answer is no. Starting with a positive initial surplus of u, we flip a fair coin repeatedly, wagering any amount b that we choose, and receiving back 2b for a head and nothing for a tail. There are two situations we want to present. In the first, we bet 1 unit each time and continue until we lose all of our initial stake. The stopping time S is the first time the surplus reaches 0. Trivially, E(X_S) = 0 ≠ u. The second case is the familiar doubling strategy. Bet 1 unit on the first toss, and double the bet each successive play. S is the first time a head comes up, and it is not hard to see that E(X_S) is u + 1, so we are sure to gain 1. The problem is that neither of these strategies are feasible in practice. (Of course the first is irrational as well.) In both cases, the amount of time we need is unbounded, and in the second case the amount of capital we need is unbounded as well. We will, however, get an affirmative answer to our question if there exists a suitable bound on a combination of the stopping time and values. The following theorem gives a precise condition. Observe first that in place of X_S, which is not defined if S = ∞, we want in general to consider X_S|S < ∞.

Theorem 23.1 (Optional stopping theorem) Suppose that {X_n} is a martingale and S is stopping time such that

(23.6)

Then

Proof. For any n,

(23.7)

Using a modification of Equation (A.29), the third term above can be written as

In view of the martingale property,

The entire third term therefore reduces to E(X_S|S ⩽ n)P(S ⩽ n). We now simply take limits as n → ∞ to reach the conclusion.

1. S is bounded. That is, for some N > 0, we have S ⩽ N.
2. The values of X_n are bounded in absolute value prior to stopping. That is, there is a constant C such that, for all n, if S > n then |X_n| ⩽ C.

Proof. In (i), the second factor in (23.6) is 0 for n ⩾ N, so (23.6) necessarily holds. In (ii), the second factor in (23.6) approaches 0 by the fact that S is finite, and the first factor is bounded in absolute value by C, so the product converges to 0.

Example 23.4 A gambler starting with an initial fortune of a units repeatedly plays an even money game against an adversary with an initial fortune of b. Each has an equal chance of winning each game. They continue the play until one is broke. What is the probability that the gambler with a units will eventually lose his initial stake before the other does? (As an equivalent formulation, we can remove any restriction from the opponents’ initial stake and instead postulate that the gambler decides to quit upon losing a or winning b.)

Solution. In Example 23.2 above, we essentially considered the case that b was infinite, that is there was no restriction on how much the opponent might lose, and we saw that ruin was certain. Here, we deal with the more realistic case that b is finite. Even a casino has some upper bound on its available wealth. Let U_k be the fortune of our gambler at time k. In view of the fairness of the game, this is indeed a martingale. Let S be the time that the gambler either loses the initial stake of a or wins b from the opponent. This is a stopping time that certainly satisfies condition (ii) of the Corollary to Theorem 23.1 since the values of U_k range from 0 to a + b. We will show later that S must take a finite value, which allows us to use the Corollary. Let π be the probability that our gambler loses his initial stake. We can invoke the Corollary to conclude that E(U_S) = a. But also, considering the two possibilities for S, we have

and by equating, we obtain

Note that π approaches 1 as b approaches ∞, verifying our conclusion following Example 23.2.

Example 23.5 Redo Example 23.4 assuming now that the probability of a win is p ≠ 1/2. (This is a classical problem, often referred to as gambler’s ruin.)

Solution. The difficulty now is that we no longer have a martingale. However, the following ingenious trick allows us to transform the process into one. To simplify notation, let q = 1 − p. Consider the process

For any n, U_{n + 1} = U_n + G_{n + 1}, where G_n takes the value 1 with probability p and − 1 with probability q. Therefore,

leading to

(23.8)‡

Now, invoking the independence of U_n and G_{n + 1},

showing that {X_n} is a martingale. We can now duplicate the calculations in Example 23.4, applied to X_n. We have that

and

Solving,

We return to the unfinished business of showing that the stopping time S of the last two examples must assume a finite value. We have a finite-state Markov Chain with states 0, 1, 2, …, a + b, corresponding to the amount held by the person who started with a. We can see, similarly to Example 18.3, that the states 1, 2, …, a + b − 1 are transient and the states 0 and a + b are both absorbing and therefore recurrent. By Theorem 18.2 the process must reach one of these two recurrent states, implying that S cannot take a value of ∞.

Here is another striking application of this idea.

Example 23.6 A gambler plays a game in which she will either win 1000 with probability p, where p < 1, or lose 1 with probability 1 − p. Suppose, however, that whenever she accumulates more than 10 000 in winnings, a companion takes everything in excess of 10 000 to spend in the casino gift shop. What is the probability that the gambler will eventually go broke?

Solution. The probability is 1. Regardless of the initial fortune or the value of p, the gambler in this case is sure to lose everything if she plays for a sufficiently long time. Once again we have a finite Markov chain with states taking values from 0 to 10 000. All states except 0 are transient, as we showed in the previous example, and we are certain to reach the one recurrent state of 0.

Remark The key to the above example is the phrase ‘a sufficiently long time’. As a practical matter, most people would be happy to play this game for a high value of p and would expect to eventually emerge a winner.

23.3.3 The adjustment coefficient

We now wish to apply Theorem 23.1 to the discrete surplus process (18.3) as interpreted in Section 23.1. We cannot, however, expect this to be a martingale. It will only be one if E(G) = 0, that is, if c = E(N)E(X). This is unrealistic, since, as we noted in previous chapters, insurers will invariably charge an amount above this expected value to guard against the possibility that the aggregate claims will be higher than expected. That is, they take c = (1 + θ)E(N)E(X) for some θ > 0. However, we can transform this process to be a martingale by the same type of procedure as we used in Example 23.4. The following definition gives the basic tool for doing this.

Definition 23.1 An adjustment coefficient of a random variable G is a positive number R satisfying

(23.9)

If R is an adjustment coefficient of a discrete random variable, then for z = e^{− R} it follows that P_G(z) = 1, so we have already essentially seen this idea in (23.23).

To justify the word ‘the’ in the title of this subsection, we will show that there cannot be two positive numbers satisfying (23.23). Let γ be the supremum of all points for which M_G( − r) is defined. (For example, if G = c − W where W ∼ Exp(β), then γ will simply be equal to β.) In many cases γ = ∞. Define a function ϕ(r) = M_G( − r) − 1 = E(e^{− rG}) − 1 on the interval [0, γ). Then

so ϕ is a convex function that takes the value 0 at the point 0 and, therefore, cannot have more than one positive root (see Figure 23.2).

An equally pertinent problem is to decide if a positive root of ϕ exists. We can show that this will almost always happen in view of the following two properties that G will satisfy in any realistic insurance context:

1. E(G) > 0;
2. P(G < 0) > 0.

As we mentioned above, (i) will hold due to the relative risk loading. Moreover, there must be the possibility of paying out more in claims than the premiums collected, or nobody would ever buy insurance. This will imply (ii).

Now (i) implies that ϕ′(0) = −E(G) < 0, so ϕ will start out negative. We must, therefore, have a positive root as long as lim_{r → γ}M_G( − r) > 1. From condition (ii), we can find a positive numbers s and δ such that P(G < −s) > δ. This means that E(e^{− rG}) > δe^rs > 1 for sufficiently large r. This guarantees a positive root whenever γ = ∞ (for example, when G is finite).

There are a few rare examples, which we will not give here, where γ is finite but lim_{r → γ}M_G( − r) ⩽ 1, so the adjustment coefficient will not exist. (This is aside from those cases where the m.g.f. does not exist in the first place, so we cannot even define the adjustment coefficient.)

It is sometimes convenient to extend the definition of the adjustment coefficient to the extreme cases. If ϕ(t) ⩽ 0 for all t, we say that R = ∞; while if ϕ(t) ⩾ 0 for all t, we say that R = 0.

The adjustment coefficient can be taken as a measure of safety. The higher R is, the less risk there is for the insurer. This is indicated by Theorem 23.4 below, which shows that as R increases, our upper bound for the probability of ruin decreases. This principle is also seen in Examples 23.2 and 23.5. These examples show that for the case where G takes the value 1 with probability p and − 1 with probability 1 − p, the adjustment coefficient is log (p) − log (1 − p), which increases as p does. The following is yet another example.

Example 23.7 Find the adjustment coefficient for a normal distribution with mean μ and variance σ².

Solution. The m.g.f. of the normal is given in (A.52). We have , so that − Rμ + σ²R²/2 = 0, and since R > 0,

(23.10)

We see that for two normals with the same mean, as the variance gets smaller, which should signify less variation and therefore less risk, the adjustment coefficient goes up.

Equation (23.10) is often used as an approximation to the adjustment coefficient for other distributions. This is justified by the following argument. The function log (M_G(t)) has first derivative equal to M′_G(t)/M_G(t), which at t = 0 equals E(G). Differentiating this latter expression yields that the second derivative at 0 is equal to Var(G). We can then write (23.9) in the form

and ignoring powers of R higher than 2 gives (23.23).

The following example will be extensively used in the continuous-time models.

Example 23.8 Suppose that G = c − ⟨N, X⟩, where N ∼ Poisson (λ). Find an expression for R in terms of X and λ.

Solution. Using (A.50) and (A.51), M_G( − r) = e^{− cr}M_{⟨N, X⟩}(r), and then using (A.42) and (21.21), this is equal to . For this to equal 1, the exponent must equal 0, and dividing by λ, we get

(23.11)

Since c = (1 + θ)λE(X), where θ is the relative risk loading, we can also write

(23.12)

23.3.4 The main conclusions

A key fact about the adjustment coefficient is that the process is a martingale. This follows by exactly the same calculation as given in Example 23.5 where we used the fact that E(e^{− RG}) = 1. We now obtain a major conclusion by applying Theorem 23.1 to this martingale. We must first verify that condition (23.6) holds.

Proof. Let μ be the mean and let σ be standard deviation of G, so that

Consider the following events referring to the situation at time n. Let A_n be the event that T > n and U_n ⩾ E(U_n)/2, and let B_n be the event that T > n and 0 ⩽ U_n < E(U_n/2). Since T > n means that U_n must be nonnegative, we see that the event T > n is a disjoint union of A_n and B_n.

We also know from Chebyshev’s inequality (A.11) that

so that P(B_n) → 0 as n → ∞. Then,

As n goes to ∞, the second term approaches zero because the first factor is bounded by 1, and the first term is less than e^{− r(u + nμ)/2} and also approaches 0.

We can now apply Theorem 23.1 to the martingale e with the stopping time T to conclude that

While stopping theorems are normally used to derive the expectation at time of stopping, in this instance we get an expression for the probability of ruin. Note that − U_T|T < ∞ is just the deficit at ruin, which we termed D(u) before, so we have the following theorem.

Note that in the case that D(u) has a constant value of 1, we recover (23.1) with z = e^{− R}.

In general, this theorem does not give us the exact value of ψ(u) since we will not know the denominator. We can say, however, that since R is positive, the denominator is greater than 1, and we conclude the following.

Theorem 23.4 If the adjustment coefficient R exists, we have

and therefore

This theorem tells us that we can make the probability of ruin as small as we like by taking the initial surplus sufficiently high, which is certainly a reasonable conclusion. It also tells us that the probability of ruin reduces exponentially as a function of initial surplus. For example, if the upper bound to the probability of ruin given by Theorem 23.4 is less than 0.05, and we double the initial surplus, we then know that the probability of ruin is less than 0.0025.

It is also sometimes useful to have a lower bound for the ruin probability. This is possible if G is bounded below by − M for some M > 0. We then know that D(u) ⩽ M, and we can conclude from Theorem 23.3 that

We conclude this section with some remarks on what happens in the rare case that the adjustment coefficient does not exist. We can still reach the second conclusion in Theorem 23.4. As long as E(G) > 0 and M_G is defined on some interval of positive length about 0, we can find β > 0 such that M_G( − β) < 1. We then have that is a supermartingale. We obtain a version of (23.7) with ⩾ replacing the first equality sign, and this is enough to derive the conclusion of Theorem 23.4 with β replacing R.

23.5.1 Calculating ruin probabilities

Suppose we have the insurance claims model

and that the following restrictions hold:

1. c = 1.
2. N takes the values 0 or 1 with probabilities 1 − q or q, respectively.
3. X takes positive integer values 1, 2, …, K, with probabilities f(1), f(2), ..., f(K), respectively.
4. 1 > qE(X), so that E(G) > 0.
5. The initial surplus u is a positive integer.

Note that (i) and (iii) simply say that all claim amounts are integer multiples of the premium, since we can always take the amount of the premium as the unit of capital.

To simplify the notation, we will first illustrate with K = 3. Clearly, the following calculations will work for any value of K. Note that G takes the values 1, 0, −1, −2 with probabilities 1 − q, qf(1), qf(2), and qf(3), respectively.

Suppose we start with u units at time 0. In the first period, the four possible outcomes for G lead to four possible values of surplus at time 1. This gives us the following set of equations, one for each value of u:

Note that for convenience we have included terms of the form ψ(i) for i < 0. These, of course, are just equal to 1. If we start with a negative amount, we are already ruined. Next, rearrange the equations above to give

Sum the first n + 1 of these equations. The left-hand side adds up to ψ(0) − ψ(n + 1). To add the right-hand side, it is convenient to add by the diagonals (running northwest to southeast) because they involve the term ψ(k) for the same value of k. Since f(1) + f(2) + f(3) = 1, all of the diagonals will sum to zero except for the three on the upper right and the three on the lower left. The third diagonal on the upper right sums to qψ(0). The first two diagonals on the upper right sum to

The three diagonals on the lower left involve terms in ψ(n + 1), ψ(n), and ψ(n − 1). These will all converge to 0 as n approaches ∞ by Theorem 23.4. Taking limits gives

and solving

(23.14)

It is instructive to write this in terms of θ, the relative risk loading. Since

(23.15)

we can substitute for E(X) in (23.14) to obtain

which tells us that

(23.16)

Now consider the general case with G taking the K values 1, 2, …, K. Formula (23.14) gives us a starting value, and all of the ruin probabilities can be calculated recursively by rearranging our first set of equations to get

(23.17)

which we can write more compactly, using the notation of Section A.12.3, as

Remark The quantity ψ(0) is of importance since it gives us our starting value. One may think, however, that it is intrinsically of little interest, since we are not likely in practice to have a situation where the initial surplus is 0. It is, however, of great significance since given any starting value u, we can interpret ψ(0) as giving the probability that we will eventually reach some point where our surplus is less than u. Similarly, D(0) is the amount by which we are less than the initial surplus, if this occurs. We will exploit this idea to great advantage in Section 23.7.

23.5.2 The distribution of D(u)

The same approach lets us deduce the distribution of D(u). For k = 1, 2, …, K − 1, let ψ_k(u) be the probability that, starting with initial surplus u, ruin eventually occurs and the value of D(u) = k. It will be convenient again to consider negative values of the argument, and we note that

Given values of ψ_k(u), we can immediately find ψ(u), as well as the distribution of D(u), since

(23.18)

and

(23.19)

We calculate ψ_k(u) by following exactly the procedure in the previous subsection. We get the same systems of equations except with ψ replaced by ψ_k. When we sum the second set of equations and take limits, we again will have everything on the right-hand side vanishing, except possibly for a finite number of diagonals on the upper right. Since these all involve negative values of the argument of ψ, these sums will also be zero except for the single diagonal involving the terms ψ_k( − k), and the sum of that will be f(k + 1) + f(k + 2) + ⋅⋅⋅ + f(K) = P(X > k). So, in place of (23.23), we get

(23.20)

From (23.23), (23.23), and (23.23), we immediately have a nice simple formula for the distribution of D(0):

(23.21)

This verifies our intuition of the previous section. In our present case, we have Y = X − 1, and we see precisely how D(0) is a type of ‘shifted to the left’ version of Y. For example, if Y takes values 1, 2, 3, with probabilities 0.2, 0.5, 0.3 respectively, then D(0) takes the values 1, 2, 3 with probabilities that are in the ratio, 1 : 0.8 : 0.3, so they are 10/21, 8/21, 3/21.

We now obtain the same recursion formula as in (23.23), except with ψ replaced by ψ_k, that is,

Example 23.9 q = 1/15, X has a constant value of 4. Calculate ψ₁(1), ψ₂(1), ψ₃(1), and ψ(1). Verify that your answer to the last agrees with that given by Theorem 23.3.

Solution. We note first that f(1) = f(2) = f(3) = 0, f(4) = 1. So

Similarly,

so that ψ(1) = 31/196, and D(1) takes the values 1,2,3 with probabilities 15/31, 15/31, 1/31, respectively. (Note that D(1) differs from D(0), which has a uniform distribution on 1, 2, 3.)

To check this by Theorem 23.3, we first solve for R, or equivalently z = e^{− R}. Since G takes the value 1 with probability 14/15 or − 3 with probability 1/15, we have

and we can verify that z = 1/2. So,

From Theorem 23.3,

as above.

23.6.1 Description of the process

We now turn to ruin calculations in the continuous case. This will be based on a compound Poisson process which is simply a process corresponding to the compound Poisson distribution, which we considered in Chapter 21. That is, instead of merely counting 1 every time an event occurs, we take an observation from some given distribution. So we have a distribution X, which we can think of as a severity distribution, and we have a Poisson process, N(t). The resulting compound Poisson process is given by

where {X_k} are independent, each has the same distribution as X, and they are independent of N(t). In other words, we can simply write

We can now provide a model for a surplus process. Suppose that an insurer’s aggregate claims up to time t are given by a compound Poisson process S(t), as above. In the discrete case, we postulated a premium of c per period. We now assume that the insurer collects premiums at a continuous rate of c per period, so in any period of length h total premiums of ch will be collected. We assume also that the insurer begins with an initial surplus of u. The compound Poisson surplus process is the process given by

(23.22)

where U(t) is the surplus at time t. Our relative risk loading is given as in the discrete case by

The probability of ruin is defined similarly to the definition in the discrete case, although we need an infimum to replace the minimum. That is,

See Figure 23.3 for a typical realization of this process. The diagonal lines have slope c, and show the increase in surplus arising from the premium payments. Downward jumps then occur whenever there is a claim.

We can approach this case by considering an approximating discrete model. Suppose that we divide up our time into periods of length h for some small h. Then, if we only view our surplus at the end of each period, we just have a discrete surplus model with the gain in each period given by

where N ∼ Poisson(λh). We will now go over the various results we obtained in the last two sections to see how they are modified for the continuous model. In many instances, we do not give rigorous proofs. However, all the results are motivated by those of the previous sections.

23.6.2 The probability of eventual ruin

We first would like to calculate the adjustment coefficient in the continuous model. It is natural to do this by calculating the adjustment coefficient of G for the approximating discrete model as given above, and then taking the limit as h approaches 0. This turns out to be extremely simple in the Poisson case. The term c/λ that we get in (23.11) is independent of h, so we get exactly the same answer as before. The adjustment coefficient is given by (23.11) or (23.23), as it was in the discrete model. The probability of ruin is now given by Theorems 23.3 and 23.4, which carry over unaltered to the surplus model with compound Poisson aggregate claims.

23.6.3 The value of ψ(0)

In light of the Poisson assumption, and Theorem 18.3, we can view our continuous-time model as a limiting case of the particular model discussed in Section 23.5. As h goes to 0, so will q and we deduce, directly from (23.23), that

This is somewhat surprising, since ψ(0) does not depend on the particular distribution of X, except through the expectation E(X), which affects θ. In other words, given any two distributions for X, as long as they have the same mean, they will produce the same value of ψ(0).

23.6.4 The distribution of D(0)

D(0) is necessarily continuous here in view of the fact that premiums are collected continuously. We can deduce, analogously to (23.23), that if f_D(0) is the density function of D(0), then

(23.23)

For later purposes, we will need the m.g.f. of this distribution. Integrating by parts,

so that

(23.24)

23.6.5 The case when is exponentially distributed

It is easy to deduce from the distribution of D(0) given above that when X ∼ Exp(β), then also D(0) ∼ Exp(β). However, much more than this is true. As a continuous analogue to (23.23), we obtain the following:

The last observation tells us that we can get an exact calculation of the ruin probability when X is exponential. We first must calculate the adjustment coefficient. From (23.23), we solve

and it is immediate that

Substituting in (23.12) gives

(23.25)

Let us verify that our formula for R makes sense intuitively, recalling that our adjustment coefficient is a measure of safety. It increases as β increases, which it should since the mean of the severity distribution is decreasing. It must be 0 in the extreme case that θ = 0, and it must increase with θ.

Note now that the denominator in Theorem 23.3 is just M_D(u)(R). When X ∼ Exp(β), D(u) has the same distribution, and from (23.25) we know the denominator is 1 + θ. We substitute in the statement of Theorem 20.3 to obtain a key result.