suppose you have a bermudan swaption with exercise times [$]t_1, ... , t_n[$] and one additional [$]t_{n+1} > t_n[$] arbitrary.Define [$]n+1[$] contracts for each [$]i=1, ... , n+1[$] paying the follwing: If the swaption is exercised on [$]t_i[$], then the [$]i[$]th contract pays [$]1[$] on [$]t_i[$], all other pay nothing. If the swaption is not exercised, then the [$](n+1)[$]th contract pays [$]1[$], all other again zero.Is it necessarily the case that[$]\sum_{i=1}^{n+1} \frac{\pi_i}{P(0,t_i)} = 1[$]with [$]\pi_i[$] denoting the npv of the [$]i[$]th contract ?

mh, I don't think so. Reducing the setup to [$]n=1[$] and denoting [$]\Omega^* = \{ \omega | \text{swaption is exercised (at } t=t_1) \}[$], [$]\Omega'=\Omega - \Omega^*[$], I can write (for any numeraire [$]N[$])[$]\pi_1 = N(0) \int_{\Omega^*} 1 / N(t_1) d\omega[$]and[$]\pi_2 = N(0) \int_{\Omega^{'}} P(t_1,t_2) / N(t_1) d\omega[$]so[$]\pi_1 / P(0,t_1) + \pi_2 / P(0,t_2) = N(0) / P(0,t_1) \int_\Omega (1_{\Omega^*} + P(t_1,t_2)/P(0,t_1,t_2) 1_{\Omega'} ) / N(t_1) d\omega [$]but [$]P(t_1,t_2)/P(0,t_1,t_2)[$] is for sure not expected to be [$]1[$] in the mean over [$]\Omega'[$], since this is the non-exercise region (so contains 'low' or 'high' rate scenarios). Numerical experiments seemto confirm, that the sum of the digitals' forward npvs is bigger than [$]1[$] for a payer swaption (so [$]\Omega'[$] contains low rate scenarios, so [$]P(t_1,t_2)/P(0,t_1,t_2)[$] will be systematically [$]>1[$]) and lowerthan [$]1[$] for a receiver swaption.Does that make sense ? If yes, is there a another notion of "exercise probability" that is (a) pricing measure indpendent and (b) sums up to [$]1[$] (excluding solutions that apply trivial adjustments to the above [$]\pi[$]s) ?

QuoteOriginally posted by: pcasperssuppose you have a bermudan swaption with exercise times [$]t_1, ... , t_n[$] and one additional [$]t_{n+1} > t_n[$] arbitrary.Define [$]n+1[$] contracts for each [$]i=1, ... , n+1[$] paying the follwing: If the swaption is exercised on [$]t_i[$], then the [$]i[$]th contract pays [$]1[$] on [$]t_i[$], all other pay nothing. If the swaption is not exercised, then the [$](n+1)[$]th contract pays [$]1[$], all other again zero.Is it necessarily the case that[$]\sum_{i=1}^{n+1} \frac{\pi_i}{P(0,t_i)} = 1[$]with [$]\pi_i[$] denoting the npv of the [$]i[$]th contract ?Rates are not my area. But since nobody has responded, here was my simple-minded translationof this into terms I understand. I go to a bank. They have two plans; one for n=0 and one for n=1 and they pay yearly.For n=0, I give them my money and they give me back 1 in a year, so the value of that contract is clear: P(0,t1) in your notationFor n=1, they explain that at the end of one year, I can receive 1 or choose to wait another year and then get 1.But, it would seem to me at the end of that first year, I would always exercise (assuming positive rates). So, that two year plan has exactly the same value to me as the one year plan. What am I missing?

The starting point is a plain vanilla swaption, i.e. a long position enables you to exercise at each [$]t_i, i\leq n[$] into a vanilla fix versus Libor payer or receiver swap, the latter starting at [$]t_i[$] plus start delay (say 2 business days). All these underlying swaps have the same maturity [$]t_m > t_n[$].I am trying to give the notion of "exercise probabilities" a well defined meaning. My attempt was to define those digital contracts which pay in case of an exercise of the original swaption, and understand their forward npv as of the respective exercise probability (which would then by construction be independent of the pricing measure). The naive approach would be to calculate simply the plain expectation of exercise indicator functions in the pricing measure (which clearly gives measure dependent results, i.e. the result will be different e.g. in a risk neutral measure or a [$]T_1[$]-forward measure or a [$]T_2[$]-forward measure etc.). However while the naive approach gives probabilities summing up to [$]1[$] (including the indicator for no exercise during lifetime of the option, for each measure), the former "more sophisticated" one seems to lack this desirable property.To explain the notation a bit further: [$]P(t_1,t_2) = P(t_1,t_2,\omega)[$] is the (stochastic) price of a zero bond at time [$]t_1[$] with maturity [$]t_2[$]. Furthermore, [$]P(t,t_1,t_2)[$] is the forward discount bond price as seen from [$]t[$], i.e. it is equal to [$]P(t,t_2)/P(t,t_1)[$].The underlying model is any reasonable interest rate model. Let's just fix a Hull White one factor model.

ok, this is more or less trivial I think. The forward digital prices mentioned below are just [$]E^{t_i} ( 1_{\text{exercise on } t_i})[$]. When adding up these for [$]i=1,...,n[$] one will not get [$]1[$] since each probability is taken in a different measure. When fixing one measure (regardless which), the sum is 1. So there does not seem to be a satisfactory definition of exercise probabilities. Either you fix some measure (but which one) and get a result adding up to [$]1[$], or you look at the digital payoff below which seems to give a natural definition, but only for isolated exercise times. Or to put it differently, don't ask for exercise probabilities extracted from pricing models, this question does not make really sense in the first place. Any further hints from someone or best practices ? The question came up in the context of liquidity planning / risk and calculation of expected cashflows.

Revisiting an old thread - were these exercise probabilities on a tree ever concisely defined? If you compute the 'naive' (unconditional) probability of exercise at each slice then is there a guarantee that they will sum up to 1? In the original example, if you have two slices at t1 and t2 then I would think that you need to compute the conditional probabilities

ExerciseProbability_t1 = P(S1 > K1)
ExerciseProbability_t2 = P(S1<K1 && S2 > K2)
NoExerciseProbability = P(S1<K1 && S2 < K2)

where S1, S2 are the swap rates at t1 and t2 and K1, K2 are the exercise boundary at t1 and t2 (I've made the assumption that the exercise boundary in both cases is above Ki, i=1,2).

If the above is true, it makes the exercise probabilities path-dependent and likely tricky to compute on a tree/lattice?