# Taking Expectation of Stopping Time and Integral Manipulation

Consider a stopping time $$\tau$$ that represents the point in time when the first credit event (e.g. default) occurs on a compact interval $$[0,T]$$.

Consider the expectation of the indicator function, $$\mathbf{1}_{\{\tau\leq T\}}$$, under a well-defined filtered probability space,$$(\Omega,\{F_t\}_{t\geq0},P)$$:

$$E_P[\mathbf{1}_{\{\tau\leq T\}}]$$

I want to vary the stopping time by fixing $$\tau=s$$ where $$s$$ varies in $$[0,T].$$ Then,

$$E_P[\mathbf{1}_{\{\tau\leq T\}}]=\int_0^T E_P[\mathbf{1}_{\{\tau=s\}}]ds.$$

My question:

(1) Is the above manipulation valid? If so, how? If not, why?

(2) Under what circumstances, would such manipulation be useful?

What is still unclear is the interpretation of the above equality's RHS.

My understanding is:

$$E_P[\mathbf{1}_{\{\tau\leq T\}}]=P(\{\omega:\tau(\omega)\leq T\}).$$

Hence, this represents the probability of the first credit event happening on $$[0,T].$$

Now, let's move one to the RHS:

$$\int_0^T E_P[\mathbf{1}_{\{\tau=s\}}]ds=\int_0^T P(\{\omega:\tau(\omega)=s\})ds.$$

So, how is this equivalent to the original equality's LHS?

To me it reads, $$P(\{\omega:\tau(\omega)=s\})$$ is the probability of the first credit event happening at time $$s$$, and we are integrating over $$s$$? I just don't understand how this yields the equivalent interpretation of the probability of the first credit event occurring on $$[0,T].$$

• Hi Frank, I tried to give an intuitive answer in the context of Credit Valuation Adjustment calculations (to give a practical example). If you were looking for a pure "mathematical" answer with regards to the conditions that have to be true for the integral to evaluate, pls let me know by responding to this comment: in that case, a more mathematical answer might be appropriate. Jun 22, 2020 at 9:31
• @JanStuller Your answer is quite nice, and thanks for responding with an intuitive example. I am looking for a more mathematical answer. For example, the original indicator function simply becomes 1 if the credit event occurs before and including $T$. Now, when we let this stopping time be fixed at $s$ but vary by the integral expression, why is this integration in the first place? I want to know exactly how the expression goes from left-hand side to the right-hand side if the equality is valid. Jun 22, 2020 at 9:59
• Ok, cool: in that case I'd suggest removing the "acceptance" of my answer below. Someone else might add a more mathematical answer with conditions. I might also add an edit to my answer later on: but until an answer fully satisfies your question, I would remove the tickmark, otherwise your question appears as "answered". Jun 22, 2020 at 10:56
• Maybe I am missing something, but you seem to imply your stopping time has a continuous distribution, in which case: $\int_0^T E_P\left(\mathbf{1}_{\{\tau=s\}}\right)ds=\int_0^TP\left(\tau=s\right)ds=0$, so your manipulation would not be valid. Jun 22, 2020 at 13:08
• @DaneelOlivaw: that's a good point. Any random variable $\tau$ that has a continuous PDF has, by definition, $\mathbb{P}(\tau = s: s \epsilon \mathbb{R})=0$. In practice though, the term does appear in CVA calcs quite a lot, whereby it is understood as the counterparty defaulting over an infinitesimal time interval $(s,s_+)$. In practice, CDS spreads have limited granularity anyway: PD can be modeled via a Poisson process with piece-wise constant intensity, where the intensity corresponds to the particular segment of the CDS curve. Jun 22, 2020 at 20:11

Let me try to answer. The term you mention in your question frequently appears in CVA (Credit Valuation Adjustment) calculations. In the context of CVA, the stopping time referring to a credit event is the point in time when a counterparty defaults (by "counterparty" I mean some financial or corporate institution which has traded a portfolio of derivatives with some bank, and therefore this financial or corporate institution is the bank's "counterparty" on this derivative portfolio). CVA is basically the market-implied cost of insuring the credit risk related to this counterparty defaulting.

The generic CVA formula can be written as below ($$Df(t)$$ is the discount factor from $$t_0$$ to $$t$$, $$V(t)$$ is the portfolio value at time $$t$$, LGD is "Loss given default". If a counterparty defaults and you can still recover "$$x$$%" of your the portfolio value, then $$LGD = 1 - x$$):

$$CVA(t |\mathbb{F_{t_0}}) = \mathbb{E_Q} \left[ \int_{s=t_0}^{s=t} Df(s)* I_{(V_s>0)} * I_{(default_s)} *LGD* V(s) ds \right] = \\ = LGD* \mathbb{E_Q} \left[ \int_{s=t_0}^{s=t} Df(s)*I_{(default_s)}* V(s)^+ ds \right] = \\ = LGD* \int_{s=t_0}^{s=t} \mathbb{E_Q} \left[Df(s)*I_{(default_s)}* V(s)^+ \right] ds = \\ = LGD* \int_{s=t_0}^{s=t} \mathbb{E_Q} \left[I_{(default_s)} \right]* \mathbb{E_Q} \left[ \tilde{V}(s)^+ \right] ds$$

Above, $$\tilde{V}(s)$$ is the discounted portfolio value.

Now the interesting term is $$\mathbb{E_Q} \left[I_{(default_s)} \right]$$, which is the expectation over an indicator function that is equal to "one" if the counterarty is in default at time $$s$$.

In my experience, many people struggle with this term. The way I like to think about this term is to tell myself that a "counterparty can only default at time $$s=t$$ if it had survived until time $$s = t_-$$, where $$t_-$$ stands for the infinitesimally earlier point in time than time $$t$$. So really, the term $$I_{(default_s)}$$ should be $$I_{(default_s\cap survival(t_0,s_-))}$$.

In words, the term you mention in your question:

$$E_P[\mathbf{1}_{\{\tau\leq T\}}]=\int_0^T E_P[\mathbf{1}_{\{\tau=s\}}]ds.$$

Is the probability that the counterparty defaults at any point in time before and including time $$T$$ (or, in more general terms, the probability that a credit event has occured before and including $$T$$).

I think the stopping time notation is not that intuitive. There's nothing wrong with the integral you wrote, but I'd probably prefer rewriting it:

$$\int_{s=t_0}^{s=T} \mathbb{P}(Default_s|Survival_{s_-})*\mathbb{P}(Survival_{s_-})ds$$.

It becomes even more intuitive if the integral is discretized into $$n$$ intervals, so that each interval has length $$t_i - t_{i-1}$$ . For each such period $$t_i - t_{i-1}$$, you can get the conditional probability of default by bootstrapping the CDS curve. So the forward CDS spreads give you (simplified slightly):

$$\frac{ CDS \left( t_{i-1},t_i \right)}{LGD} = \mathbb{P} \left(Default\left( t_{i-1},t_i \right)|Survival \left( t_0,t_{i-1} \right) \right)$$

And:

$$1 - \frac{ CDS \left( t_0,t_{i-1} \right)}{LGD} = \mathbb{P} \left(Survival \left( t_0,t_{i-1} \right) \right)$$.