Drawdown Formulas

November 21, 2025

The aim of this article is to record and add some references on the recent post by Gappy discussing a number of drawdown formulas, see the original post [1] ( here) - primarily to act as a keepsake for myself in the future. Some other useful references on this topic include [2,3,4] .

Definitions #

Assume you have a process (e.g. a capital stream) $X = (X_{t})_{t}$ described as a drifted Brownian motion with drift $\mu > 0$ and volatility $\sigma > 0$, i.e. such that

where $(B_{t})_{t > 0}$ denotes the standard Brownian motion with zero drift and unit variance. Then, the running maximum, and drawdown processes can be defined as

where we note that the drawdown process $(D_{t})_{t}$ is non-negative.

We introduce the following stopping time, for when the processes $X_{t}$ first hit some value $x \in \mathbb{R}$, which will be the emphasis of our discussions below

Remark. Here, for a process $(X_{t})_{t}$ that describes capital, $\mu$ can be interpreted as the expected daily PnL, and $\sigma$ as the daily volatility.

In such a scenario, define the Sharpe ratio across a time horizon $T$ (in business days) as $\operatorname{SR}_{T} := \frac{\mu}{\sigma} \sqrt{T}$ and a period specific volatility as $\sigma _{T} := \sigma \sqrt{T}$. In addition, when considering a drawdown $- b$, for some $b \geqslant 0$, define the vol-normalised drawdown as $b_{\sigma} := \frac{b}{\sigma}$. Furthermore, for brevity, we write $\operatorname{SR}_{1}$ as $\operatorname{SR}$.

Questions of Interest #

We now concern ourselves with the following questions (slightly rephrased to improve clarity for my part). We begin with questions relating to the process $(X_{t})$ rather than the drawdown process $(D_{t})$.

Question 1: Hitting Probability at Any Time #

What is the hitting probability that the process $X_{t}$ hits a given limit at any point in the future – i.e. for $b > 0$, what is $\mathbb{P} \left( X_{t} \leqslant - b, \text{ for some } t > 0 \right)$?

This is a classical problem in stochastic analysis, and can be treated in a number of ways. We provide here a simple proof utilising some existing theory.

The hitting time probability can be restated into the following more tractable form

therefore $\mathbb{P} \left( X_{t} \leqslant - b, \text{ for some } t > 0 \right) =\mathbb{P} (\tau _{b} < \infty)$. Using the monotone converge theorem, we can restate the problem once again into investigating the Laplace transform of $\tau _{b}$

The Laplace transform of $\tau _{b}$ is well studied, and an explicit expression for this can be found in the reference book [5] (Equation 2.0.1, pg. 295), or also accompanied with a proof in [6] (Theorem 8.3.2 pg. 346). In particular, the references above state that for a drifted Brownian motion with drift $\nu$ and unit variance, i.e. $\mathrm{d} Y_{t} = \nu \mathrm{d} t + \mathrm{d} B_{t}$ with $Y_{0} = 0$, the Laplace transform of the hitting probability $\eta _{x} := \inf \lbrace t \geqslant 0 : T_{t} \leqslant - x \rbrace$ for $x > 0$ is

Letting $X_{t} = \sigma Y_{t}$ with $\mu = \sigma \nu$, we obtain (using the transformation $b = \sigma x$)

Putting together, and taking the limit as $r \rightarrow 0^-$ then yields

precisely as Gappy described. What this tells you is that if your drift is negative, you will almost surely hit the negative boundary, and with exponentially decreasing probability hit the negative boundary if your drift is positive. For $\mu > 0$, rewriting the above equation in terms of Sharpe, we see that

i.e. the chance of hitting a lower boundary from a given starting point of zero decreases exponentially with Sharpe.

Question 2: Hitting Probability Within a Given Time #

What is the hitting probability of a given drawdown limit within a given interval – i.e. for $b, T > 0$, what is $\mathbb{P} \left( X_{t} \leqslant - b, \text{ for some } t \in [0, T) \right)$?

In Question 1, we determined (or referred to the fact) that the Laplace transform of the stopping time $\tau _{b}$ can be described as

By uniqueness of the Laplace transform and distribution (also by uniqueness of the Moment Generating Functions, which is equivalent to the Laplace transform), $\tau _{b}$ is a random variable described by the Inverse Gaussian distribution $\text{IG} \left( \frac{b}{\mu}, \left( \frac{b}{\sigma} \right)^2 \right)$, see [7] , which has conditional CDF

with $\Phi$ denoting the standard normal distribution.

Knowing this distribution, we can also determine a number of other facts relating to this hitting time. Restating the original probability in terms of the hitting time $\tau _{b}$, we have

where we have restated the last equation in volatility units. One can check that as $T \rightarrow \infty$, the right hand side of the equation approaches $e^{- \frac{2 \mu b}{\sigma^2}}$, the eventual hitting probability.

Remark. The probability of hitting the barrier $- b$ within a finite $T$ is given in two terms, where

Question 3: Distribution of Drawdown Values #

We now move on to discuss drawdown probabilities, which are slightly more complicated, as they are path dependent, as the drawdown process $(D_{t})_{t}$ is derived from the joint distribution $(M_{t}, X_{t})_{t}$.

What is the distribution of drawdown values?

Recall that the drawdown process $(D_{t})_{t}$ measures the current drop from the historical maximum, which is tracked by the process $(M_{t})_{t}$. By manipulating, and integrating out the joint density, it can be determined that the probability that the drawdown at time $t < \infty$ is greater than $b > 0$ is

and in particular, as time $t \rightarrow \infty$

i.e. $\mathbb{P} (D_{\infty} \leqslant b) = 1 - e^{- \frac{2 \mu b}{\sigma^2}}$, and hence $D_{\infty} \sim \operatorname{Exp} \left( \frac{2 \mu}{\sigma^2} \right)$. We have two remarkable things here, one is that the probability of a drawdown exceeding a certain threshold at a given time $t$ is precisely the hitting time of the process $X_{t}$ of the barrier $- b$, and secondly that the complicated path dependent drawdown process, at large times, converges to the stationary Exponential distribution (even though $X_{t}$ itself is not stationary), see [8] . The intuition here is that under the positive drift $\mu > 0$, $X_{t}$ drifts upwards, and at every time, it continues to hit new heights, which resets $D_{t}$ to zero, making it memoryless. Furthermore, in the sense of a capital process, a larger $\operatorname{SR}$ makes drawdowns exponentially less likely.

Since the infinite time drawdown process follows an Exponential distribution with parameter $\lambda = \frac{2 \mu}{\sigma^2}$, it directly follows (from the results regarding the distribution) that in the infinite time limit, the average drawdown is

where the last equation represents the value in volatility units.

Question 4: Expected Recovery Time #

What is the expected recovery time from a given drawdown?

Assume that at time $t^{\ast}$, we are currently at a drawdown $- b$ for some $b > 0$, i.e. $D_{t^{\ast}} = b$, and $M_{t^{\ast}} = \sup _{0 \leqslant s \leqslant t_{0}} X_{s}$. We want to determine the recovery time from this drawdown, i.e.

Rewriting in terms of $X_{t}$, we have

where the final line shows that $\rho (b)$ is simply the first passage hitting time of a fresh Brownian motion $\tilde{X}_{t}$ with $\mathrm{d} \tilde{X}_{t} = \mathrm{d} \mu + \sigma \mathrm{d} B_{t}$ with $\tilde{X}_{0} = 0$. As we did for Questions 1 and 2, we can determine that as $\mu$ is positive, the probability of recovery is one in infinite time (as now the hitting boundary is above the starting point), and that $\rho _{b} \sim \operatorname{IG} \left( \frac{b}{\mu}, \left( \frac{b}{\sigma} \right)^2 \right)$. With this, the expected time for recovery can be computed as

What is interesting is that the expected recovery time is independent of the volatility of the process, even though one would intuitively think that the volatility has some action here. Instead, what you find is that the volatility instead dictates the variance of the recovery time

Something else one can notice here is the relation of the above identity to physics, i.e. if $\mu$ denotes the general speed of travel, and $b$ the distance one needs to traverse, then the time taken can be calculated as $\frac{b}{\mu}$, which is precisely $\mathbb{E} (\rho _{b})$.

Question 5: Expected Time in Drawdown #

What is the expected time spent in a drawdown?

This question is a little ambiguous. In explicit terms, the process $X_{t}$ is almost surely always in a drawdown, and therefore across a time $[0, T]$, the expected time spent in a drawdown is $T$.

However, what Gappy discusses here is not the above - instead, it is, given a random point in time, what is the expected time I should wait to recover from my current drawdown, i.e. the expected remaining recovery time.

From the above, we determined that the expected time required to recover to the previous high, given that the current drawdown depth is $b$ is $\mathbb{E} (\rho _{b})$. Let $T_{\text{obs}}$ denote a random variable chosen uniformly across $(0, T)$ as $T \rightarrow \infty$ (with $T$ large enough such that $(D_{t})_{t}$ resembles the stationary distribution). At this moment, the current drawdown depth can be described as

For the sake of simplicity, assume that $D_{\text{obs}} \sim D_{\infty}$, i.e. that the drawdown is distributed according to a stationary distribution. Define now $\tau _{\text{remain}}$, the duration required for the process $X_{t}$ to return to the level of the maximum observed at time $T_{\text{obs}}$, i.e.

Then, we wish to determine $\mathbb{E} \left( \tau _{\text{remain}} \right)$, which using the law of total expectation, we can determine to be

More specifically, the expected time to wait to exit a given drawdown that you may find yourself in is inversely proportional to the square of the Sharpe ratio. For example, assume you have a strategy that has an annualised Sharpe of $\xi$, i.e. $\operatorname{SR}_{260} = \xi$, then, at any given drawdown, your expected recovery time (in business days) is

which is 520bd (2y) for a $\xi = 0.5$ strategy (quite a considerable time).