Stop-Loss Conditioning and the Resulting Return Distribution

Summary

The document considers a drifted Brownian-motion payoff and asks for its terminal distribution conditional on a lower stop-loss barrier never being reached. Its answer treats the resulting payoff as a truncated normal: retain terminal values above the barrier and normalize their normal density by the probability of landing in that retained region. It also gives the general density form for truncation between lower and upper bounds.

This provides a compact way to describe the terminal payoff among paths that finish without triggering the stop, under the stated setup. The example gives a normal distribution with unit drift and volatility over one time unit, and a stop at minus one; the answer expresses the conditional density using the corresponding truncation probability. A key limitation is that this conditions on the terminal value being above the barrier, rather than modeling whether a continuous path touched the barrier earlier. For a continuously monitored stop in a Brownian process, those events differ, so the proposed truncation does not in general capture the true distribution conditional on no stop being hit.

Key ideas

• A terminal normal payoff conditioned to lie within bounds has a normalized truncated-normal density.
• The conditional density is obtained by dividing the original density by the probability mass inside the retained range.
• For a continuously monitored stop, avoiding a barrier throughout the path is not equivalent to finishing above it.
• The answer’s truncation formula does not generally describe the pathwise no-hit condition posed in the question.

Tags

# Return distribution with stop loss


# Return distribution with stop loss












Assume there is a investment with payoff going like a brownian motion, i.e. $dS=\mu dt+\sigma dW$, for simplicity, setting $\mu= \sigma=1$. At $t=1$, the payoff distribution is $P=Normal(1,1)$. If we set a stop loss at $-1$, meaning we will liquidate our investment whenever its payoff hits $-1$. We can calculate the probability of triggering stop loss, which is

$Prob(SL)=2\int_{\infty}^{-1}{Normal(1,1,x)dx}\approx{}0.05$.

The question is, What's the payoff distribution given the stop loss isn't triggered?

I think we need to solve the following differential equation:

$u_t=\frac{1}{2}\sigma^2 u_{xx}+\mu u_x$

$u(-1,t)=0, u(x,0)=\delta(x)$

## Answer by Richard Hardy (score 3)

https://quant.stackexchange.com/a/77095

The payoff distribution will be the truncated normal distribution. In the general case of truncated from both sides, $a\leq X\leq b$, the density is $$ f(x;\mu,\sigma,a,b)=\frac{1}{\sigma}\frac{\phi(\frac{x-\mu}{\sigma})}{\Phi(\frac{b-\mu}{\sigma})-\Phi(\frac{a-\mu}{\sigma})} $$ where $\phi(x)=\frac{1}{\sqrt{2\pi}}e^{-x^2}$. In your case, that translates to \begin{aligned} f(x;1,1,1,\infty) & =\frac{1}{1}\frac{\phi(\frac{x-1}{1})}{\Phi(\frac{\infty-1}{1})-\Phi(\frac{-1-1}{1})} \\ &=\frac{\phi(x)}{1-\Phi(-2)} \\ &\approx 1.02328\cdot \phi(x). \end{aligned} (Writing $\Phi(\infty-1)$ is probably abuse of notation, but you get what I mean.) For more details about the truncated normal distribution, see Wikipedia.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.