Question
A research firm launches many independent trading strategies, all built under the same simple return model. Each new strategy is high quality with probability 1/2 and low quality otherwise. Its persistent expected monthly return, in percentage points, is \(Q=2\) if high and \(Q=0\) if low. In each month its actual return is
\[
R_t=Q+\varepsilon_t,
\]
where each \(\varepsilon_t\) is independently +1 or −1 percentage point, equally likely. Monthly noises are independent of quality and of each other.
After month 1 the firm closes every strategy with \(R_1<1\%\). Only survivors are mentioned in its report. Calculate: (i) the probability a launch survives, and the high-quality fraction among survivors; (ii) the reported average month-1 return of survivors; and (iii) the expected month-2 return of survivors versus that of a completely new launch. Finally split the excess of the survivors' reported month-1 mean over a new launch's expected month-2 return into a persistent-quality selection effect and a temporary-noise effect.
Get hint
Hint 1
List the four equally likely (quality, first-month noise) combinations. Mark which pass the 1% survival threshold.
Check your answer
Solution
The four equally likely combinations of latent quality and first-month noise are:
\[
\begin{array}{c|c|c|c}
Q&\varepsilon_1&R_1\ (\%)&\text{survives }R_1\ge1\%?\\\hline
2&+1&3&\text{yes}\\
2&-1&1&\text{yes}\\
0&+1&1&\text{yes}\\
0&-1&-1&\text{no}
\end{array}
\]
There are three equally likely surviving cases, so \(P(S)=3/4\). Two of those cases have high quality, giving \(P(Q=2\mid S)=2/3\), while \(P(Q=0\mid S)=1/3\).
The reported month-1 mean among survivors is
\[
E[R_1\mid S]=(3+1+1)/3=\boxed{5/3\%}.
\]
The future noise \(\varepsilon_2\) is independent of the survival selection and has conditional mean zero, so
\[
E[R_2\mid S]=E[Q\mid S]+E[\varepsilon_2\mid S]
=2\cdot\tfrac23+0\cdot\tfrac13+0
=\boxed{4/3\%}.
\]
For a new, unselected strategy, the high- and low-quality types are equally likely, so \(E[R_2]=E[Q]=2(1/2)=\boxed{1\%}\).
Decompose the difference between reported survivor returns and a new strategy's future expectation:
\[
\underbrace{E[R_1\mid S]-E[R_2]}_{5/3-1=2/3\ \text{pp}}
=\underbrace{E[Q\mid S]-E[Q]}_{4/3-1=1/3\ \text{pp}}
+\underbrace{E[\varepsilon_1\mid S]}_{5/3-4/3=1/3\ \text{pp}}.
\]
The first third of a percentage point is persistent selection bias: survivors are more likely to be genuinely high quality and thus also do better on average next month. The second third is temporary noise selection: a low-quality strategy survives only when it has a lucky first month, and that luck does not repeat in expectation. Even observing the future survivor mean \(4/3\%\) would overstate a new launch's mean by 1/3 pp, because the survivors are not a representative launch cohort.
Assumptions
Each new strategy independently has \(Q=2\%\) or \(Q=0\%\) persistent expected monthly return with probability 1/2 each.
The monthly noise is independent and symmetric ±1 percentage point; all noises are independent across periods and strategies.
The rule keeps strategies with month-1 realised return at least 1% and closes those below; survivors alone are reported.
All averages are population expectations in percentage-return units, not a finite-sample estimate with sampling error.
Final Answer
Survival probability 3/4; high quality among survivors 2/3. Survivors' month-1 mean 5/3%; their expected month-2 return 4/3%, versus 1% for a new launch. The 2/3-percentage-point reported-to-new gap splits into 1/3 pp persistent quality and 1/3 pp lucky month-1 noise.