P&L, Sharpe Ratio & Risk

Medium3–4 hrsP&LSharpe RatioDrawdownRisk Management

Overview

P&L analysis and risk metrics are how you evaluate whether a strategy actually works. The Sharpe ratio is the most important single number in systematic trading — it tells you how much return you get per unit of risk. The t-statistic tests whether the Sharpe is statistically significantly different from zero. In practice, quant firms require t-stat > 3 (much stricter than the academic 1.96 threshold) because they test many strategies.

How to Recognize

→'Is this strategy statistically significant?'
→Given returns, compute performance metrics
→Sizing a position based on a signal's strength
→Questions about when to stop trading a losing strategy

Step-by-Step Approach

1.Sharpe = (mean return − risk-free rate) / std of returns × √252 (annualized)
2.t-stat of Sharpe ≈ Sharpe × √N (N = number of observations)
3.Position size ∝ signal strength / volatility (Kelly-type sizing)
4.Maximum drawdown = largest peak-to-trough decline in equity curve
5.Strategy is 'significant' when t-stat > 3.0 (quant standard)

Key Formulas

SR = \frac{E[R] - r_f}{\sigma_R} \times \sqrt{252}

(annualized daily Sharpe)

t_{SR} \approx SR \times \sqrt{N}

(statistical significance)

\text{Kelly fraction} = \frac{\mu}{\sigma^2}

(continuous version)

\text{Max Drawdown} = \max_{t} \left(\max_{s \leq t} V_s - V_t\right) / V_{peak}

Worked Examples

0Quick Example

Strategy: 100 daily returns, mean=0.05%, std=1%. Sharpe = 0.05%/(1%/√252) = 0.05/(0.063%) × √252... daily SR = 0.05/1 = 0.05, annualized = 0.05 × √252 = 0.79.

Problem

A strategy produces daily returns over 252 trading days with mean 0.08% and standard deviation 1.2%. What is the annualized Sharpe ratio? Is it statistically significant?

Solution

Daily mean = 0.08% = 0.0008. Daily std = 1.2% = 0.012.

Assume risk-free rate ≈ 0 for simplicity.

Daily Sharpe = 0.0008/0.012 = 0.0667.

Annualized Sharpe = daily Sharpe × √252 = 0.0667 × 15.87 ≈ 1.06.

Is it significant? t-stat = SR × √N = 1.06 × √252 ≈ 1.06 × 15.87 ≈ 16.8.

t-stat >> 3: very significant! But note this is for 1 year. If we only had 1 month (21 days): SR_annual ≈ 1.06, t = 1.06 × √21 ≈ 4.85. Still significant.

Answer

SR ≈ 1.06, t-stat ≈ 16.8 over 252 days: highly significant. Rule of thumb: a Sharpe of 1.0 over 1 year of daily data gives t ≈ SR·√T ≈ 1.0·15.87 ≈ 16.

Common Mistakes

!
Not annualizing: daily Sharpe is 16× smaller than annual Sharpe (divide by √252). Always specify the period.
!
Thinking high Sharpe in-sample guarantees future performance. Overfitting can make any strategy look great historically.
!
Using the wrong number of observations for the t-stat. If you have monthly data (not daily), N = 12/yr, not 252.
!
Forgetting that Sharpe doesn't capture tail risk. A strategy with perfect Sharpe can have huge drawdowns if returns aren't normal.

Practice Problems

Click "Show Answer" to reveal

A fund has annualized return of 15% and annualized volatility of 10%. Risk-free rate = 3%. What is the Sharpe ratio?

Hint: Sharpe = (excess return) / volatility = (15% − 3%) / 10%.

How many daily observations do you need to achieve t-stat = 3 with an annualized Sharpe of 0.8?

Hint: t = SR × √N. SR daily = 0.8/√252.

Back to Study Plan

Adverse Selection Put-Call Parity