Relating Discrete Return Volatility to Continuous-Time Volatility

Summary

The document asks whether volatility computed from discrete returns can be converted to a continuous-time volatility estimate without access to the underlying return data. It compares two calculations on a series of monthly returns: an estimate motivated by Itô’s lemma and a sample standard deviation. The reported values are close, but the question does not establish a general conversion rule or explain all assumptions behind the calculations.

The motivation is to use an annualized standard deviation to construct confidence bounds for a geometric Brownian motion-style expression. That expression uses continuous log volatility, so distinguishing it from a discrete-return standard deviation matters when choosing the volatility input. The document provides no answer, derivation, or evidence about the accuracy of a conversion across sampling intervals or return definitions. It therefore raises a useful measurement issue but leaves the method unresolved; a reliable conversion would require specifying how returns and volatility are defined and how the data are sampled.

Key ideas

• The document contrasts volatility estimates from discrete returns with a continuous-time estimate motivated by Itô’s lemma.
• The two reported estimates are close for the presented monthly-return sample.
• Continuous log volatility is used in the stated geometric Brownian motion confidence-bound expression.
• The document does not derive a general conversion rule or specify the assumptions needed to apply one.

Tags

# Possible to convert between continuous & discrete volatility without underlying data?


# Possible to convert between continuous & discrete volatility without underlying data?












Is there a way to convert between continuous and discrete volatility without the actual return data?

For example, calculating on 12 monthly returns:

```
{0.03, 0.91, 0.05, -0.17, -1.64, 0.79, -1.41, 0.69, 1.08, 0.42, 1.29, 1.66}
```

from Ito's Lemma, volatility is

$\sigma =\sqrt{\frac{(dS)^2}{S^2 dt}}$

```
σ = 0.0352148
```

whereas using standard deviation

$\sigma =\sqrt{\frac{m}{T-1}\sum _{t=1}^T (r[t]-R)^2}$

```
σ = 0.035048
```

The reason for asking is I have annualised standard deviation and wish to compute confidence bounds for `b = -2, -1, 1, 2`. However $\sigma$ here is the continuous log version:

$y_0 e^{\left(a-\frac{\sigma ^2}{2}\right) t+b \sigma \sqrt{t}}$

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