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Forecasting Mean Price from Log Returns Requires Distributional Information

Article Quant Q&A · Author: QMath

Summary

Exponentiating a forecast of cumulative log returns produces a price forecast associated with the conditional median in general, not the conditional mean. To obtain the expected future index level, the relevant quantity is the conditional expectation of the exponential of the return, multiplied by the current index level. A forecast of the return mean alone does not determine that expectation.

The answer gives alternatives: estimate a full conditional return distribution and calculate the expectation analytically or by simulation; under normal returns, include half the conditional variance in the exponent; or use a second-order Taylor approximation involving both the conditional mean and variance. Another option is to model simple returns, whose conditional mean maps directly to the expected price ratio. The normal formula and Taylor expression rely on their stated assumptions or approximation, while distribution-based methods require a credible model. The document does not compare these approaches empirically or prescribe a forecasting model.

Key ideas

  • Exponentiating expected log returns generally gives a conditional median price forecast rather than a conditional mean.
  • Expected future price depends on the conditional expectation of the exponentiated return.
  • Under normally distributed log returns, the mean price calculation uses both the return mean and variance.
  • A second-order Taylor approximation also adds a variance adjustment, with approximation error possible.
  • Modeling simple returns provides a direct route from their conditional mean to expected price.

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Full text
# Effect of back-transforming forecasted mean of log returns to get forecasted mean of price


# Effect of back-transforming forecasted mean of log returns to get forecasted mean of price












When trying to forecast time series, say forecasting the level of a stock index so we can forecast the future values of an option, it tends to be helpful to analyze the log returns versus the original levels as these are additive across time, can be more numerically stable, etc.

Assuming that we try to predict the conditional mean of the log returns, when we back-transform this to the index level by exponentiating the multi-period log return (sum of log returns over a time interval), as documented widely, we do not generally get the conditional mean, but instead get the conditional median.

Is there a way around this if our goal is to predict the conditional mean of the index at a future time, $T$, without using methods that seem to make strict distributional assumptions on our data/introduce more uncertainty through assumptions such as approximations that try to convert the conditional mean of the log returns back to the conditional mean of index level by applying normality assumptions, etc.?

## Answer by fes (score 3, accepted)

https://quant.stackexchange.com/a/76132

Assume the stock index is given by $S_t$ and you form a forecasting model for the log-returns $r_{t+1}=\log(S_{t+1}/S_{t})$. You are then interested in the expected next period stock index level

$$\mathbb{E}_t[S_{t+1}]=\mathbb{E}_t[\exp(r_{t+1})]S_{t}$$

If your forecasting model gives a distribution for $r_{t+1}$ you can solve the above analytically or using simulation. For example in the special case of a normal distribution for $r_{t+1}$ we have

$$\mathbb{E}_t[\exp(r_{t+1})]S_{t}= \exp(\mathbb{E}_t [r_{t+1}]+0.5\mathbb{V}ar_{t}[r_{t+1}])S_{t}$$

Alternatively, you can apply a second order Taylor approximation to get

$$\mathbb{E}_t[\exp(r_{t+1})]S_{t} \approx (1+\mathbb{E}_t[r_{t+1}]+0.5\mathbb{V}ar_{t}[r_{t+1}])S_{t}$$

so here you need both the conditional mean and variance of the log-returns to solve for the expected next period stock index level.

The final alternative is to instead formulate the forecasting model for simple returns instead of log-returns, in which case you can solve for the expected stock index level simply as $$\mathbb{E}_t[\frac{S_{t+1}}{S_{t}}]S_{t}$$

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

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