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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.?

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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 $$\mathbb{E}_t[\frac{S_{t+1}}{S_{t}}]S_{t}$$

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  • $\begingroup$ Thank you for the concise and informative answer! For clarification on my end: 1. What is meant by the $t$ in the subscript of $\mathbb{E}_t$? I've usually seen this to mean expectation taken with respect to a distribution, but I don't think I have seen this usage. 2. What definition for simple returns are you using in the final alternative you propose? $\frac{S_{t+1}}{S_t}$, $\frac{S_{t+1} - S_t}{S_t}$, or a different one? $\endgroup$
    – QMath
    Jul 15 at 21:35
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    $\begingroup$ @QMath $\mathbb{E}_t$ means you are forming the expectation/forecast at time $t$. Formally, it is shorthand for $\mathbb{E}[ \cdot | \mathcal{F}_t]$, where $ \mathcal{F}_t$ is the time $t$ information set. $\endgroup$
    – fes
    Jul 16 at 5:18
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    $\begingroup$ I would define simple returns as $\frac{S_{t+1}-S_t}{S_t}=\frac{S_{t+1}}{S_t}-1$. But from this it is easy to solve for $\mathbb{E}_t[\frac{S_{t+1}}{S_t}]$. $\endgroup$
    – fes
    Jul 16 at 5:20
  • $\begingroup$ Great, those points both make sense, thanks again! :) $\endgroup$
    – QMath
    Jul 16 at 9:21

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