Say, I have hourly returns $r_1,r_2,...,r_T$, where $r_t$ = $ln(p_t)$ - $ln(p_0)$ for $t = 1...T$. So what is the value of $E[r_t]$? Would $r_T$ be the $\prod{(r_t)}$?

Basically $r_t$s are the returns w.r.t to a fixed point $t_0$. My question then is how can I mathematically prove that the $E[r_t]$ is $r_T$ or it isn't?

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    $\begingroup$ What is your model? You cannot prove anything without any additional assumptions. $\endgroup$ Jan 29, 2012 at 22:40
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    $\begingroup$ Could you show an example (even just one example) of where you tested this beforehand? $\Pi{(r_t)}$ would be "double counting" partial returns since you define all returns each from the same fixed point. $\endgroup$ Jan 29, 2012 at 23:13
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    $\begingroup$ Yes, @chrisaycock, I wouldn't mind double counting. If the definition of $r_t$ was $ln(p_t)$ - $ln(p_{t-1})$, the product would be obvious but if I define it from a fixed point, what would that mean? $\endgroup$
    – Antelope
    Jan 30, 2012 at 9:39
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    $\begingroup$ @moderators, Please cancel my edit. I was not able to see to redo my edit. $\endgroup$
    – vinux
    Feb 1, 2012 at 9:22
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    $\begingroup$ Are you looking for an unconditional expectation or an expectation as of some specific time? Why would $E[r_t]=r_T$? This question sounds rather confused, and as currently stated appears to be "not a real question." $\endgroup$ Feb 6, 2012 at 21:26

3 Answers 3


It seems that your real question is: is the PFP (Price Formation Process) diffusive from intraday to weekly sampling rate?

It is a very good question since on intraday, some academics found some multifractal features into intraday returns, meaning that the PFP is not a Geometric Brownian Motion at small scales (even considering stochastic volatility).

You have for instance successful modellings of the PFP using pint processes, and especially Hawkes ones (that are not diffusive and even not Markovian): Modeling microstructure noise with mutually exciting point processes by: E. Bacry, S. Delattre, M. Hoffmann, J. F. Muzy (forthcoming in Quant. Finance). They obtained some formula to express characteristics of the diffusive limit of such processes with respect to ones of the underlying Hawkes process, like the large scale diffusive volatility: $$\sigma=\frac{2\mu}{1-||\phi||_1}\,\frac{1}{(1+||\phi||_1)^2}$$ (with $\phi$ the kernel of the Hawks process linking its stochastic intensity with its realizations and $\mu$ is the deterministic part of its intensity).

But also more "classical" multifractal approaches: Modelling fluctuations of financial time series: from cascade process to stochastic volatility model by: J. F. Muzy, J. Delour, E. Bacry in Euro. Phys. Journal B, Vol. 17 (2000), pp. 537-548. In such cases, a classical "Hurst exponent" allows zooming in or out.


Isn't this a simple mathematical rule?

$$\Delta r_{t}=r_{t} - r_{t-1} = ln(p_{t}) - ln(p_{0}) - ln(p_{t-1}) + ln(p_{0})=ln(\frac{p_{t}}{p_{t-1}})$$ i.e. logarithmic or continuously compounded return. As a result: $$E(\Delta r_{t})=\frac{1}{T}\sum_{t=1}^{T}\Delta r_{t} = \frac{1}{T}\sum_{t=1}^{T}ln(\frac{p_{t}}{p_{t-1}})=\frac{1}{T}ln(\prod_{t=1}^{T}\frac{p_{t}}{p_{t-1}})=\frac{1}{T}ln(\frac{p_{T}}{p_{0}})=\frac{1}{T}r_{T}$$

is hourly expectation.

Or $$\frac{p_{T}}{p_{0}}\frac{p_{T-1}}{p_{0}}...\frac{p_{2}}{p_{0}}\frac{p_{1}}{p_{0}}=(\frac{p_{T}}{p_{0}})(\frac{p_{T-1}}{p_{0}}...\frac{p_{2}}{p_{0}}\frac{p_{1}}{p_{0}})$$ and $$ln[\frac{p_{T}}{p_{0}}\frac{p_{T-1}}{p_{0}}...\frac{p_{2}}{p_{0}}\frac{p_{1}}{p_{0}}]= ln(\frac{p_{T}}{p_{0}}) + ln[\frac{p_{T-1}}{p_{0}}...\frac{p_{2}}{p_{0}}\frac{p_{1}}{p_{0}}]$$ or $$ln[\frac{p_{T}}{p_{0}}\frac{p_{T-1}}{p_{0}}...\frac{p_{2}}{p_{0}}\frac{p_{1}}{p_{0}}] - ln[\frac{p_{T-1}}{p_{0}}...\frac{p_{2}}{p_{0}}\frac{p_{1}}{p_{0}}]= ln(\frac{p_{T}}{p_{0}})$$ and if we assume convergence $$T\cdot E(r_{t}) - (T-1)\cdot E(r_{t})\approx r_{T}$$

  • $\begingroup$ The last line of your post concerns me. What if my returns are $X$ bps for each of five hours, and then 0 bps for the sixth hour? Expectation is slightly less than $X$, but your claim is that $r_T$ should be precisely this number, even though we know it is 0 bps! $\endgroup$ May 4, 2012 at 19:59
  • $\begingroup$ I mentioned the following condition "if we assume convergence", otherwise I certainly may be wrong. $\endgroup$
    – rtybase
    May 8, 2012 at 15:23

If you assume GBM, then

Expected Change in price = Stock * Drift * change in time

The Weiner term disappears.

This implies, E[r(T)] = drift * T

Use hourly return to estimate drift and plug it in the formula.

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    $\begingroup$ What does GBM have to do with anything in this question?! The OP is asking whether accumulating intraday returns defined from a fixed point would lead to the end-of-day's return. Your answer doesn't make the slightest bit of sense. $\endgroup$ Feb 12, 2012 at 18:59
  • $\begingroup$ I do not like to run into argument with any body but I do not see how this does not bake sense if you think through the question. E[r(T)] = S * meu * T, if this is calculated ex ante and if the hourly return are know, ex post, r(T) = exp(r(0) + ... + r(T)) = exp(r(0) * ... * exp(r(T)). This is not an expected value. $\endgroup$ Feb 12, 2012 at 21:33
  • $\begingroup$ Since expected values are mentioned I safely assumed it is the former form. E[r(t)] = S * meu * t. If you down voted me please undo this as I do not see this as wrong under certain assumptions. $\endgroup$ Feb 12, 2012 at 21:38
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    $\begingroup$ Why don't you put your comments in your answer??? $\endgroup$
    – SRKX
    Feb 12, 2012 at 22:07
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    $\begingroup$ Why are you commenting this on your own answer? These look like comments that should be on the question. $\endgroup$ Feb 12, 2012 at 22:08

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