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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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What is your model? You cannot prove anything without any additional assumptions. – Alexey Kalmykov Jan 29 at 22:40
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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. – chrisaycock♦ Jan 29 at 23:13
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? – Antelope Jan 30 at 9:39
@moderators, Please cancel my edit. I was not able to see to redo my edit. – vinux Feb 1 at 9:22
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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." – Tal Fishman♦ Feb 6 at 21:26
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2 Answers

up vote 1 down vote

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 $$T\cdot E(r_{t}) - (T-1)\cdot E(r_{t})\approx r_{T}$$

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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! – chrisaycock♦ May 4 at 19:59
I mentioned the following condition "if we assume convergence", otherwise I certainly may be wrong. – rtybase May 8 at 15:23
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up vote -3 down vote

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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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. – chrisaycock♦ Feb 12 at 18:59
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. – Suminda Sirinath Salpitikorala Feb 12 at 21:33
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. – Suminda Sirinath Salpitikorala Feb 12 at 21:38
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Why don't you put your comments in your answer??? – SRKX Feb 12 at 22:07
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Why are you commenting this on your own answer? These look like comments that should be on the question. – chrisaycock♦ Feb 12 at 22:08
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