I am very confused on how to derive the attached equation (15).
Would someone be kind enough to walk me through the proof?
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$\begingroup$ Use the fact that $\mathbb{V}[r_{t,t+k}] = \mathbb{E}[r_{t,t+k}^2] - \mathbb{E}[r_{t,t+k}]^2$ and use equations (6), (12) and (13) to compute these expectations. $\endgroup$– JejeBelfortOct 6, 2017 at 7:16
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$\begingroup$ Hello, Thank you so much for your response @JejeBelfort!! If you expand the PI term from n=1 to k, and subtract 1. You are left with a long equation of r terms multiplied with each other which you then square. how do you simplify this proof and see the pattern? also where does the V(of R_t) come from in the answer? E(r^2)-E(r)^2 from the expansion I mentioned? $\endgroup$– jungsun65813215Oct 6, 2017 at 7:50
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$\begingroup$ Since the $r_{t+n}$ are uncorrelated, you can write $\mathbb{E}\left[\prod_{n=1}^k r_{t+n}\right] = \prod_{n=1}^k \mathbb{E}\left[ r_{t+n}\right]$ $\endgroup$– JejeBelfortOct 6, 2017 at 11:31
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$\begingroup$ Why delete the attachment? The problem was: if the one day returns are iid $\mu,\sigma^2$, what is the variance of k-day returns? The original attachment is here imgur.com/a/t5OqN $\endgroup$– nbbo2Oct 6, 2017 at 21:53
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$\begingroup$ @molly Don't vandalize your own question. $\endgroup$– Bob Jansen ♦Oct 7, 2017 at 13:59
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1 Answer
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$\newcommand{\E}{\mathbb{E}}$ $\newcommand{\V}{\mathbb{V}}$
First note that
\begin{eqnarray} \E[(r_{t + n} + 1)^2] &=& \E[r_{t+n}^2 + 2r_{t + n} + 1] = \E[r_{t + n}^2] + 2\mu + 1 \\&=& (\V[r_{t+n}] + \E^2[r_{t+n}]) + 2 \mu + 1 \\ &=& \sigma^2 + (\mu^2 + 2\mu + 1) = \sigma^2 + (\mu + 1)^2 \tag{1} \end{eqnarray}
Now, since $r_{t+n}$ are independent random variables we have
\begin{eqnarray} \V[r_{t,t+k}] &=& \V\left[\prod_{n=1}^k (r_{t+n} +1) - 1\right] = \V\left[\prod_{n=1}^k (r_{t+n} +1)\right]\\ &=& \E\left[\left(\prod_{n=1}^k (r_{t+n} +1)\right)^2\right] - \E^2\left[\prod_{n=1}^k (r_{t+n} +1)\right] \\ &=& \E\left[\prod_{n=1}^k (r_{t+n} +1)^2\right] - \left(\E\left[\prod_{n=1}^k (r_{t+n} +1)\right]\right)^2 \\ &=& \prod_{n=1}^k \E[(r_{t+n} +1)^2] - \left(\prod_{n=1}^k \E[r_{t+n} +1]\right)^2 \\ &\stackrel{(1)}{=}& \prod_{n=1}^k [\sigma^2 + (\mu + 1)^2] - \prod_{n=1}^k(\mu + 1)^2 \\ &=&[\sigma^2 + (\mu + 1)^2]^k - (\mu + 1)^{2k} \end{eqnarray}
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$\begingroup$ Thank you so much for working through that in such basic terms!! I really appreciate your time and help. $\endgroup$ Oct 6, 2017 at 16:42