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Given: the standard deviation is 15.5 cents per gallon in one set of quoted prices and a standard deviation of $5.5 per barrel in another set of quoted prices.

There are 33.5 gallons in 1 barrel.

How do I get these two standard deviations on an equivalent level so I can compare them and do an accurate analysis? Thanks!

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    $\begingroup$ What does it mean to "get these two standard deviations on an equivalent level"? Why do you need to compare them? What do you want to do with your data in the end? $\endgroup$ – gung - Reinstate Monica Oct 8 '14 at 2:05
  • $\begingroup$ ultimately, I am trying to obtain an optimal hedge ratio based off of the two standard deviations: correlation*(SD1/SD2). I can't do this when the units don't line up. $\endgroup$ – HyungPark Oct 8 '14 at 2:20
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    $\begingroup$ You aren't allowed to put both datasets in the same units? Prior to analysis?! That is, can't you multiply each case in the "cents per gallon" by 33.5 to normalize the "cents per gallon" dataset into "dollars per barrel" (same unit of dataset 2), and conduct the analysis from that? $\endgroup$ – Ramalho Oct 8 '14 at 2:24
  • $\begingroup$ Do you have the original sets of quoted prices? $\endgroup$ – gung - Reinstate Monica Oct 8 '14 at 2:33
  • $\begingroup$ I do not have the original sets. $\endgroup$ – HyungPark Oct 8 '14 at 2:56
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In general, if $X$ and $Y$ are random variables and $X=a\cdot Y$, then $\text{sd}(X)=|a|\cdot\text{sd}(Y)$. Hence if your new variable should be the price of 33.5 gallons of the commodity, the standard deviation of such variable should be 33.5 times 15.5 cents, i.e. roughly 5.2 USD.

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    $\begingroup$ "and X=a⋅Y, then sd(X)=a⋅Y"... there's a missing "sd" in the second equation there. I'll edit. $\endgroup$ – Glen_b Oct 8 '14 at 6:20
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    $\begingroup$ Technically that should be $|a|\cdot\text{sd}(Y)$ $\endgroup$ – James Oct 8 '14 at 10:41

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