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I'm using a molecular biological method which requires the following normalization

$$Q = \frac{A}{\sqrt[k]{\prod_{i=1}^k B_i}}.$$

How can I find the standard deviation of $Q$, given that $A$ and $B_i$ have known standard deviations? Unfortunately my math training is not good enough for me to solve this. Is there some kind of standard approach to solve this?

EDIT

The logarithm of the above expression is given by: $$ \log(Q)=\log(A)-\frac{1}{k}\cdot \sum_{i=1}^{k}\log(B_{i}) $$

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  • $\begingroup$ You probably don't want to do this. Why not analyze $\log(Q)$? Very likely it will have better statistical behavior. Some evidence for this is that its standard deviation is readily expressed in terms of the standard deviations of $\log(A)$ and the $\log(B_i)$. $\endgroup$ – whuber Jul 11 '13 at 15:50
  • $\begingroup$ Thanks for the edit and the reply. It looks easier now, I'll have a look into it. $\endgroup$ – Eekhoorn Jul 11 '13 at 17:31
  • $\begingroup$ No, that is not possible in that way, I think. I don't know if the re-transformation of the SD to the linear scale is possible at all. @whuber what do you think? $\endgroup$ – COOLSerdash Jul 12 '13 at 18:50
  • $\begingroup$ @COOLS Yes, it is possible to convert back (and easy if the logarithms have Normal distributions). But the point of my original comment was that it's probably not even useful to transform back to the linear scale in the first place. $\endgroup$ – whuber Jul 12 '13 at 18:54
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If $(A, B_{1},\ldots, B_{k})$ are uncorrelated (i.e. $\mathrm{Cov}(A,B_{i})=0$ for all $i$ and $\mathrm{Cov}(B_{i},B_{j})=0$ for all $i\neq j$), then the standard deviation of $\log(Q)$ is given by: $$ \mathrm{SD}(\log(Q))=\sqrt{\mathrm{SD}(\log(A))^{2} + \left(\frac{1}{k}\right)^{2}\cdot \sum_{i=1}^{k}\mathrm{SD}(\log(B_{i}))^{2}} $$

Derivation

Let's define $\mathrm{Var}(X)$ as the variance of $X$ and $\mathrm{SD}(X)$ as the standard deviation of $X$, where $\mathrm{SD}(X)=\sqrt{\mathrm{Var}(X)}$. Further, we can make use of the following properties of variances where $(X, Y, X_{1}, \ldots, X_{i})$ are uncorrelated variables and where $a$ denotes a constant: $$ \begin{align} \mathrm{Var}(aX) &= a^{2}\mathrm{Var}(X)\\ \mathrm{Var}(X+Y) &= \mathrm{Var}(X) + \mathrm{Var}(Y)\\ \mathrm{Var}(X-Y) &= \mathrm{Var}(X) + \mathrm{Var}(Y)\\ \mathrm{Var}(X-aY) &= \mathrm{Var}(X) + a^{2}\mathrm{Var}(Y)\\ \mathrm{Var}\left(\sum_{i}X_{i}\right) &= \sum_{i}\mathrm{Var}(X_{i})\\ \end{align} $$ Our equation has the basic form $X-a\sum_{i}Y_{i}$, where $X=\log(A), a=\frac{1}{k}$ and $Y_{i}=\log(B_{i})$. Now we can put all pieces together: $$ \mathrm{Var}(\log(Q))=\mathrm{Var}(\log(A)) + \left(\frac{1}{k}\right)^{2}\cdot \sum_{i=1}^{k}\mathrm{Var}(\log(B_{i})) $$ And because the standard deviation is just the square root of the variance, we can substitute the variance by the squared standard deviation and take the sqare root on both sides and get the formula above.

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  • $\begingroup$ Thanks. Where can I read up on the procedure you used to get this equation? $\endgroup$ – Eekhoorn Jul 11 '13 at 18:47
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    $\begingroup$ That is correct assuming $(A, B_1, \ldots, B_k)$ are uncorrelated. Biologue, see stats.stackexchange.com/questions/31177 for an account of properties of variances and search our site for even more information. $\endgroup$ – whuber Jul 11 '13 at 18:56
  • $\begingroup$ This might be a trivial question, but can I just do $$10^{\mathrm{SD}(\log (Q))}$$ to get the SD of $Q$? If not, how can I convert back to the linear SD? $\endgroup$ – Eekhoorn Jul 12 '13 at 12:44

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