1
$\begingroup$

Is there any opportunity to create such interval where a variable ($\{\ln(X_i)\}^n_{i=1}$) is the fraction of prices for two periods?

$$ X_i = \frac{price.new_i}{price.old_i} $$

Please, look at my attempt below. Is everything correct?

$\endgroup$
2
$\begingroup$

Here is my attempt

Geometric mean: $$ ( \prod_{i=1}^{n} X_i ) ^{1 / n} = Const. $$

I use the excellent tip of Dmitrij Celov at Confidence interval for geometric mean, who stated that the geometric mean can be transformed to an arithmetic mean of logarithms (if strictly positive), we have

$$ \ln{(\prod_{i=1}^{n}X_i)^{1 / n}} = \ln{(Const.)} \\ \overline{\ln(X)} =\frac{ \sum_{i=1}^{n}\ln{X_i} }{n} \\ $$

where $\overline{\ln(X)}$ arithmetic mean of $\{\ln(X_i)\}^n_{i=1}$.

Let's construct confidence interval $$ P [\ln(X_i) \in (\overline{\ln(X)} \pm Z_{\alpha/2} \cdot \frac{sd}{ sqrt (n) })] = 95 \% $$ where

$$ sd = sqrt{\frac{\sum_{i=1}^{n}(\ln{X_i} - \overline{\ln(X)})^2}{n-1}} $$

Then for the "real" values we exponent it

$$ P [\exp(\ln(X_i) \in \exp(\overline{\ln(X)} \pm Z_{\alpha/2} \cdot \frac{sd}{sqrt (n)})] = 95 \% \Leftrightarrow \\ P [X_i \in \exp({\overline{\ln(X)}} \pm {Z_{\alpha/2} \cdot \frac{sd}{sqrt (n)}})] = 95 \% $$

Or, without mess, $$ P (\ln{X_i} \in [L; U]) = 95 \% \Leftrightarrow \\ P (X_i \in [\exp(L); \exp(U)]) = 95 \%. $$

$\endgroup$
  • 1
    $\begingroup$ If I understand what you're saying, you assume that $ln(X_i) \sim N$ and build a confidence interval for it the ususal way one would do with a Normal r.v. Your finishing expression is right, but when you say that you "exponent something" you are not being very clear on what you do and, in fact, you are stating something that is wrong. Again your last statement is right, but it seems that intuition got you there, you are not justifying it. $\endgroup$ – Cristián Antuña Jan 28 '15 at 19:21
  • $\begingroup$ @CristiánAntuña okay, thanks, can you help to make it true? $\endgroup$ – Vladimir Iashin Jan 28 '15 at 19:28
  • $\begingroup$ I'll use the answer space, since it is bigger than the comments one. $\endgroup$ – Cristián Antuña Jan 28 '15 at 19:35
  • $\begingroup$ @CristiánAntuña What, precisely, is the "wrong" statement in this answer? $\endgroup$ – whuber Jan 28 '15 at 20:54
  • $\begingroup$ These events are not equivalent: $\{ ln(X_i) \in (\overline{ln(X)} + b)\}$ and $\{ e^{ln(X_i)} \in (e^{\overline{ln(X)}} + b)\}$ and, then, in general won't have the same probability. $\endgroup$ – Cristián Antuña Jan 28 '15 at 21:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.