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?

$\endgroup$

1 Answer 1

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, $$ P (\ln{X_i} \in [L; U]) = 95 \% \Leftrightarrow \\ P (X_i \in [\exp(L); \exp(U)]) = 95 \%. $$

$\endgroup$
6
  • 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. $\endgroup$ Jan 28, 2015 at 19:21
  • $\begingroup$ @CristiánAntuña okay, thanks, can you help to make it true? $\endgroup$
    – vdi
    Jan 28, 2015 at 19:28
  • $\begingroup$ I'll use the answer space, since it is bigger than the comments one. $\endgroup$ Jan 28, 2015 at 19:35
  • $\begingroup$ @CristiánAntuña What, precisely, is the "wrong" statement in this answer? $\endgroup$
    – whuber
    Jan 28, 2015 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. $\endgroup$ Jan 28, 2015 at 21:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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