5
$\begingroup$

Just learning some stats, so please forgive if this is simple but I couldn't find a good explanation.

Let $X \sim \mathcal{N}(\mu,\sigma^2)$ and $Y = e^X$. To find an approximately 95% confidence interval, note \begin{align*} P(a \leq Y \leq b) & = P(a \leq e^X \leq b) \\ & = P(\log a \leq X \leq \log b) \\ & = P\left(\frac{\log a - \mu}{\sigma} \leq Z \leq \frac{\log b - \mu}{\sigma}\right) \\ & = \frac{1}{\sqrt{2\pi}} \int_{\frac{\log a - \mu}{\sigma}}^\frac{\log b - \mu}{\sigma} e^{-z^2/2} dz \\ & \triangleq 0.95, \end{align*} for which we know \begin{align*} \frac{\log b - \mu}{\sigma} & \approx 2 \iff b = e^{\mu + 2\sigma}, \\ \frac{\log a - \mu}{\sigma} & \approx 2 \iff a = e^{\mu - 2\sigma}. \end{align*} Then, my understanding of a confidence interval (CI) would lead me to believe 95% of the values of $Y$ should lie within the interval $$ [e^{\mu + \sigma^2/2} - e^{\mu - 2\sigma},e^{\mu + \sigma^2/2} + e^{\mu + 2\sigma}], $$ where $e^{\mu + \sigma^2/2}$ is the mean of $Y$. Is this correct? Specifically, when we speak of a " 95% confidence interval," do we mean that 95% of the values lie within the mean of the random variable, or another average like median or mode?

Finally, to clear up a source of confusion on notation. For a normally-distributed random variable $X \sim \mathcal{N}(\mu,\sigma^2)$, the variance $\sigma^2$ is also the square of the standard deviation (SD) $\sigma$, for which an approximate 95% confidence interval is $[\mu - 2\sigma, \mu + 2\sigma]$. Similarly for a lognormally-distributed random variable $Y = e^X$, its variance is given by $(e^{\sigma^2} - 1) e^{2\mu + \sigma^2}$, and I believe its standard deviation would again just be the square root of this (by definition), namely $\left(\sqrt{e^{\sigma^2} - 1}\right) e^{\mu + \sigma^2/2}$. But now we don't have that an approximate 95% confidence interval is $[mean - 2*SD, mean + 2*SD]$ since the pdf of $Y$ is not symmetric.

So, is the $mean \pm SD$ property for a confidence interval only valid for normal random variables?

$\endgroup$
6
$\begingroup$

Is this correct?

No.

i) This isn't a confidence interval you're calculating (since those are for parameters or functions of them), nor is it really a prediction interval, a tolerance interval, or any of the more common statistical intervals ... since for starters it's based on known population values, not on a sample.

ii) You already calculated the limits of an interval that includes 95% of the probability; it's $(a,b)$, not $(\mu-a,\mu+b)$.

do we mean that 95% of the values lie within the mean of the random variable

No. The mean is a single value. How can 95% of a continuous distribution lie "within" a single value?

But now we don't have that an approximate 95% confidence interval is [mean−2∗SD,mean+2∗SD] since the pdf of Y is not symmetric.

Just because the density isn't symmetric doesn't of itself mean that a symmetric interval can't include 95% of the probability.

It doesn't include 95%, as it happens, though it's often fairly close to 95% for unimodal distributions. However, while it works pretty well for $\pm 2\sigma$, that doesn't always carry over nearly as well to other numbers of sds not close to 2.

So, is the mean±SD property for a confidence interval only valid for normal random variables?

(Again, keeping in mind that it's not a confidence interval)

Well, actually, for normal random variables, 95% of the distribution is within 1.96 sd's of the mean and 95.4% is within 2 sd's of the mean.

Those numbers are calculated from the normal distribution function; $\Phi(1.96)-\Phi(-1.96)=0.9500$ and $\Phi(2)-\Phi(-2)=0.9545$.

$\endgroup$

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.