7
$\begingroup$

Suppose $X$ is distributed $N(\mu, \sigma^2)$ where $\mu \neq 0$.
Can I use the Delta Method to say that $log(X)$ ~ $N(log(\mu), \sigma^2/\mu^2)$?

$\endgroup$
  • 1
    $\begingroup$ What if $\mu$ is negative? Is $\log(\mu)$ still defined as a real number? $\endgroup$ – Dilip Sarwate Apr 8 '14 at 21:37
  • 3
    $\begingroup$ Even when $\mu/\sigma$ is huge, so that (arguably) we could neglect the chance of a negative value, the expectations are wrong: $X=\exp(\log(X))$ would have a mean of $\exp(\log(\mu) + \sigma^2/(2\mu^2)),$ which is strictly greater than $\mu$. It is true that lognormal distributions with very small geometric standard deviations are approximately Normal, provided the parameters are appropriately chosen. $\endgroup$ – whuber Apr 8 '14 at 22:03
15
$\begingroup$

It is not the case.

For $\log(X)$ to be normal, $X$ must be lognormal.

(Consider: if $Z=\log(X)$ is normal, then $X=\exp(Z)$ ... and when you exponentiate a normal random variable, what you get is called a lognormal random variable.)

More generally, taking logs "pulls in" more extreme values on the right (high values) relative to the median, while values at the far left (low values) tend to get stretched back. So if it's symmetric before taking logs, it will be relatively left skew after. This is a simple consequence of the shape of the function $\log(x)$:

enter image description here

(the line is tangent to the curve. In general it doesn't necessarily go close to the origin, that's just an artifact of the particular value of $m$ in this case)

Values very close to the median (indicated by an $m$ in the plot) will experience an approximately linear rescaling (the dashed blue line). Values far above $m$ will be pulled back toward $m$ relative to that rescaling experienced by the middle values, while values far below $m$ will be pulled further away from $m$, relative to that linear rescaling.

As a result, values at an equal distance, $d$ above and below $m$ before transformation will not be equally distant from it afterward - the transformed value above will be closer to $\log(m)$ than the transformed value below it will be. This would happen for every value of $d$.

So symmetric $X$ implies asymmetric $\log(X)$.


Now let's talk not about normality, but approximate normality. (For simplicity let's assume that the distribution is such that the values are going to be essentially always positive - i.e. if the original values were normal, the chance of a negative value is extremely low.)

There is one situation where approximately normal values tend to still be approximately normal after transformation.

That's when the standard deviation is very small compared to the mean (low coefficient of variation).

If you look at the above diagram, consider values on the x-axis in a very narrow band around $m$. The pulling-in/stretching-out effect is minimal (the black curve doesn't have room to move far away from the blue tangent line), and so the shape still looks normal.

Here's an example: the top plot is a set of approximately normal data (the Q-Q plot shows a fairly straight line), and its log is also approximately normal (the Q-Q plot still shows a fairly straight line). That's because the coefficient of variation in the original values was pretty small (somewhere around 0.2 I think) - the nonlinear transformation was still nearly linear in the narrow range of values around the middle.

enter image description here

In this situation, the delta method does indeed tend to be useful at giving approximate values for the mean and variance of the log-values, though it would not actually be the distribution of the log of an exactly normal random variate.

| cite | improve this answer | |
$\endgroup$
4
$\begingroup$

No. $X$ may be negative. Therefore, $\log(X)$ will not return a real number with positive probabiity. The normal distribution is only defined on the real line. QED.

| cite | improve this answer | |
$\endgroup$
3
$\begingroup$

No. counter example: $x\sim\mathcal{N}(-1,1)$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ The question was whether $\log(X)$ CAN be normally distributed if $X$ is normally distributed. This example doesn't prove that statement false, so it's not a counterexample. You'd need to show that it is impossible for $\log(X)$ to be normal if $X$ is normal (which it obviously is; e.g., see Sycorax's answer). $\endgroup$ – Do not reinstate Monica Aug 4 at 13:02
3
$\begingroup$

If $X$ is normally distributed, can $\log(X)$ also be normally distributed?

Theoretically: No

Other answers have stated that it is not possible. Indeed, theoretically, it is not possible. If $X$ is normally distributed, then $X$ can have negative values and $\log(X)$ does not exist for negative input.

In practice: Yes

However, in practice, one might deal with a distribution that is only approximately normal distributed, and which has a domain $X>0$. For such a distribution it might be still interesting to imagine what will happen when we take the transform $\log(X)$.


The Delta method

The delta method approximates the transformation $\log(X)$ by linearization. This approximation will work well when the difference around the mean (the point where you linearize) is not too large.

The image below illustrates the transform for different coefficient of variation of $X$ (See for similar images here).

You can see how the image on the left is not so close to a linear transform and the resulting histogram of $Y = \log(X)$, drawn in the margin, does not resemble so much a normal distribution and is skewed. However, the image on the right is closer to a linear transform and the transformed variable does resemble reasonably a normal distribution.

linearization, Delta method, with different scales

The values of $\mu$ and $\sigma$ in the last/right image show that the Delta method works for that case:

> ymean
[1] 2.995604
> log(xmean)
[1] 2.995921
>  
> ysig
[1] 0.02521176
> xsig/20
[1] 0.02519255
> 

More precise than the Delta method

The Delta method will be less accurate when the $\sigma/\mu$ is larger because the approximation with linearization is less accurate.

The image below demonstrates this. It shows simulations of 10k points for $Y = \log(X)$ where $X \sim N(\mu_X = 1,\sigma_X = CV)$ where $CV$ has been varied (values $X<0$ were removed).

simulations with varying CV

The red broken curve illustrates that the mean of $Y = \log(X)$ can be reasonably approximated by inverting the formula for the mean of a log-normal distributed variable.

  • If $Y = \exp(X)$ or $X = \log(Y)$, where $X \sim N(\mu_X,\sigma_X^2)$ then $\mu_Y = \exp(\mu_X + 0.5 \sigma_X^2) $, and the inverse $\mu_X \approx \log(\mu_Y)-0.5 \sigma_X^2$.

  • We could also do the same for the relatioship for the variance of a log-normal distribution but that is a bit awkward expression, so we simplify things a bit and fill in the Delta approximation $\sigma_X \approx \sigma_Y/\mu_Y$.

So we end up with

$$\mu_X \approx \log(\mu_Y)-0.5 \sigma_Y^2/\mu_Y^2$$

this is the red curve in the above plot and it seems to correspond well with the data.

Practical application:

In this question:

Monte carlo results becomes more skewed with more sampling

one deals with a logarithm of $X$ where it is the percent of monthly return of an investment. The mean is 1.01 and the sd = 0.04 such that the coefficient of variation is very small.

In that question, the delta method works, but the more precise method even better.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

If $X$ is normally distributed, can $log(X)$ also be normally distributed?

Yes. It is possible. And indeed it is true if and only if $X\sim\mathcal{N}(\mu,0)$, with $\mu > 0$, in which case $log(X) \sim\mathcal{N}(log(\mu),0)$.

Note: "Can" is different than "must".

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I didn't know normally distributed variables could take on infinite or undefined values (e.g., $\log(x)$ when $x\leq 0$) with non-zero probability. $\endgroup$ – Do not reinstate Monica Aug 4 at 13:05
  • $\begingroup$ @Do not reinstate Monica Thanks for your comment. i just edited my answer to specify $\mu > 0$, rather than my originally stated $\mu \ge 0$ $\endgroup$ – Mark L. Stone Aug 4 at 15:28
  • $\begingroup$ Regardless of the value of $\mu$, $P(X \leq 0) > 0$. $\endgroup$ – Do not reinstate Monica Aug 4 at 15:36
  • 1
    $\begingroup$ Isn’t the variance parameter in a normal strictly positive? $\endgroup$ – Dimitriy V. Masterov Aug 5 at 4:41
  • 1
    $\begingroup$ I didn't downvote your answer, but I think calling a degenerate random variable "normally distributed with a variance of zero" is debatable and, in any case, seems to completely miss the point of the question. $\endgroup$ – Do not reinstate Monica Aug 5 at 16:00

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.