I have a random variable $X(a) = \log(a)$ where a is normal distributed $\mathcal N(\mu,\sigma^2)$. What can I say about $E(X)$ and $Var(X)$? An approximation would be helpful too.
$\begingroup$
$\endgroup$
2
gung - Reinstate Monica
145k88 gold badges400 silver badges708 bronze badges
asked Apr 30, 2013 at 15:42
-
5$\begingroup$ I think the question was about the "inverse" of the log-normal, i.e. where a normal rv A leads to log-normal X = exp(A), the questioner was asking about the distribution of X = log(A), which is undefined (due to sometimes requiring the log of a negative number). There may be some results for a truncated normal, but they're likely to be messy. $\endgroup$– Martin O'LearyApr 30, 2013 at 15:56
-
3$\begingroup$ rocksportrocker, as @Martin O'Leary points out, it is not mathematically possible to have such a variable $X$, because $\log(a)$ is undefined for negative values. At a minimum you need to truncate $a$ at some non-negative value. Could you tell us why you believe $a$ might be Normal? $\endgroup$– whuber ♦Apr 30, 2013 at 16:08
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
4
If we consider "approximation" in a fairly general sense we can get somewhere.
We have to assume not that we have an actual normal distribution but something that's approximately normal except the density cannot be nonzero in a neighborhood of 0.
So let's say that $a$ is "approximately normal" (and concentrated near the mean*) in a sense that we can handwave away the concerns about $a$ coming near 0 (and its subsequent impact on the moments of $\log(a)$, because $a$ doesn't 'get down near 0'), but with the same low order moments as the specified normal distribution, then we could use Taylor series to approximate the moments of the transformed random variable.
For some transformation $g(X)$, this involves expanding $g(\mu_X + X-\mu_X)$ as a Taylor series (think $g(x+h)$ where $\mu_X$ is taking the role of '$x$' and $X-\mu_X$ takes the role of '$h$') and then taking expectations and then either computing the variance or the expectation of the square of the expansion (from which can be obtained the variance).
The resulting approximate expectation and variance are:
$\text{E}\left[g(X)\right]\approx g(\mu_X) +\frac{g''(\mu_X)}{2}\sigma_X^2$ and
$\text{Var}\left[g(X)\right]\approx \left(g'(\mu_X)\right)^2\sigma^2_X$
and so (if I didn't make any errors), when $g() = \log()$:
$$\text{E}\left[\log(a)\right]\approx log(\mu_a) -\frac{\sigma_a^2}{2\mu_a^2}$$
$$\text{Var}\left[\log(a)\right]\approx \sigma^2_a/\mu_a^2$$
* For this to be a good approximation you generally want the standard deviation of $a$ to be quite small compared to the mean (low coefficient of variation).
-
7$\begingroup$ Because the Taylor series for log has a relatively small radius of convergence, caution is advised in applying these approximations. $\endgroup$– whuber ♦May 1, 2013 at 6:00
-
$\begingroup$ @whuber for an expansion around the mean, I think this would correspond to the advice that the "standard deviation of $a$ should be quite small compared to the mean" that my answer ends with -- if I am missing some further issue that that advice doesn't cover I should fix my answer. $\endgroup$– Glen_bMay 1, 2013 at 6:05
-
3$\begingroup$ The approximation for the mean works pretty well for $\mu/\sigma \gt 1.5$ and that for the variance works pretty well for $\mu/\sigma \gt 2.5$ or so. $\endgroup$– whuber ♦May 1, 2013 at 6:30
-
$\begingroup$ In any case it is certainly worth being clear that we're indirectly relying on the convergence of $\ln(1+x)$ (since $\ln(\mu + y-\mu) = \ln[\mu\{1 + (y - \mu)/\mu\}] = \ln(\mu) + \ln[1 + (y - \mu)/\mu]$). Thanks also for the suggested explicit values; if anything perhaps I am slightly overcautious when I use it. Two valuable comments. $\endgroup$– Glen_bMay 1, 2013 at 6:39