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Let $X \sim N(\mu, \sigma^2)$, where $\mu = -800$ and $\sigma = 76$

Let $Y = \exp(X)$, so Y has a lognormal distribution, $E(Y) = \exp(\mu + \sigma^2/2) = \exp(2088)$, which is a very large number.

However, if I first sample from $X$, and then take $\exp()$ of the sampled points, I got all near 0's. This is understandable since $X$ has a big negative mean and 0 is 10 standard deviations away.

But how should I understand that of $E(Y)$ being a very large number and a sample of all near 0 points? Because a very large variance?

What is the implication of the use of lognormal then?

EX:

sample <- rnorm(100000, -800, 76)

summary(exp(sample))

 Min.      1st Qu.     Median       Mean    3rd Qu.       Max. 

0.000e+00  0.000e+00   0.000e+00 1.170e-208  0.000e+00 1.169e-203
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    $\begingroup$ It's because $X$ will occasionally take on moderate-sized positive values and these become enormous when exponentiated, whereas the large negative values get bounded from below. I don't think it's accurate to say $E(Y) \cong \infty$. It's a large finite number, which is quite far away from infinity. $\endgroup$
    – dsaxton
    Dec 8 '15 at 19:47
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    $\begingroup$ $\exp\{2088\}$ is not "near infinity", no number is near infinity. It is infinitely far away from infinity. It's just a very, very big number. $\endgroup$
    – jbowman
    Dec 8 '15 at 19:57
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    $\begingroup$ In the applications I have seen, which cover engineering, environmental, financial, physical, demographic, and economic ones, most lognormal distributions used in models and fit to data have $\sigma$ less than $1$ and almost always less than $5$ or so. What are you trying to model with such an astronomically large value of $\sigma$? $\endgroup$
    – whuber
    Dec 8 '15 at 19:58
  • $\begingroup$ @jbowman @ dsaxton Yes, I will edit it as very large number $\endgroup$
    – mac
    Dec 8 '15 at 20:03
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    $\begingroup$ If you're not actually using it, then what exactly is your question? That you have chosen an unrealistic combination of parameters and found that unrealistic values result is no surprise and has nothing to do with the "practical use of the lognormal" distribution. $\endgroup$
    – whuber
    Dec 8 '15 at 21:35
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Note that although the mean is large, the 99.99th percentile is almost 0

 exp(-800+qnorm(0.9999)*76)
 [1] 2.068655e-225

which is to say, only one observation in ten thousand will exceed $2.069\times 10^{-255}$, an extremely small number.

Note that for the log of this random variable, $0$ is $10.526$ standard deviations above the mean. This means that there's only a probability of $3.3\times 10^{-26}$ of the lognormal variable exceeding $1$.

The mean is large because there's a (very) small chance of incredibly large observations. Even though those chances are astronomically small, the values the variable can exceed with those small probabilities are so astoundingly large that the mean is still very big. If you go out far enough, he balance between rarity and size eventually becomes dominated by size, making the mean large.

This the kind of thing you can get with a sufficiently skew variable - even one whose moments all exist!

How big is the (third moment) skewness? Quite large. It's very close to $\exp(8664)$ -- roughly a $5$ with 3762 $0$'s after it.

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  • $\begingroup$ Thank you! How did you get the $3.3 *10^{-26}$? It seems R cannot go that far $\endgroup$
    – mac
    Dec 9 '15 at 2:47
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    $\begingroup$ Sure it can. pnorm(0,-800,76,lower.tail=FALSE) or pnorm(800/76,lower.tail=FALSE) (see ?pnorm). Indeed, even the approximation $1-\Phi(x) \approx \phi(x)/x$ is adequate when you're that far into the tail. $\endgroup$
    – Glen_b
    Dec 9 '15 at 2:57
  • $\begingroup$ beautiful, and the approximation...cool stuff:) BTW, 1- pnorm(0, -800, 76) will only give 0. $\endgroup$
    – mac
    Dec 9 '15 at 3:39
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    $\begingroup$ Subtracting something that's almost 1 from 1 to get a tail area is a very inaccurate way to find the result; you can't get a good answer that way; much better to find the tail area in one go. The ratio of survivor function to density in the standard normal, $\frac{1-\Phi(x)}{\phi(x)}$ is known as Mills' ratio; numerous papers give useful approximations for largish values of $x$, (and/or bounds). The approximation (1/x) is a simple upper bound for the ratio that is also a good approximation for large x, and multiplying through by the density gives the approximation I mentioned earlier. $\endgroup$
    – Glen_b
    Dec 9 '15 at 5:41

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