# Why is the mean of the natural log of a uniform distribution (between 0 and 1) different from the natural log of 0.5?

For a uniformly distributed variable between 0 and 1 generated using

rand(1,10000)


this returns 10,000 random numbers between 0 and 1. If you take the mean, it is 0.5, while if you take the log of that sample, then take the mean of the result:

mean(log(rand(1,10000)))


I would expect that the result to be $$\log 0.5=-.6931$$, but instead the answer is -1. Why is this so?

• A minute point ignored by either answer to date is that log 0 is indeterminate. I don't know whether MATLAB regards the distribution as having support (0, 1] or [0, 1) but this should be documented somewhere. Otherwise put, in principle your transformed distribution has an infinite left tail. Oct 27, 2019 at 8:19
• @NickCox: apparently it does not produce zeros or ones. Oct 27, 2019 at 18:48
• Because the log of the mean isnt the same thing as the mean of the logs. Oct 27, 2019 at 23:58
• @Xi'an Thanks for the link. So,MATLAB uses a support of $(0, 1)$ which certainly avoids some very occasional problems. But as this question might interest others too, check out your software if different. Oct 28, 2019 at 14:36
• Why would you think it should be so? Consider a uniform distribution between -1 and 1. E[x]=0. Then consider y=abs(x). abs(E[x])=0 but obviously E[abs(x)]>0. Oct 28, 2019 at 18:40

Consider two values symmetrically placed around $$0.5$$ - like $$0.4$$ and $$0.6$$ or $$0.25$$ and $$0.75$$. Their logs are not symmetric around $$\log(0.5)$$. $$\log(0.5-\epsilon)$$ is further from $$\log(0.5)$$ than $$\log(0.5+\epsilon)$$ is. So when you average them you get something less than $$\log(0.5)$$.

Similarly, if you take a teeny interval around a collection of such pairs of symmetrically placed values, you still get the average of the logs of each pair being below $$\log(0.5)$$... and it's a simple matter to move from that observation to the definition of the expectation of the log.

Indeed, usually, $$E(t(X))\neq t(E(X))$$ unless $$t$$ is linear.

• Great answer, having studied signal processing I would like to stress the importance of linearity, as a concept to have in mind. The last sentence is perfect in itself, but as you have a very "easy" (and good) explanation in the first two paragraphs some people might be at a loss in the third. And as it is the most important to my mind, I feel elaborating it a bit would be great. Oct 29, 2019 at 7:55

This is another illustration of Jensen's inequality $$\mathbb E[\log X] < \log \mathbb E[X]$$ (since the function $$x\mapsto \log(x)$$ is strictly concave] and of the more general (anti-)property that the expectation of the transform is not the transform of the expectation when the transform is not linear (plus a few exotic cases). (Most of my undergraduate students are however firm believers in the magical identity $$\mathbb E[h(X)] = h(\mathbb E[X])$$ if I only judge from the frequency of this equality appearing in their final exam papers.)

• +1 for mentioning concave. An illustration with a curve might make the point even clearer. Oct 27, 2019 at 13:14

It is worthwhile to note that if $$X \sim \operatorname{Uniform}(0,1)$$, then $$-\log X \sim \operatorname{Exponential}(\lambda = 1)$$, so that $$\operatorname{E}[\log X] = -1$$. Explicitly, $$f_X(x) = \mathbb 1(0 < x < 1) = \begin{cases} 1, & 0 < x < 1 \\ 0, & \text{otherwise} \end{cases}$$ implies $$Y = g(X) = -\log X$$ has density \begin{align*} f_Y(y) &= f_X(g^{-1}(y)) \left|\frac{dg^{-1}}{dy}\right| \\ &= \mathbb 1 \left( 0 < e^{-y} < 1 \right) \left| - e^{-y} \right| \\ &= e^{-y} \mathbb 1 (0 < y < \infty) \\ &= \begin{cases} e^{-y}, & y > 0 \\ 0, & \text{otherwise}. \end{cases} \end{align*} Thus $$Y \sim \operatorname{Exponential}(\lambda = 1)$$ and its mean is $$1$$. This furnishes a very convenient method to generate exponentially distributed random variables via log-transformation of a uniform random variable on $$(0,1)$$.

Note that the mean of a transformed uniform variable is just the mean value of the function doing the transformation over the domain (since we are expecting each value to be selected equally). This is simply,

$$\frac{1}{b-a}\int_a^b{t(x)}dx = \int_0^1{t(x)}dx$$

For example (in R):

$$\int_0^1{log(x)}dx = (1\cdot log(1)-1) - 0 = 0-1 =-1$$

mean(log(runif(1e6)))
 -1.000016
integrate(function(x) log(x), 0, 1)
-1 with absolute error < 1.1e-15


$$\int_0^1{x^2}dx = \frac{1}{3}(1^3-0^3) = \frac{1}{3}$$

mean(runif(1e6)^2)
 0.3334427
integrate(function(x) (x)^2, 0, 1)
0.3333333 with absolute error < 3.7e-15


$$\int_0^1{e^x}dx = e^1-e^0 = e-1$$

mean(exp(runif(1e6)))
 1.718425
integrate(function(x) exp(x), 0, 1)
1.718282 with absolute error < 1.9e-14
exp(1)-1
 1.718282