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

Question

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?

Answer 1

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.

Answer 2

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.)

Answer 3

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 $(0,1)$.

Answer 4

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] -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)
[1] 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] 1.718425
integrate(function(x) exp(x), 0, 1)
1.718282 with absolute error < 1.9e-14
exp(1)-1
[1] 1.718282