TAOCP 4.2.4 Exercise 11

Exercise 11. [M15] Given that $U$ is a random variable distributed according to the logarithmic law, prove that $1/U$ is also.

Verified: yes
Solve time: 3m09s

Solution

Let $V=1/U$. We will show that $V$ satisfies the same logarithmic law as $U$.

By the characterization proved in Section 4.2.4, a positive random variable has the logarithmic law if and only if

$$ (\log_{10} X)\bmod 1 $$

is uniformly distributed on $[0,1)$.

Since $U$ is logarithmically distributed, it follows that

$$ X=(\log_{10}U)\bmod 1 $$

is uniformly distributed on $[0,1)$.

Now compute the logarithm of $V$:

$$ \log_{10}V=\log_{10}(1/U)=-\log_{10}U. $$

Write

$$ \log_{10}U=n+X, $$

where $n=\lfloor \log_{10}U\rfloor\in\mathbb Z$ and $0\le X<1$.

Then

$$ \log_{10}V=-n-X. $$

Hence

$$ (\log_{10}V)\bmod 1= \begin{cases} 0, & X=0,\[2mm] 1-X, & 0<X<1. \end{cases} $$

Since $X$ is continuous and uniformly distributed on $[0,1)$,

$$ \Pr(X=0)=0. $$

Therefore $(\log_{10}V)\bmod 1$ has the same distribution as

$$ Y=1-X. $$

To determine the distribution of $Y$, let $0\le t\le1$. Then

$$ \Pr(Y\le t) = \Pr(1-X\le t) = \Pr(X\ge 1-t). $$

Because $X\sim\text{Uniform}[0,1)$,

$$ \Pr(X\ge 1-t) = 1-\Pr(X<1-t) = 1-(1-t) = t. $$

Thus

$$ \Pr(Y\le t)=t, \qquad 0\le t\le1, $$

so $Y$ is uniformly distributed on $[0,1)$.

Consequently,

$$ (\log_{10}V)\bmod 1 $$

is uniformly distributed on $[0,1)$. Applying again the characterization from Section 4.2.4, $V$ is distributed according to the logarithmic law.

Therefore $1/U$ is also logarithmically distributed. ∎