In a popular heuristic argument in favor of the Collatz conjecture, one calculates a kind of "geometric expected value" for the ratio $C(x)/x$, where $x$ is odd, and

\begin{align*} C(x) := \frac{3x+1}{2^{\upsilon_2 (3x+1) }} \end{align*}

is next odd number after $x$ in the sequence obtained by repeatedly applying the Collatz function

\begin{align*} T(n) := \begin{cases} n/2 &\text{if $n$ is even,}\\ 3n+1 &\text{if $n$ is odd} \end{cases} \end{align*}

to $x$. The argument is that the quantity

\begin{align*} \frac{C(x)}{x} = \frac{3+\frac{1}{x}}{2^{\upsilon_2 (3x+1) }} \approx \frac{3}{2^{\upsilon_2 (3x+1)}} \end{align*} will equal $3/2$ with probability $1/2$, will equal $3/4$ with probability $1/4$, and will - generally - equal $3/2^k$ with probability $1/2^k$, and so the "expected value" of $C(x)/x$ is calculated as the following infinite product, which converges to $3/4 < 1$:

\begin{align*} \prod_{k=1}^\infty \left(\frac{3}{2^k}\right)^{\frac{1}{2^k}}. \end{align*}

My question: Why don't we calculate the expected value in the "usual" (ie. arithmetic, as opposed to geometric) way, ie. as

\begin{align*} \mathbb{E}\left(\frac{C(x)}{x}\right) = \sum \hspace{0.1cm} \text{"outcome"} \cdot \text{"probability of this outcome"} = \sum_{k=1}^\infty \frac{3}{2^k} \cdot \frac{1}{2^k}? \end{align*}

If we did this, the expected value would be different, namely equal to $1$.

  • $\begingroup$ A "geometric" approach may be more attractive since it helps explain why the Collatz Conjecture seems to converge to 1 and it takes into account the incremental consequences of $n/2$. The value 3/4 more clearly explains why on average the Conjecture converges, where as the value 1 does not give that sort of insight. $\endgroup$ Commented Aug 25, 2017 at 14:48

1 Answer 1


Composition of ratios is achieved by multiplication whereas composition of differences is achieved by addition.

If you want to use the arithmetic mean then you must be working with the difference between one term and the next rather than the ratio of one term to the next.

Also, when using ratios and geometric arguments, there is the useful property that fixed size components such as $+1$ shrink to zero as you move into greater numbers, while they can accumulate when using sums.


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