Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 124 / 5 / b / Solution

Write $y=N^{1/\phi (N)}$, $L={\sum }_{p\le y}|g(p)|$, and $A={\sum }_{p\le y}g(p)/p$. Put $I_p(n)=1_{p\mid n}$ and $Y_p=I_p-1/p$. The strongly additive arithmetic function identity is $g-A={\sum }_{p\le y}g(p)Y_p$.

For any tuple $p_1,\dots ,p_j$, expand ${\prod }_{r=1}^j(I_{p_r}-1/p_r)$ by choosing an indicator in some positions and the constant in the others. If the selected positions contain distinct primes whose product is $d$, thenFor the corresponding product of the independent Bernoulli random variables $X_p$, the expectation is exactly $1/d$. This includes repeated primes, since their indicator functions are idempotent, and the empty selection has zero error. Summing absolute values of all expansion coefficients bounds the difference bywhen $y\ge 2$. Expanding the $j$th power and summing the absolute coefficient weights gives

It remains to center at the actual expectation. Since ${\mathbb{E}}_N{I}_p=\lfloor N/p\rfloor /N$, we have $|{\mathbb{E}}_{N}g-A|\le L/N$. For each $n$, both $g(n)-A$ and $g(n)-{\mathbb{E}}_{N}g$ have absolute value at most $L$, because all their indicator coefficients have absolute value at most one. The identity $u^j-v^j=(u-v){\sum }_{r=0}^{j-1}{u}^{j-1-r}{v}^r$ therefore bounds the change in the central moment by $j{L}^j/N$.

Finally, independence factors each term of the independent central moment. If $q_1,\dots ,q_w$ are the distinct primes of the tuple and $a_i$ their multiplicities, the desired comparison becomesIf $y<2$, then $g=0$ and both sides vanish, so the formula remains valid. The argument also covers complex coefficients $g(p)$; a real-valued assumption is needed for the later normal distribution conclusion, not for this centered moment comparison for additive arithmetic functions.