= Truncated Euler-product lower bound
For a finite set of <primes> choose $0<g(p)<1$, put $k(p)=g(p)/(1-g(p))$, and let $W=\prod_p(1+k(p))$. Give each <squarefree integer> $d$ on these primes probability $k(d)/W$. Inclusion of each <prime> is represented by a <Bernoulli random variable> with parameter $g(p)$; these are <independent random variables>, so $\mathbb E[\log d]=\sum_pg(p)\log p$. If this is at most $\theta\log L$ with $0<\theta<1$, the <Markov inequality> gives
$$
\sum_{d\leq L}k(d)\geq(1-\theta)W.
$$
This turns a full <Euler product> into a lower bound for the truncated normalizing sum of a <Selberg sieve>.
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