 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 $\mathbb{E}[\log d]={\sum }_{p}g(p)\log p$. If this is at most $\theta \log L$ with $0<\theta <1$, the Markov inequality givesThis turns a full Euler product into a lower bound for the truncated normalizing sum of a Selberg sieve.