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With $n$ fixed, the asymptotic family is the scaled sumAn unscaled fixed random variable alone has no diverging parameter for a large deviation principle; this scaling is the continuation of part (a). Suppose first $0<p_i<1$ and put $a_i=-\log (1-p_i)$. Independence and the product large-deviation principle give the vector $(X_1/L,\dots ,X_n/L)$ rateAddition is continuous, so the contraction principle for large deviations givesFor $s\ge 0$ every feasible vector has cost at least $s{a}_{\ast }$, where $a_{\ast }={\min }_i{a}_i$. Equality holds by assigning all scaled excess to an index attaining $a_{\ast }$. For $s<0$ the constraint set is empty. ThusEquivalently, the coefficient is $-\log (1-p_{\ast })$ with $p_{\ast }={\min }_i{p}_i$. On the exponential scale the rare large sum is carried by a variable having the slowest geometric tail. Several equal slowest tails can change polynomial prefactors, but not this rate function. This is the large deviations of a fixed sum of geometric random variables.

As a direct check of the tail exponent, for each $0<\theta <a_{\ast }$ the moment-generating function of the sum is finite, so the Chernoff bound gives an upper rate at most $-\theta s$ for $s>0$. Letting $\theta \uparrow a_{\ast }$ gives $-a_{\ast }s$. For a lower neighborhood bound, keep all other variables at fixed finite values and let a slowest-tail variable take an integer near $Ls$; its probability has rate $-a_{\ast }s$. If some $p_i=1$, their deterministic contributions disappear after division by $L$, so minimize over the nondeterministic variables. If all $p_i=1$, the rate is zero only at zero and infinite elsewhere.