= Solution
Use the positive-integer <geometric distribution>, $\mathbb P(X=k)=p(1-p)^{k-1}$ for $k\geq1$, initially with $0<p<1$. Put $a=-\log(1-p)>0$. The precise <good rate function> on $\mathbb R$ is
$$
\boxed{I(x)=
\begin{cases}
ax,&x\geq0,\\
+\infty,&x<0.
\end{cases}}
$$
The <large-deviation speed> is $L\to\infty$. We prove both bounds of the <large deviation principle> for $X^L=X/L$.
For $x>0$, the geometric tail gives the exact expression
$$
\mathbb P(X^L\geq x)
=(1-p)^{\lceil Lx\rceil-1},
\qquad
\lim_{L\to\infty}L^{-1}\log\mathbb P(X^L\geq x)=-ax.
$$
Let $F$ be a <closed set>. If it does not meet $[0,\infty)$ its probability is zero. If it contains zero, $\limsup L^{-1}\log\mathbb P(X^L\in F)\leq0=-\inf_FI$. Otherwise, if it meets the nonnegative half-line, its least possible nonnegative value $r$ is positive: a sequence in $F$ approaching zero would force zero into the closed set. Its event is contained in $\{X^L\geq r\}$, giving
$$
\limsup_{L\to\infty}L^{-1}\log\mathbb P(X^L\in F)
\leq-ar=-\inf_FI.
$$
For an <open set> $G$ containing $x>0$, choose positive integers $k_L$ with $k_L/L\to x$. Eventually $k_L/L\in G$, and
$$
\liminf_{L\to\infty}L^{-1}\log\mathbb P(X^L\in G)
\geq\lim_{L\to\infty}L^{-1}\log[p(1-p)^{k_L-1}]
=-ax.
$$
If $0\in G$, the event $X=1$ alone gives lower exponential rate zero; alternatively the probability of a neighborhood of zero tends to one. Optimizing over $x\in G$ proves the open-set lower bound. Sets containing no nonnegative point require only the trivial lower bound $-\infty$.
The <rate function> is <lower semicontinuous> at zero as well as elsewhere, and $\{I\leq r\}=[0,r/a]$ for finite $r\geq0$ is <compact>. Hence the <large deviation principle> has a <good rate function>. This establishes the <large deviations of a scaled geometric random variable>. For $p=1$, $X=1$ deterministically and the rate instead is zero at $x=0$ and infinite elsewhere; the displayed logarithmic coefficient must not be treated as an ordinary finite number in that case.
Back to article page
