Negative moment of the number of arrivals in a renewal process Suppose $\{X_1, \ldots, X_n,\ldots\}$ are i.i.d. non-negative random variables with $\mathbb{E}[X_1]= \mu <\infty$, and $N(t):= \sup\{k: \sum_{i=1}^k X_i < t\}$. Notice that $X_i$ may not have higher order moments. The classical renewal theorem tells us
$$
\frac{\mathbb{E}[N(t)]}{t}\rightarrow \frac{1}{\mu}
$$
as $t\to \infty$. I am interested in whether:
$$
\mathbb{E}\left[\frac{t}{\max\{N(t),1\}}\right]\to \mu.
$$
as $t\to \infty$? Or at least is
$$
\limsup_{t\to \infty}
   \mathbb{E}\left[\frac{t}{\max\{N(t),1\}}\right]
\leq C\mu
$$
for some constant $C$?
(From Jensen's inequality one can easily see $
\liminf_{t\rightarrow\infty } \mathbb{E}[\frac{t}{N(t)}]\geq \mu.)
$
Thanks!
 A: This is not possible, unfortunately.
Let us play a little bit:
$$
\mathbb{E}\Big[\frac{t}{\max(1,N(t))}\Big] = t\int_0^1  \mathbb{P}(N(t)^{-1}\ge x)dx\\ = t\int_1^\infty \mathbb{P}(N(t)\le u)\frac{du}{u^2}
= \int_1^\infty t\,\mathbb{P}(S_{\lfloor u\rfloor }\ge t)\frac{du}{u^2}.
$$
By the law of large numbers and the Fatou lemma, for any $a>1/\mu$,
$$
\liminf_{t\to\infty} \int_{t/\mu}^{\infty}t\,\mathbb{P}(S_{\lfloor u\rfloor }\ge t)\frac{du}{u^2}  = \liminf_{t\to\infty}\int_{1/\mu}^{\infty}\mathbb{P}(S_{\lfloor vt\rfloor }\ge t)\frac{dv}{v^2}\\ \ge \int_{1/\mu}^{\infty}\liminf_{t\to\infty} \mathbb{P}\Big(\frac{S_{\lfloor v t\rfloor }}{t}\ge 1\Big)\frac{dv}{v^2} \ge \int_a^\infty \frac{dv}{v^2} = \frac1a.
$$
Letting $a\to 1/\mu$, we get
$$
\liminf_{t\to\infty}  \int_{t/\mu}^{\infty}t\,\mathbb{P}(S_{\lfloor u\rfloor }\ge t)\frac{du}{u^2} \ge \mu.
$$
So it remains to make $$
t \int_1^{t/\mu} \mathbb{P}(S_{\lfloor u\rfloor}\ge t)\frac{du}{u^2} \ge\sum_{k=1}^{\lfloor t/\mu\rfloor -1} \frac{t\,\mathbb{P}(S_{k}\ge t)}{(k+1)^2}
$$ as large as possible. There are many ways to do this. One way is to construct a distribution with  $\limsup t\,\mathbb{P}(X_1\ge t)$ as large as possible, which would immediately gives the desired result. This possible, but a bit tricky.
So let's assume that $t\mathbb{P}(X_1\ge t) \to 0$ and write by the inclusion-exclusion formula, for $k\le t/\mu$ and $t$ large enough
$$
t\,\mathbb{P}(S_k \ge t)\ge t\,\mathbb{P}(\max(X_1,\dots,X_k)\ge t) 
\ge  kt\,\mathbb{P}(X_1\ge t) - kt^2\, \mathbb{P}(X_1\ge t)^2\\ \ge k t\, \mathbb {P}(X_1\ge t) \big(1-  t \mathbb {P}(X_1\ge t)/\mu\big) \ge k t\, \mathbb {P}(X_1\ge t)/2.
$$
Then,
$$
t \int_1^{t/\mu} \mathbb{P}(S_{\lfloor u\rfloor}\ge t)\frac{du}{u^2} \ge {t\,\mathbb{P}(X_{1}\ge t)}\sum_{k=1}^{\lfloor t/\mu\rfloor -1}
\frac{k}{2(k+1)^2}\sim \frac{t\log t}2  \mathbb P(X_1\ge t).
$$
Now let
$$
\mathbb{P}(X_1 = 2^{2^n}) = \frac{K}{2^{2^n} n\log^2 n}, n\ge 2.
$$
Then, for $t = 2^{2^n}$,
$$
t\log t \, \mathbb{P}(X_1 \ge  2^{2^n}) \ge \frac{C 2^n}{n \log^2 n},
$$
so with this distribution we even get
$$
\limsup_{t\to\infty}  \int_1^{t/\mu} t\,\mathbb{P}(S_{\lfloor u\rfloor}\ge t)\frac{du}{u^2} = +\infty,
$$
and hence
$$
\limsup_{t\to\infty} \mathbb{E}\Big[\frac{t}{\max(1,N(t))}\Big] = +\infty.
$$
