Finite in probability implies finite expectation Let $T_n$ be a random variable with $T_n=X_1+...+X_n$ where the $X_i$'s are iid. Further we set $N(t)=max\{ n: T_n\leq n\}$.
If $\Pr(N(t)<\infty)=1$, does this implies $\mathbb{E}[N(t)]<\infty$?
I think intuitively yes. Since every $N(t)$ is finite, then in the mean it is also finite. Is that correct?
Thank you for any help!!!
 A: I'm afraid that your claim is not correct. It is entirely possible for a variable $X$ to be "finite in probability", meaning $P(X<\infty)=1$, while having infinite mean. One example is the absolute value of a Cauchy distributed variable (http://en.wikipedia.org/wiki/Cauchy_distribution), which is finite in probability while having infinite mean.
An explicitly constructed example can be obtained as follows. Consider a variable $X$ which takes its values in the naturals $1,2,\ldots$, and which satisfies
$$
  P(X = n) = \frac{1}{n^2}\frac{6}{\pi^2}.
$$
We then have
$$
  \sum_{n=1}^{\infty} P(X = n) = \frac{6}{\pi^2}\sum_{n=1}^\infty\frac{1}{n^2} = 1,
$$
by classical infinite series results (http://en.wikipedia.org/wiki/Basel_problem), so $X$ is well-defined. Also, $P(X<\infty) = 1$. However,
$$
  E(X) = \sum_{n=1}^\infty nP(X = n)
       = \frac{6}{\pi^2}\sum_{n=1}^\infty \frac{1}{n} = \infty,
$$
as the harmonic series diverges (http://en.wikipedia.org/wiki/Harmonic_series_(mathematics)). Thus, this variable $X$ is "finite in probability", taking only finite values, while having infinite mean.
A: No, it is not correct. Take, for example, $X\sim\text{Cauchy}(0,1)$. Then
$$
\Pr\{|X|<\infty\}=1,
$$
but $\operatorname E|X|=\infty$.
