I am reading a book about ruin theory and I got stuck in a certain part of proof of ruin probability formula.

Namely, given $U(t)$ - a classical risk process, i.e. $$U(t)=u_{0}+c\cdot t-S(t)$$ where $S(t)$ has a compound Poisson distribution with parameters $(\lambda t, F)$ and $u_0$, c - constants and assuming that ruin didn't occur ($\forall_{t} \ U(t)\geq 0$),

author claims that $$\lim_{t\to \infty}\mathbb{P}\left(U(t)\leq \mathbb{E}[U(t)]-\sqrt{Var[U(t)]}\cdot t^{1/6}\right)=0 \ \ \ (\star)$$

Explaining this result, he recalls Chebyshev's inequality.

But yet, how we can use Chebyshev's inequality here? In Chebyshev's inequality, an inequality sign is opposite ($"\geq"$ instead of $"\leq"$). I am not able to prove $(\star)$ myself, so I begin to doubt if this limit really tend to zero.


1 Answer 1


The random variable $X=-(U(t)-\mathbb{E}[U(t)])/\sqrt{Var[U(t)]}$ is centered with variance $1$ hence $$\mathbb P(X\geqslant t^{1/6})\leqslant\mathbb P(X^2\geqslant t^{1/3})\leqslant \mathbb E(X^2)/t^{1/3}=1/t^{1/3}.$$


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