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Suppose $X$ is a random variable on $\mathbb R_+$ with finite mean, i.e. $\mathbb E X <+\infty$.

Let $F_X(t)$ be its c.d.f. and $\mathcal{L}_X(\cdot)$ its Laplace transform, i.e. $\mathcal{L}_X(s)=\int_0^\infty e^{-s t} d F(t)$

Can one conclude immediately that, as $s \to 0$, $\log \mathcal{L}_X(s) \approx -s \mathbb E X + o(s^2) $ ?

If not, suppose now that $X$ has finite moments of all orders. Can one now conclude that, as $s \to 0$, $\log \mathcal{L}_X(s) \approx -s \mathbb E X + o(s^2) $ ?

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1 Answer 1

up vote 4 down vote accepted

Since the random variable $X$ is assumed to have finite mean, the derivative of the Laplace transform $$ \mathcal{L}_X^\prime(s) = -\mathbb{E}\left(X \mathrm{e}^{-s X}\right) $$ is finite at $s=0$. Hence for small $s$, $\mathcal{L}_X(s) = 1 - \mathbb{E}(X) s + \mathcal{o}(s)$. Taking the logarithm: $$ \log \mathcal{L}_X(s) = \log(1 - \mathbb{E}(X) s + \mathcal{o}(s)) = -s \mathbb{E}(X) + \mathcal{o}(s) $$

If moments of higher order exists, series approximation of the higher order can be constructed. Check out the cumulant generating function. Notice that $\mathcal{L}_X(s) = \mathcal{M}_X(-s) = \exp\left(\mathcal{K}_X(-s)\right)$, where $\mathcal{M}_X$ is the moment generating function, and $\mathcal{K}_X$ stands for the cumulant generating function of $X$.

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