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Suppose we have a r.v. $X\ge 0$ and a constant $c>0$. Then we look at the Laplace Transform:

$$E[e^{-c X}]$$

We can suppose that this has a closed form $f(c)$. My question is now, why is it true, that for $c\to 0$ we can conclude that

$$\lim_{c\to 0}E[e^{-cX}]=\lim_{c\to 0}f(c)\Rightarrow P[X<\infty]=\lim_{c\to 0}f(c)$$

I do not see while letting $c$ go to $0$ we obtain the probability that $X$ is not infinity.

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1 Answer 1

up vote 3 down vote accepted

Because, for every $x$ in $[0,+\infty]$, $\mathrm e^{-cx}\to\mathbf 1_{x\lt\infty}$ when $c\to0^+$.

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