# Laplace Transform of a Truncated Random Variable

Let $B$ be a random variable, and $S=\mbox{min}(1,B)$. Can you help me see why the laplace stieltjes transform of $S$ is given by $$E[e^{-\alpha S}]=1-\alpha\int_{0}^{1}e^{-\alpha y}P(B\geq y)dy$$

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This is the integrated form (with respect to the distribution of $B$) of the equality $$1-\mathrm e^{-\alpha\min\{1,x\}}=\int_0^1\alpha\mathrm e^{-\alpha y}\,[x\geqslant y]\,\mathrm dy,$$ valid for every $x\geqslant0$. To prove the equality holds, note that the RHS is $$\int_0^{\min\{1,x\}}\alpha\mathrm e^{-\alpha y}\,\mathrm dy=\left[-\mathrm e^{-\alpha y}\right]_{y=0}^{y=\min\{1,x\}}=1-\mathrm e^{-\alpha\min\{1,x\}}.$$