Laplace transform identity for exponential random variable

Question

Suppose $X$ is a nonnegative random variable and $Y_s$ is an exponentially distributed random variable with intensity $s$ independent of $X$. Prove the Laplace transform identity

$$E(e^{-sX}) = Pr(X < Y_s)$$

Answer 1

We start with the right hand side of the equation, $P(X < Y_s)$. In words, this expression asks ''for all values of $X$, what is the probability that $Y_s$ is greater than it?'' This can be written as

\begin{align*} P(X < Y_s) &= \int_0^{\infty} P(Y_s > x) f_x(x) dx\\ &=\int_0^{\infty} (1 - F_{Ys}(x)) f_{x}(x) dx\\ &=\int_0^{\infty} (1 - (1 - e^{-sx})) f_{x}(x) dx\\ &=\int_0^{\infty} (1 - 1 + e^{-sx})) f_{x}(x) dx\\ &=\int_0^{\infty} e^{-sx} f_{x}(x) dx\\ &=E[e^{-sX}] \end{align*}

The second line follows from recognizing that $P(Y_s > x)$ is the right-tail probability. The third line follows from plugging in the cumulative distribution function for an exponential random variable. The final line follows from the definition of the expectation operator.