Let $X$ and $Y$ be two independent random variables. $X \sim U[a, b]$ and $Y \sim \exp(\lambda)$ where $a$ and $b$ are all positive real numbers such that $a < b<\infty$. I would like to find $\mathbb{P}[X<Y]$. Here is my trial using the result found here:
\begin{align*}
\mathbb{P}(X<Y)&=\int_{-\infty}^{\infty}(1-F_{Y}(x))dF_{X}(x)\\
&=\int_{-\infty}^{\infty}(1-F_{Y}(x)) f_X(x) dx\\
&= \int_{0}^{\infty}(1-(1-e^{-\lambda x})) f_X(x) dx\\
&= \int_{0}^{\infty}e^{-\lambda x} f_X(x) dx\\
&= \int_{0}^{a}e^{-\lambda x} f_X(x) dx + \int_{a}^{b}e^{-\lambda x} f_X(x) dx + \int_{b}^{\infty}e^{-\lambda x} f_X(x) dx\\
&=\int_{a}^{b}e^{-\lambda x} f_X(x) dx \\
&= \int_{a}^{b}e^{-\lambda x} \left( \frac{1}{b-a} \right) dx\\
&= \frac{e^{-\lambda a}-e^{-\lambda b}}{\lambda (b-a)}
\end{align*}
Numerical Evaluation
Let $a=1, b=10, \lambda=1$. The result of $\mathbb{P}(X<Y)$ is $3.3105 \notin [0,1] $ and thus surely is wrong. Why is that? What am I missing?
Best Answer
You appear to have made an error in your numerical example, since $\dfrac{e^{-1}-e^{-10}}{9}\approx 0.04$. Your probability formula is $$\dfrac{1-\lambda a+\frac{\lambda}{2}a^2-\cdots -1+\lambda b-\frac{\lambda^2}{2}b^2+\cdots}{\lambda(b-a)}=1-\frac{\lambda}{2}(a+b)+\cdots, $$so it doesn't seem liable to violate unitarity.