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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?

Lod
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1 Answers1

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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.

J.G.
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