Assume we have two random variables \(X\) and \(Y\) and we are interested in computing the probability of the event \(X<Y\). This is equivalent to computing the expected value of the indicator \[ \begin{align} \mathbb{P}(X<Y)=\mathbb{E}[\mathbb{1}_{X<Y}] & =\int _{\Omega} \mathbb{1}_{X<Y}(\omega)d\mathbb{P}(\omega) \\ & =\int_{Y}\int_{X}\mathbb{1}_{X<Y}f_{X,Y}(x,y)~dxdy \\ & = \int_{Y}\int _{X<Y} f_{X,Y}(x,y)~dxdy. \end{align} \] ##### Example Take an example from birth-death processes where times until the next births and deaths when the population has \(i\) individuals are given by the independent exponential random variables \[ B(i)\sim\mathcal{E}(\lambda_{i}) \quad\&\quad D(i)\sim\mathcal{E}(\mu_{i}). \] The population increases by 1 when \(B(i)<D(i)\) which occurs with probability \[ \begin{align} \mathbb{P}(B(i)<D(i)) & =\mathbb{E}[\mathbb{1}_{B<D}]=\int_{D}\int_{B<D}f_{B,D}(s,t)~dsdt \\ & = \int_{0}^\infty \int_{0}^tf_{B}(s)f_{D}(t)~dsdt\quad(\text{ind.}) \\ & = \int_{0}^\infty \int_{0}^t \lambda_{i}\exp(-\lambda_{i}s)\mu_{i}\exp(-\mu_{i}t)~dsdt \\ zz \end{align} \]
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