# Find the probability of one expression

Could you please help me to calculate the probability as follows \begin{align} \rho=\mathsf{Pr}\left\{\frac{X_1}{X_2} >\frac{Y_1+a}{Y_2 + a}\right\} \end{align} where $X_1,X_2, Y_2$, and $Y_2$ are independent exponential distributed random variables with mean $\Omega_{X_1}, \Omega_{X_2}, \Omega_{Y_1}$, and $\Omega_{Y_2}$, respectively and $a$ is a positive constant.

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What are your attempts? I would also suggest you to use another notation for means, otherwise integrals will be messy. – Ilya Aug 23 '12 at 9:44
Many thanks Ilya for your comment. I have edited my question. – Tran Tam Aug 23 '12 at 9:54

I will change the notations. Let $\mu_k = \mathsf{E}(X_k)$ and $\lambda_k = \mathsf{E}(Y_k)$ for $k=0, 1$.

Lemma 1. For $\alpha > 0$, $\mathsf{Pr}\left( X_1 > X_2 \alpha \right) = \frac{\mu_1}{\mu_1 + \alpha \mu_2}$.
Proof:
$$\mathsf{Pr}(X_1 > \alpha X_2) = \mathsf{E}\left( \mathsf{Pr}(X_1 > \alpha X_2 |X_2) \right) = \mathsf{E}\left(\mathrm{e}^{\frac{\alpha}{\mu_1} X_2}\right)$$ The latter is the moment generating function of $X_2$ at $t=\frac{\alpha}{\mu_1}$, yielding the result. $\square$

Lemma 2. $\mathsf{E}\left(\mathrm{e}^{-t Y_k}\right) = \frac{1}{1 + t \lambda_k}$.
Proof:
$$\mathsf{E}\left(\mathrm{e}^{-t Y_k}\right) = \int_0^\infty \frac{1}{\lambda_k} \mathrm{e}^{-\frac{x}{\lambda_k}} \mathrm{e}^{-t x} \mathrm{d} x = \frac{1}{\lambda_k}\frac{1}{t+ \frac{1}{\lambda_k}} = \frac{1}{1+t \lambda_k} .\quad \square$$

Now: $$\mathsf{Pr}\left( \frac{X_1}{X_2} > \frac{Y_1+a}{Y_2+a} \right) = \mathsf{E}\left( \mathsf{Pr}\left( \left. X_1 > X_2 \frac{Y_1+a}{Y_2+a} \right| Y_1, Y_2 \right) \right) \stackrel{\text{Lemma 1}}{=} \mathsf{E}\left( \frac{\mu_1}{\mu_1 + \mu_2 \frac{Y_1+a}{Y_2+a} } \right)$$ The latter expectation can be evaluated using $\frac{1}{x} = \int_0^\infty \mathrm{e}^{-x t} \mathrm{d} t$: $$\mathsf{E}\left( \frac{\mu_1}{\mu_1 + \mu_2 \frac{Y_1+a}{Y_2+a} } \right) = \mathsf{E}\left( \frac{\mu_1 (Y_2+a)}{\mu_1 (Y_2+a) + \mu_2 (Y_1+a)} \right) = \int_0^\infty \mathsf{E} \left( \mu_1 (Y_2+a) \mathrm{e}^{- t (\mu_1 (Y_2+a) + \mu_2 (Y_1+a))} \right)\mathrm{d} t$$ The expectation under the integral sign is easy to evaluate using: $$\mathsf{E} \left( \mu_1 (Y_2+a) \mathrm{e}^{- t (\mu_1 (Y_2+a) + \mu_2 (Y_1+a))} \right) = -\left. \frac{\mathrm{d}}{\mathrm{d} s} \mathsf{E} \left( \mathrm{e}^{- s \mu_1 (Y_2+a) - t \mu_2 (Y_1+a))} \right) \right|_{s=t} \stackrel{\text{indep.}}{=} -\left. \frac{\mathrm{d}}{\mathrm{d} s} \mathsf{E} \left( \mathrm{e}^{- s \mu_1 (Y_2+a)} \right) \mathsf{E} \left( \mathrm{e}^{- t \mu_2 (Y_1+a))} \right) \right|_{s=t}$$ Now using Lemma 2, $$\mathsf{E} \left( \mu_1 (Y_2+a) \mathrm{e}^{- t (\mu_1 (Y_2+a) + \mu_2 (Y_1+a))} \right) = -\left. \frac{\mathrm{d}}{\mathrm{d} s} \left(\mathrm{e}^{- s \mu_1 a - t \mu_2 a} \frac{1}{1 + s \mu_1 \lambda_2} \frac{1}{1 + t \mu_2 \lambda_1} \right) \right|_{s=t} = \mathrm{e}^{-a t (\mu_1 + \mu_2)} \frac{\mu_1 ( a +\lambda_2 + a t \lambda_2 \mu_1)}{ (1 + t \lambda_1 \mu_2 )( 1 + t \lambda_2 \mu_1)^2}$$ It now remains to evaluate the integral with respect to $t$: $$\mathsf{Pr}\left( \frac{X_1}{X_2} > \frac{Y_1+a}{Y_2+a} \right) = \int_0^\infty \mathrm{e}^{-a t (\mu_1 + \mu_2)} \frac{\mu_1 ( a +\lambda_2 + a t \lambda_2 \mu_1)}{ (1 + t \lambda_1 \mu_2 )( 1 + t \lambda_2 \mu_1)^2} \mathrm{d} t$$ The result involves exponential integrals: $$\mathsf{Pr}\left( \frac{X_1}{X_2} > \frac{Y_1+a}{Y_2+a} \right) = \frac{\mu_2 \left(\lambda_1 \lambda_2 \mu_1 + a \left(\lambda_2 \mu_1 - \lambda_1 \mu_2\right)\right)}{(\lambda_1 \mu_2 - \lambda_2 \mu_1)^2} f \left( \frac{a\left(\mu_1+\mu_2\right)}{\mu_1 \lambda_2} \right) - \frac{\mu_1 \left(\lambda_1 \lambda_2 \mu_2 + a \left(\lambda_1 \mu_2- \lambda_2 \mu_1\right)\right)}{(\lambda_1 \mu_2 - \lambda_2 \mu_1)^2} f \left( \frac{a\left(\mu_1+\mu_2\right)}{\mu_2 \lambda_1} \right) + \frac{\lambda_2 \mu_1}{\lambda_2 \mu_1 - \mu_2 \lambda_1}$$ where $f(x) = \exp(x) \operatorname{Ei}(-x)$. The result can be obtained in a symbolic computer algebra systems, like Mathematica. The result can be obtained by using the partial fraction decomposition for the rational factor of the integrand. Here is the result, with verification by Monte-Carlo and by quadratures:

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Many thanks for your precious help! Sasha! – Tran Tam Aug 23 '12 at 15:59