Thu 24 March 2016

Probability that one RV is greater than another

Written by Hongjinn Park in Articles

Here I'm interested in the $P(X>Y)$, where X and Y are independent with the same distribution but different parameters.

Normal RVs

Let $X$ and $Y$ be independent RVs with $X \sim Normal(\mu_X, \sigma_X^2)$ and $Y \sim Normal(\mu_Y, \sigma_Y^2)$ respectively. Then,

$$P(Y< X) = P(Y-X < 0)$$

and note that $Y-X \sim Normal(\mu_Y - \mu_X, \sigma_X^2 + \sigma_Y^2)$, therefore

$$P(Y< X) = P(Y-X < 0) = P \left( \frac{Y-X - (\mu_Y - \mu_X)}{\sqrt{\sigma_X^2 + \sigma_Y^2}} < \frac{0- (\mu_Y - \mu_X)}{\sqrt{\sigma_X^2 + \sigma_Y^2}} \right) = \Phi \left( \frac{\mu_X - \mu_Y}{\sqrt{\sigma_X^2 + \sigma_Y^2}} \right)$$

Exponential RVs

Let $X$ and $Y$ be independent Exponential RVs with rate parameters $\lambda_1$ and $\lambda_2$ respectively. Then,

\begin{align*} P(Y < X) & = \int_{0}^\infty \int_{0}^x\lambda_1 e^{-\lambda_1x} \lambda_2 e^{-\lambda_2y} \, dy \, dx \\ & = \lambda_1 \lambda_2 \int_{0}^\infty e^{-\lambda_1x} \int_{0}^x e^{-\lambda_2y} \, dy \, dx \\ & = \lambda_1 \lambda_2 \int_{0}^\infty e^{-\lambda_1x} \left[ \frac{-1}{\lambda_2} e^{-\lambda_2y} \right]_{y=0}^{y=x} \, dx \\ & = \lambda_1 \lambda_2 \int_{0}^\infty e^{-\lambda_1x} \frac{-1}{\lambda_2} \left[e^{-\lambda_2y} \right]_{y=0}^{y=x} \, dx \\ & = -\lambda_1 \int_{0}^\infty e^{-\lambda_1x} \left( e^{-\lambda_2x} - 1 \right) dx \\ & = -\lambda_1 \int_{0}^\infty e^{-\lambda_1x} e^{-\lambda_2x} - e^{-\lambda_1x} dx \\ & = -\lambda_1 \int_{0}^\infty e^{-(\lambda_1 + \lambda_2)x} - e^{-\lambda_1x} dx \\ & = -\lambda_1 \left( \left[ \frac{-1}{\lambda_1+\lambda_2}e^{-(\lambda_1 + \lambda_2)x} \right]_{0}^\infty - \left[ \frac{-1}{\lambda_1}e^{-\lambda_1x} \right]_{0}^\infty \right) \\ & = -\lambda_1 \left(\frac{1}{\lambda_1+\lambda_2} - \frac{1}{\lambda_1}\right) \\ & = -\frac{\lambda_1 }{\lambda_1+\lambda_2} + 1 \\ & = -\frac{\lambda_1 }{\lambda_1+\lambda_2} + \frac{\lambda_1+\lambda_2}{\lambda_1+\lambda_2} \\ & = \frac{\lambda_2 }{\lambda_1+\lambda_2}\\ \end{align*}


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