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**lllll** Assume $\displaystyle X_1$ and $\displaystyle X_2$ are independent random variables, both following a exponential distribution. Calculate $\displaystyle P\{X_1<X_2\}$

The way it's done in the book

$\displaystyle P\{X_1 < X_2\} = \int_0^\infty P\{X_1 < X_2|X_1 = x\} {\color{blue}\lambda_1 e^{-\lambda_1x}} \ dx$ (a)

$\displaystyle = \int_0^\infty P\{ x< X_2\} {\color{blue}\lambda_1 e^{-\lambda_1x}} \ dx$ (b)

$\displaystyle = \int_0^\infty e^{-\lambda_2x}{\color{blue}\lambda_1 e^{-\lambda_1x}} \ dx$ (c)

I'm wondering is why are they putting the pdf of exponential function (the part in blue)?

in part (b) would is it $\displaystyle \int_0^\infty \frac{P\{X_1 < X_2, X=x\}}{P\{X_1=x\}} {\color{blue}\lambda_1 e^{-\lambda_1x}} \ dx$ $\displaystyle \Longrightarrow \int_0^\infty \frac{P\{X_1 < X_2\} P\{X=x\}}{P\{X_1=x\}} {\color{blue}\lambda_1 e^{-\lambda_1x}} \ dx$ $\displaystyle \Longrightarrow \int_0^\infty P\{ x< X_2\} {\color{blue}\lambda_1 e^{-\lambda_1x}} \ dx$

in part (c) isn't it $\displaystyle P\{ x< X_2\} = \lambda_2 e^{-\lambda_2x}$, therefore wouldn't it be: $\displaystyle \int_0^\infty {\color [cmyk] {0.75,0,0.75,0.45}\lambda_2}e^{-\lambda_2x}{\color{blue}\lambda_1 e^{-\lambda_1x}} \ dx$

why isn't there a $\displaystyle {\color [cmyk] {0.75,0,0.75,0.45}\lambda_2}$?