I have two random variables, $X_1$ and $X_2$, both of which have exponential distributions, and $\lambda_1 = 1$, $\lambda_2 = 2$. The question is then, what is the probability $P(X_1 > 2X_2)$. So I have two distribution functions:

$$f(x_1) = e^{-x_1}$$

$$f(x_2) = 2e^{-2x_2}$$

Looking at the integral I get

$$\int \int_{x_1 > x_2}f_1(x_1)f_2(x_2) dx_1 dx_2$$

Combining this all I come to

$$\int_0^\infty \int_{2x_2}^\infty 2e^{-(x_1 + x_2)}$$

This is my concern. Is the lower limit on my second integral actually $2x_2$ or should it be $2x_1$. To me $2x_2$ makes more sense, and leads to a result of $\frac{1}{2}$. However having a lower limit of $2x_1$ leads to a result of $\frac{1}{5}$ and I have seen this elsewhere. Can anybody perhaps shed some light on my issue and help me out with my confusion?


It is

$$\int_0^\infty \int_{2x_2}^\infty 2e^{-(x_1+x_2)}\, dx_1 \, dx_2$$

Let $x_2$ takes value from $0$ to $\infty$. After which, fixing $x_2$, what are the values that $x_1$ can take to satisfy the domain of interest. We want $x_1 >2x_2$, hence the lower limit should be $2x_2$.

  • $\begingroup$ Thanks very much for clearing that up. $\endgroup$ – user25758 Jul 28 at 15:41

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