0
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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 $2x_2$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.