I have the following probability-density function: $$ f(x,y) = 2e^{-x-2y} $$ I need to calculate the following things:
- The marginal densities of $x$ and $y$
- The probability that $x>y$ i.e. $P(x>y)$
I tried the following:
- My textbooks says that the marginal probability density function of $x$ is the integral with bounds from $[-\infty, \infty]$ with respect to $dy$.
So I first integrate $2e^{-x-2y}$ with respect to $dy$ $$ 2e^{-x}\times (\frac{-1}{2}\times e^{-2y}) $$
Second I need to fill in the integration bounds, but if i plug in $-\infty$, the equation will also go to infinity.
What do I need to do to solve this problem?
- I don't know how to begin :(
If anybody could give me feedback, thanks in advance.
Ter
$P(X>Y)=\int_0^{\infty} \int_y^{\infty} f(x,y)dxdy$. We have $\int_y^{\infty} f(x,y)dx=(2e^{-2y})(e^{-y})$ so the answer is $\int_0^{\infty} 2e^{-3y}=\frac 2 3 $.