Let $P= \frac{1}{4}\lambda^2_{|(0,2)^2}$ be the bivariate uniform distribution on $(0,2)^2$, where $\lambda$ denotes the Lebesgue measure.
Calculate the probability $P(A)$ of $A= \{(x,y) \in (0,2)^2 : y ≤ exp(-x)\}$.
How do I even approach this problem? I want to understand every little step.
I know that $P(A) = \int \mathbb{1}_A dP$. Is it possible to express this in the form of $\int \int \cdots dx dy$? Do I have to show that the two dimensions are independent? Do I have to use Fubini's theorem?
Approach:
$\int \mathbb{1}_A d\left(\frac{1}{4}\lambda^2_{|(0,2)^2}\right) = \int \mathbb{1}_{(0,exp(-x)) \times (0,x)} (x,y) \ d\left(\frac{1}{4}\lambda^2_{|(0,2)^2}\right) = \int_0^1 \int_0^{exp(-x)} y \cdot x \ dy \ dx$
Am I allowed to write it like this?
You note correctly that $P(A) = E[1_{A}] = \int_{\mathbb{R^2}}1_{A}dP$. To evaluate this integral we first compute $dP = \frac{1}{2}\lambda d\lambda$ (check out Radon-Nikodym) and finally $$\int_{\mathbb{R^2}}1_{A}dP = \frac{1}{2}\int_0^2x\int_0^21_{y\leq e^{-x}}ydydx = \frac{1}{4}\int_0^2xe^{-2x}dx = \frac{1}{4}(\frac{1}{4}-\frac{5}{4}e^{-4})$$ To justify this you can use an "easier" version of Fubini's theorem, namely Tonelli's theorem, which holds for non-negative functions.