probability density function is given by $f(x_1,x_2)=(-1.5)(1-x_2)$ if $0 \leq x_1 \leq x_2 \leq 2$.
I need to find $P(x_1\leq 0.75, x_2 \geq 0.5)$. I'm not sure how to bound the double integral.
Is it maybe $\int_{0.5}^2 \int_0^{0.75}$ ?
I'm unsure mostly because $x_1 \leq x_2$, but the integral bounds above leave room for overlap.
Best Answer
The question is ambiguous and pdf is valid if you choose different $k$ for $y \leq 1$ and $y \geq 1$.
Probability density function $f(x, y) = k (1-y), 0 \leq x \leq y \leq 2$
i) For $0 \leq y \leq 1$
$\displaystyle \int_0^1 \int_0^y k (1-y) \ dx \ dy = \frac{k}{6}$
ii) For $1 \leq y \leq 2$
$\displaystyle \int_1^2 \int_0^y k(1-y) \ dx \ dy = -\frac{5k}{6}$
That leads to two different pdf's for $0 \leq y \leq 1$ (with $k = 1$) and for $1 \leq y \leq 2$ (with $k = -1$). It could have been simply defined as $|1-y|$.
To find probability for the shaded region, split the integral as,
$\displaystyle \int_{0.5}^{0.75} \int_0^y (1-y) \ dx \ dy + \int_{0.75}^{1} \int_0^{0.75} (1-y) \ dx \ dy + \int_{1}^{2} \int_0^{0.75} (y-1) \ dx \ dy$