Compute $P(8- \frac12X_1+X_2<0)$ for $(X_1,X_2)$ i.i.d. uniform on $[-8;8]$

Question

Random variables $X_{1}$ and $X_{2}$ are stochastically independent, uniformly distributed in a range $[-8;8]$. Let $g(x_{1},x_{2}) = 8- \frac{1}{2}x_{1}+x_{2}$. What is the probability that $g(X_1,X_2)<0$ ?

My solution looks like this. Firstly I limit $x_{2}$:

$x_{2}< \frac{1}{2}x_{1}-8$

Main equation that I use to calculate the probability:

$P = \int_{x_{1}=- \infty }^{x_{1}= \infty} \int_{x_{2}=- \infty }^{x_{2}= \frac{1}{2}x_{1}-8 } f_{x_{2}}(x_{2})dx_{2} f_{x_{1}}(x_{1})dx_{1}$

Where

$\int_{x_{2}=- \infty }^{x_{2}= \frac{1}{2}x_{1}-8 } f_{x_{2}}(x_{2})dx_{2}=F_{x_{2}}( \frac{1}{2}x_{1}-8 )= \frac{\frac{1}{2}x_{1}-8-(-8)}{8-(-8)}= \frac{1}{32}x_{1}$

So

$P = \int_{x_{1}=- \infty }^{x_{1}= \infty}\frac{1}{32}x_{1}f_{x_{1}}(x_{1})dx_{1} =\frac{1}{32} E(X_{1})=\frac{1}{32} * \frac{-8+8}{2}=0$

I know that the result should be $\frac{1}{16}$. I would be thankful for pointing out where I do the mistake.

Answer 1

The lower limit and the upper limit of $x_2$ are $-8$ and $\frac12x_1-8$ respectively. That means that $\frac12x_1-8\geq -8\Rightarrow \frac12x_1 \geq 0\Rightarrow x_1\geq 0$ Therefore the calculation is

$$P(8- \frac{1}{2}x_{1}+x_{2}<0)=\int_{0}^8 \frac{1}{16} \frac{x_1}{32} \, dx_1 $$

Answer 2

When you evaluate

$$F_{X_2} \left( \frac12 x_1 - 8 \right),$$

You have assumed that $-8 \leq \frac12x_1-8 \leq 8$, which is not true.

If $\frac12 x_1 - 8 \leq -8$, $F_{X_2} \left( \frac12 x_1 - 8 \right)=0$