Question

4. The probability density function for the random variables X1 and X2 is given by
f(x,y) = 1/2 ; x1+x2 <= 2, x1>= 0, x2>=0
= 0 ; otherwise
a) Find f(x1 | x2) . Are X1 and X2 independent?
b) Calculate the probability that X1>2X2

Answer 1

a)

Recall that \[ \large f(x_1|x_2)=\frac{f(x_1,x_2)}{f_{X_2}(x_2)}\]

And \[\large f_{X_2}(x_2)=\int_{-\infty}^{\infty}f(x_1,x_2)\,dx_1\]

Sine you're integrating w.ith respect to \(x_1\), the bounds of integration follow from noticing that \(x_1+x_2\le 2 \implies x_1 \le 2-x_2\). Now combined with \(x_1\ge 0\) you get:
\(0\le x_1 \le 2-x_2 \)
Here:
\[\large \begin{align} f_{X_2}(x_2)&=\int_{0}^{2-x_2} \frac{1}{2} \, dx_1\\
&= \left. \frac{1}{2}x_1\right|_{0}^{2-x_2} \\
\, \\

&=1-\frac{x_2}{2} ,0\le x_2\le 2\end{align} \]

So, Now you just divide the joint pdf by the marginal pdf to find \(f(x_1|x_2)\).

For independence:
We know 2 random variables will be independent if you can write the joint PDF as a product of the marginal densities, that is \(f(x_1,x_2)=f_{X_1}(x_1)f_{X_2}(x_2)\), and that the joint support can be written as a Cartesian product, that is the product of the support of the marginal densities will be rectangular).

Given how you find \(f_{X_2}(x_2)\) above, it is clear the the product of the marginal densities \(\ne 1/2\), the joint PDF. You can also notice that the supports do not form a Cartesian product (the supports are dependent).
------------------------
b) When asking for a probability like this, you note that in general:
\[ \large P\left( (X_1,X_2)\in D\right)=\iint\limits_{(x_1,x_2)\in D\cap A}f(x_1,x_2)\,dx_1\,dx_2\]
where \(A\) is the support for the joint PDF of \(X_1, X_2\), that is \(A=\{(x_1,x_2)|f(x_1,x_2)>0 \}\) and \(D\) is usually just the region imposed by some given inequality.

So \[\large P(X_1>2X_2)=\iint\limits_{x_1>2x_2\cap A}\frac{1}{2}\,dx_1\,dx_2\]
Drawing a figure always helps to know the region of integration. The region of integration will be imposed by the bounds of the support of the joint PDF, and by the given inequality, so the region must satisfy these bounds:
\( x_1+x_2\le 2\implies x_2\le 2-x_1\\
x_1\ge 0\\
x_2 \ge 0\\
x_1 > 2x_2 \implies x_2<x_1/2\)

Now you can integrate this w.r.t x1 or x2 first, but clearly it will be easier to do it w.r.t x1 first since you won't have to decompose the double integral into a sum of 2 double integrals. Notice that if you solve for the intersection point, it occurs at the point \((4/3, \, 2/3)\)
\[\Large \begin{align}P(X_1>2X_2)&=\int_{x_2=0}^{x_2=2/3}\int_{x_1=2x_2}^{x_1=2-x_2} \frac{1}{2}\,dx_1\,dx_2 \\
&=...\\
\, \\

&=\frac{1}{3} \end{align} \]