I am struggling finding the right bounds on integration when I try solve the following:
$X_1 \sim U[0,1]$; $ X_2\sim U[1,2]$.
$Y_1 = X_1 X_2$
$Y_2 = X_1$
I want to find $f_{Y_1}$.
I have gotten to the point where I have:
$x_1 = y_2$.
$x_2=\frac{y_1}{y_2}$.
$f_{Y_1,Y_2} (y_1,y_2)= \frac{1}{y_2}$.
$x_1 \geq 0: y_2\geq0$
$x_1\leq1: y_2\leq1$
$x_2\geq1: y_2\leq y_1$
$x_2\leq2: y_2\geq \frac{y_1}{2}$
Putting this together I have for:
$0\leq y_1 \leq 1$ -> $0 \leq \frac{y_1}{2} \leq {y_2} \leq y_1$
$1\leq y_1 \leq2$ -> $ \frac{y_1}{2} \leq {y_2} \leq 1$
Which to me suggests that
$f_{Y_1}(y_1) = \int_{\frac{y_1}{2}}^{y_1} \frac{1}{y_2} dy_2 + \int_{\frac{y_1}{2}}^1\frac{1}{y_2} dy_2 = ln(y_1)-2ln(\frac{y_1}{2}) = ln(4y_1)$
However, $\int_0^2 f_{Y_1}(y_1)dy_1$ does not integrate to 1 which indicates that I have messed something up. I would appreciate any pointer to where I have gone wrong.
I found my answer to $f_{Y_1}$ but I had to go another route.
I instead used:
$F_{Y_1}(y_1) = \int_{x_2=1}^1Pr(X_1<\frac{y_1}{x_2})f_{X_2}dx_2$.
For $0\leq y_1\leq1$: $y_1\int_1^2\frac{1}{t}dt = y_1ln(2)$.
For $1\leq y_1\leq2$: $\int_1^{y_1}dt + y_1\int_{y_1}^2\frac{1}{t}dt = (y_1-1)+y_1ln(\frac{2}{y})$.
So
$f_{Y_1}(y_1)=\begin{cases} ln(2) \text{ for }y_1\in[0,1] \\ ln(\frac{2}{y_1}) \text{ for } y_1\in[1,2] \end{cases}$