In the following exercise I got two different distributions for $Z.$ I want to know where my mistake is. Every hint or comment is appreciated.
The exercise goes as follows:
Let $(X,Y)$ be a random vector with values in $\mathbb{R}^2$ such that it has a joint density given by: $f(x,y)=\frac{1}{x}\exp(-x)\chi_{\{0
where $\chi_{\{0 is the indicator fct. on $\{0 Let $Z:=\frac{X}{Y}$. Compute the distribution of $(X,Z)$ and $Z$.
Now my computations:
First computation: \begin{align*} \mathbb{E}[f(X,Z)]&=\int_{\{0
Where I used the change of variables $\phi:\mathbb{R}_{>0}\times\mathbb{R}_{>1} \rightarrow \{(x,y) \in \mathbb{R}^2 : 0
Hence $d\mathbb{P}_{(X,Z)}=\frac{1}{z^2}\exp(-x)\cdot \chi_{\mathbb{R}_{>0}\times\mathbb{R}_{>1}}$.
Thus $\mathbb{P}(Z\leq \alpha)=\int_{1}^{\alpha}\frac{1}{z^2}\int_{0}^{\infty}\exp(-x)dxdy=\int_{1}^{\alpha}\frac{1}{z^2}dy$, leading to $d\mathbb{P}_{Z}=\frac{1}{z^2}\chi_{\mathbb{R}_{>1}}.$
Second computation: \begin{align*} \mathbb{P}(Z\leq \alpha)=\mathbb{P}(\frac{X}{Y}\leq \alpha)&=\int_{\{\frac{X}{Y}\leq \alpha\}}\frac{1}{x}\exp(-x)\cdot \chi_{\{0 Hence $Z$ is uniformly distributed on $(0,1)$. Now, where is my mistake? (I am always uneasy when doing change of variables so I fear there is my problem...). Thanks in advance.