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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_{\{00}\times\mathbb{R}_{>1}}f(x,z)\frac{1}{x}\frac{x}{z^2}\exp(-x)d(x,y). \end{align*}

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.

1 Answers 1

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I don't see how you got from ∫χ{0 < y < x}⋅χ{x ≤α y}d(x,y) to ∫dydx⋅χ{0<α<1} where first integral is over R and the second is from 0 to αx. As y goes from 0 to αx how is the constraint x≤αy maintained?

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    Ah I see then \int_{0}^{\infty}...\int_{\{0 Then I differentiate and get $\frac{1}{\alpha^2}\chi_{\alpha \geq 1}$, the above density. (The $\chi_{\alpha \geq 1}$ comes in because $\frac{x}{\alpha} \leq x$) Thanks a lot.2012-06-07