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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: $$f(x,y)=\frac{1}{x}\exp(-x)\chi_{\{0<y<x\}}$$ where $\chi_{\{0<y<x\}}$ is the indicator fct. on $\{0<y<x\}.$

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<y<x\}}f(x,\frac{x}{y})\frac{1}{x}\exp(-x)d(x,y)\\ &=\int_{\mathbb{R}_{>0}\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 $\phi:\mathbb{R}_{>0}\times\mathbb{R}_{>1} \rightarrow \{(x,y) \in \mathbb{R}^2 : 0<y<x\}; (x,z) \mapsto (x,\frac{x}{z}).$

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<y<x\}}d(x,y)\\ &=\int_{\mathbb{R}^2}\frac{1}{x}\exp(-x)\cdot \chi_{\{0<y<x\}}\cdot \chi_{\{x\leq \alpha y\}}d(x,y)\\ &=\int_{0}^{\infty}\frac{1}{x}\exp(-x)\int_{0}^{\alpha x}dydx\cdot \chi_{\{0<\alpha<1\}}\\ &=\alpha\chi_{\{0<z<1\}} \end{align*}

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.

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1 Answer 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 $\int_{0}^{\infty}...\int_{\{0<y<x\}}\chi_{\{x\leq\alpha y\}}dydx=\int_{0}^{\infty}...\int_{0}^{x}\chi_{\{\frac{x}{\alpha} \leq y\}}dydx=\int_{0}^{\infty}...\int_{\frac{x}{\alpha}}^{x}dydx\cdot \chi_{\alpha \geq 1}=\int_{0}^{\infty}\exp{(-x)}(1-\frac{1}{\alpha})dx\chi_{\alpha \geq 1}=(1-\frac{1}{\alpha})\cdot \chi_{\alpha \geq 1}$ 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. –  AndreasS Jun 7 '12 at 19:58

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