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I have this problem Let $X$ and $Y$ be random variables that have a joint density function given by,

Find the density function of $X - Y$.

My problem is, that I don´t how to choose the integrals intervals

$$\int_{0}^{\infty}\int_{y}^{y+z}8(y+z)y\,dy\,dz$$

Is this statement correct?

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  • $\begingroup$ No. Your inner integral has both $y$ and $z$ mentioned in its bounds, so that is clearly erroneous. $\endgroup$ – Graham Kemp Jun 30 at 6:44
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First do a drawing. Then realize that

$$F_Z(z)=\int_{0}^{1+z} 8x dx \int_{x-z}^{1}y dy$$

This because:

  1. $z \in [-1;0]$

  2. Using the CDF method you get

$$F_Z(z)=\mathbb{P}[X-Y \leq z]=\mathbb{P}[Y \geq X-z]$$

Then the area to be integrated is the purple one, as shown in the following picture

enter image description here

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  • $\begingroup$ Thanks, for your help. Could you tell me why doesn´t FZ(z) have an expression like this one fX,Y(ξ,ξ−a)dξ $\endgroup$ – rosita galixx Jul 2 at 4:19
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Well, firstly note that the support $0\leq X\lt Y\leq 1$, means $-1\leq X{-}Y\lt 0,$ and $-(X{-}Y)\lt Y\leq 1$.

Then we just use the Jacobian transformation, to find the joint pdf, and integrate this to find the required marginal.

$$\begin{align}f_{\small X-Y,Y}(z,y)&=\begin{Vmatrix}1&1\\1&0\end{Vmatrix}~f_{\small X,Y}(z+y,y)\\[1ex]&=8(z+y)y~\mathbf 1_{\small 0\leq z+y\lt y\leq 1}\\[1ex]&=8(z+y)y~\mathbf 1_{\small 0\lt -z\leq y\leq 1}\\[4ex]f_{\small X-Y}(z)&={8\int_\Bbb R (z+y)y~\mathbf 1_{\small 0\lt -z\leq y\leq 1}~\mathrm d y}\\[1ex]&={8~\mathbf 1_{\small -1\leq z\lt 0}\cdot\int_{-z}^1 (z+y)y~\mathrm d y}\\[1ex]&~~\vdots\end{align}$$

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  • $\begingroup$ thanks, but I have doubts already, I just want to understand better. Sorry It´s a little bit confusing for me :( $\endgroup$ – rosita galixx Jul 2 at 5:41
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It looks like you are integrating over a triangle given by points $(x, y)=(0, 0)$, $(0, 1)$, and $(1, 1)$. Your solution should cover the whole region to be correct.

Then consider that the probability that $X-Y$ equals a given quantity $a$ as lines intersecting the triangle with slope 1 and $x$-intercept of $a$.

Then the integrated probability that $X$ and $Y$ lie on that line would be the following integral

$$ P(X-Y=a)=\int_{a}^{1} f_{X,Y}(\xi, \xi-a)d\xi $$

Here $\xi$ is a variable of integration which represents the $x$ coordinate on the prescribed line. With $a$ taking values from 0 to 1 the whole region is covered.

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  • $\begingroup$ Thanks for your help, could u show me how can I express fX,Y(ξ,a−ξ)dξ, this part, this is to know if what I am thinking is correct $\endgroup$ – rosita galixx Jun 30 at 4:40
  • $\begingroup$ Because the integral is only placed on the region where the function is nonzero then $f_{X,Y}(\xi, a-\xi)=8\xi(a-\xi)$ $\endgroup$ – eschavez Jun 30 at 5:03
  • $\begingroup$ I'm sorry, i misread your original question, I need to edit my response $\endgroup$ – eschavez Jun 30 at 5:09

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