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Consider two random variables with the following joint PDF: $$ f_{X,Y}(x,y) = \begin{cases} 2, & x > 0, y > 0, x + y < 1 \\ 0, & \text{otherwise} \end{cases} $$

I need to find the PDF of $U = Y - X$.

I'm having trouble determining the range of the distribution of $U$.

I know $U = Y - X$, so $Y = U + X$, and the partial derivative of $Y$ with respect to $U$ is $1$. Therefore, I know that the joint distribution of $U$ and $X$ is $2|1| = 2$. I know that $0 < x < 1$, because that is given in the problem. However, I'm struggling to determine the range for $U$. Once I have the range for $U$, I know how to find the distribution of $U$.

Björn Friedrich
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mmm
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1 Answers1

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Okies, $U=Y-X \iff Y= X+U$ so we require the convolution: $$f_{Y-X}(u) ~=~ \int_\Bbb R f_{X,Y}(x,x+u)\operatorname d x$$

Now the $X,Y$ support is: $$(X,Y)\in\{(x,x+u):x>0, x+u>0, 2x+u<1\}$$

Rearranged this becomes: $$(X,Y)\in\Big\{(x,x+u): \big(-1{<}u{<}0, -u{<}x{<}\tfrac{(1-u)}2\big)\vee\big(0{\leqslant}u{<}1, 0{<}x{<}\tfrac{(1-u)}2\big)\Big\}$$

Therefore the convolution is:

$$f_{Y-X}(u)~ =~ \int_{-u}^{(1-u)/2} 2\cdot\mathbf 1_{u\in(-1;0)}\operatorname d x+ \int_{0}^{(1-u)/2} 2\cdot\mathbf 1_{u\in[0;1)}\operatorname d x $$

Graham Kemp
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