Let $X, Y$ ~ Uniform(0,1) be independent. Find the PDF for $X - Y$ and $X/Y$ using transformations

Question

I know that their joint PDF is also 1 given $0 < x < 1, 0 < y < 1$ but I'm just really confused how we can find the support of $X-Y$, if I'm not mistaken there should be 2 cases, one where $X < Y$ and this giving us $-1 < x-y < 0$ and the other one giving when $X>Y$ giving us a range of $0 <= x-y < 1 $

I'm not really sure how to move forward or visualize their support.

Any help would be appreciated

Possible hint here: https://math.stackexchange.com/questions/1673598/probability-distribution-of-the-difference-of-two-uniform-variables — Đào Minh Dũng, Oct 04 '20 at 14:50
Let Z=-Y and get the distribution for X+Z using convolution. — herb steinberg, Oct 04 '20 at 16:38
Is there a way to do this by transformations? I'm unfamiliar with the 'convolution' term — Wallace, Oct 04 '20 at 17:17

score 0 · Answer 1 · answered Oct 05 '20 at 00:34

Let $W=X+Z$, then the density function $P_W(w)=\int P_Z(w-x)P_X(x)dx$ where $P_Z(z)=1$ for $-1\le z\le 0$ and $P_X(x)=1$ for $0\le x \le 1$. Both $=0$ otherwise. The integral then is $\int_{max(0,w)}^{min(1,1+w)}dx$ which leads to $P_W(w)=1+w$ for $-1\le w\le 0$ and $P_W(w)=1-w$ for $0\le w \le 1$.

Let $X, Y$ ~ Uniform(0,1) be independent. Find the PDF for $X - Y$ and $X/Y$ using transformations

1 Answers1