# Sum of two independent uniform distribution

I am doing a game theory question where $$X$$ is uniform over $$[0,100]$$ and $$Y$$ is uniform over $$[10,110]$$, where $$X$$ and $$Y$$ are the respective valuations of two people . I want to find the density function for $$Z=X+Y$$ in order to calculate some probabilities.

I've seen many solutions where X and Y follow a uniform distribution over $$[0,1]$$ but I don't seem to be able to replicate it using this range.

I have drawn a diagram and calculated some probabilities, e.g. $$Pr(X+Y>60)=7/8$$ but I would like to check them and calculate harder probabilities using the convolution of the two densities. Any help would be greatly appreciated!

• Hint: on the $xy$ grid $[0,100] \times [10,110]$, draw the level curves of $y=z-x$ for different $z$ values. This will tell you the regions to integrate over. The joint density will be $f_X(x)f_Y(y)=f_{X,Y}(x,y)$ by independence. Mar 27, 2020 at 15:34
• Intuitively for me the regions are $[0,10]$, $[10,200]$ and $[200,210]$ but why is it a joint density, I thought it would be a convolution of the x and y function (which I'm not sure how to compute) Mar 27, 2020 at 21:30

Write $$X=100A,\,Y=10+100B,\,C=A+B$$ so $$Z=10+100C$$. Then $$C$$ has an $$n=2$$ Irwin–Hall distribution, which can be obtained as per @zugzug's comment. In terms of Iverson brackets$$f_C(c)=|1-c|[c\in[0,\,2]],\,f_Z(z)=\frac{1}{100}|1-(z-10)/100|[z\in[10,\,210]].$$