What is the pdf of $X_{1}+X_{2}$, given that $X_{1} $ and $ X_{2} $ are iid uniform $ (\theta,\theta+1)$?

Question

$ X_{1},X_{2}$ are iid uniform$(0,1)$. I need to find the probability of $ X_{1}+X_{2}>c $. The joint pdf of $X_{1}, X_{2} $ is uniform $(0,1)$ and the $X_{1}+X_{2}$ should also be uniform$(0,2)$ but the result is not consistent with the result of the mgf method. Also, help with the method of finding the limits of the probability of $X_{1}>C-X_{2}$.

Answer 1

To get the density for $X_1+X_2$ you use convolution. If $X_1$ has density $f_1$ and $X_2$ has density $f_2$, then the density for $X_1+X_2$ is $g$, where $$ g(x) = \int_{-\infty}^{+\infty} f_1(x-t) f_2(t)\;dt $$ Doing this computation in case $f_1 = f_2 = \mathbf1_{[0,1]}$, I leave to you. You will need to split your intgral into cases: where $f_1(x-t) f_2(t)=1$, that is $0\le t \le 1$ and $0 \le x-t\le 1$; and the rest where $f_1(x-t) f_2(t)=0$.

The result is not uniform on $[0,2]$. We may suspect this from other cases in probability. Rolling two dice, each with uniform distibution on the finite set $\{1,2,3,4,5,6\}$ does not give us uniform distribution for the sum on $\{2,3,\cdots, 12\}$. We know that $7$ is most likely, and $2,12$ are least likely.