If X and Y are iid from uniform(theta,theta + 1), and the P(X+Y) > k = 0.05, how do you find K?
Attempt: I know it's a double integration for both X and Y but I am not sure why.
If X and Y are iid from uniform(theta,theta + 1), and the P(X+Y) > k = 0.05, how do you find K?
Attempt: I know it's a double integration for both X and Y but I am not sure why.
No double integration is necessary (though of course the problem can be solved that way too).
Note that we can write the density function for $Z = X+Y$ as $f_Z(z) = \int_{-\infty}^\infty \mathbf{1}(z-2\theta - x)\mathbf{1}(x)\,dx = \int_0^1 \mathbf{1}(z-2\theta-x)\,dx$ where $\mathbf{1}$ is the indicator function for $[0,1]$.
Now integrand is non-zero iff $0 < z-2\theta -x < 1$, i.e. $z-2\theta -1 < x < z-2\theta$.
If $2\theta < z < 2\theta + 1$ then $f_Z(z) = \int_{0}^{z - 2\theta} \,dx = z-2\theta $ (we don't actually care about this case).
If $2\theta+ 1 < z < 2\theta + 2$ then $f_Z(z) = \int_{z-2\theta-1}^{1} \,dx = 2 - z+ 2\theta $
Therefore if $2\theta+1 < k < 2\theta+2$ then $P(Z > k) = \int_{k}^{2\theta +2} 2 - z+ 2\theta\,dz = \frac{1}{2}(2-k+2\theta)^2$
From this you can solve $P(Z>k)=0.05$ for $k$.
(Hopefully no mistakes)