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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.

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But why is that? –  lord12 Oct 19 '11 at 19:53
    
I retract my answer and point you towards: www-stat.stanford.edu/~susan/courses/s116/node114.html The way I wrote it should work, but there was something subtle I was missing. –  Mike Wierzbicki Oct 19 '11 at 20:22
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2 Answers

up vote 2 down vote accepted

No double integration is necessary (though of course the problem can be solved that way too).

  • Sketch the plane with coordinate axes and draw on it the square region (of unit area) over which the joint density is nonzero. Figure out the value of the joint density on this square region based on the given information.
  • Persuade yourself that if $B$ is any region in the plane, then $$ P\{(X, Y) \in B\} = \text{Area of}~ (B \cap \text{square region where pdf is nonzero}) $$
  • Draw the line $x + y = k$ (Hint: it will cross the axes at $(k, 0)$ and $(0,k)$.
  • Verify that for $2\theta \leq k \leq 2(\theta + 1)$, the line will divide the square into two regions, at least one of which will be a triangle
  • Verify that for $2\theta + 1 \leq k \leq 2\theta + 2$ the upper region is a triangle of area $\frac{1}{2}(2\theta + 2 - k)^2$
  • Persuade yourself that you want this triangle to have area $0.05$
  • Solve for $k$
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+1. I tried the double integration way and kept getting stuck on my limits of integration (Il-Bhima's solution is basically what I tried to do). Drawing a pic of the problem is often a useful trick to try -- it can really simplify the problem. –  Mike Wierzbicki Oct 19 '11 at 22:01
    
@MikeWierzbicki It is my opinion that drawing a sketch as an aid to understanding is always a good thing to do, though I say this with some trepidation in this math forum since some mathematicians prefer that all these things be done without any diagrams. Even in Il-Bhima's solution, if the density of $Z$ is first sketched, then finding $$P\{Z > k\} = \text{area under density to the right of} ~k\} = \text{area of triangle}$$ is very easy, hardly needing integration to calculate. –  Dilip Sarwate Oct 19 '11 at 22:10
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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)

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