# Sum of random variables uniformly distributed (0,1) and (0,2)

I'm trying to get $P(0.9<Y<=1.8)$ for the sum of 2 random and uniform values x1,x2 (so that y=x1+x2) where $x1$~$u(0,1)$ and $x2$~$(0,2)$ and I'm trying to do the convolution for it. Seems like $\int\limits_a^b\int_0^2 xf(x)\,\mathrm{d}x$ where a=0.9, b=1.8 and which seems like a logical way to start. I'm not comfortable with convolution but I'm trying to understand the step-by-step reason that this is the proper equation, and how the problem can be solved. I'd like to understand as much about this as possible so I'd like to also compare that problem to finding the $P(0.9<Y<=1.8)$ for just $x1$~$u(0,1)$ where the values summed are independent but still over the same (0,1) area, and also compare it to $P(0.9<Y<=1.8)$ for $exp(2)$ where lambda is 2, which I'm also not really understanding the summed distribution.

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How is $Y$ related to $x_1$ and $x_2$? It is not clear from your description. – Daryl Nov 10 '12 at 22:37
sum of two random variables is y. – Seyhmus Güngören Nov 10 '12 at 22:42
that's correct, and also added to the above. y=x1+x2 – stackuser Nov 10 '12 at 22:44

The answer to your question is $\frac{179}{400}$. First you need to make the convolution of two $p.d.f.$s. However note that this operation is valid only for the independent random variables $X_1$ and $X_2$. I assume they are independent and continue with the solution.

The result of the convolution will be on the positive $y$ axis, having non zero values between $0$ and $3$. The density of $Y$ is a linear increasing function from $0$ to $1/2$ for $y\in [0,1]$, a constant function $p_Y(y)=1/2$ when $y\in [1,2]$ and a linear decreasing function from $1/2$ to $0$ when $y\in [2,3]$.

What remains to do is to find the area under this $p.d.f.$. If you draw and calculate the area either with integrals or using some simple geometric relations, you will find that the area under the curve w.r.t. the square is $0.4$ and the area with respect to the left triange is $19/400$, which adds up to $\frac{179}{400}$.

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thank you!! so it sounds like the graph is a triangle from 0 to 2 peaking at 1/2. not sure what the limits of integration are here, sounds like a single integral though. if this can be shown in latex, it would be great for visualizing. sorry it tells me i can't upvote until i get to 15 but i would if i could. – stackuser Nov 11 '12 at 1:36
@stackuser: There's a very elegant solution to that in this case. If this answer answers your question, you can accept it by checking the little checkmark next to it (see this faq and this one). That gives you $2$ reputation points, exactly what you're missing to be able to upvote :-) – joriki Nov 11 '12 at 8:08
done and done. that works! there's so much for a newcomer to learn about the stack-exchange rules. – stackuser Nov 11 '12 at 8:27

Here is a blog post describing something similar to the problem you are attempting to solve. Convolution is the way to go.

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