I'm trying to get $P(0.9
Sum of random variables uniformly distributed (0,1) and (0,2)
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0How is $Y$ related to $x_1$ and $x_2$? It is not clear from your description. – 2012-11-10
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0sum of two random variables is y. – 2012-11-10
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0that's correct, and also added to the above. y=x1+x2 – 2012-11-10
2 Answers
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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0thank 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. – 2012-11-11
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0@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](http://meta.math.stackexchange.com/questions/3286) and [this one](http://meta.math.stackexchange.com/questions/3399)). That gives you $2$ reputation points, exactly what you're missing to be able to upvote :-) – 2012-11-11
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0done and done. that works! there's so much for a newcomer to learn about the stack-exchange rules. – 2012-11-11
Here is a blog post describing something similar to the problem you are attempting to solve. Convolution is the way to go.