Uniform distribution of two random variables

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Homework Statement:

given X an Y uniformly distributed on [0,1] independent find the density of S=X+Y

Relevant Equations:

Fs(Z)=P[X+Y<=Z]
i did not get how the professor came to such result. In particular:

in order to evaluate
P[x+y<=z] solved a double integral of the joint density. What i am not getting is did i choose the extreme of integration in order to get as result ##\frac {z^2} {2}##
 

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  • #2
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  • #3
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in order to evaluate
##P\left[x+y<=z\right] ##solved a double integral of the joint density.

X is gong from 0 to 1 and same holds for Y.
integrating x from 0 to z and y from 0 to z-x i get

##\frac {z^2} {2}##

i think the reasoning should be this.
 
  • #4
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he divided the problem in two parts the above is for z between 0 and 1. the other is for z between 1 and 2, for which an other integral has been computed
i do not understand how this 2 came out
 
  • #5
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he divided the problem in two parts the above is for z between 0 and 1. the other is for z between 1 and 2, for which an other integral has been computed
i do not understand how this 2 came out
If both x and y are numbers in the interval [0, 1], then z = x + y will be in the interval [0, 2]. Is that what you're asking? The double interval arises by working with the region in the plane [0, 1] X [0, 1].
 
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  • #7
etotheipi
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You might think about it by considering the joint distribution ##f_{X,Y}(x,y) = f_X(x)f_Y(y)##. You want $$P(Z \leq z) = P(X + Y \leq z) = P(Y \leq z - X)$$ You can try sketching this on the plane ##[0,1] \times [0,1]##, and calculate this probability using a double integral or much more simply by considering areas (be careful about different cases!). Then convert cdf ##\rightarrow## pdf. I have the feeling that this is what your professor was driving at.

Alternatively, you can use the idea of a convolution $$f_Z(z) = \int_{-\infty}^{\infty} f_X(z-y) f_Y(y) dy$$but I think this approach has more room for error.
 
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haruspex
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in order to evaluate
##P\left[x+y<=z\right] ##solved a double integral of the joint density.

X is gong from 0 to 1 and same holds for Y.
integrating x from 0 to z and y from 0 to z-x i get

##\frac {z^2} {2}##

i think the reasoning should be this.
That is outlining your work, not showing it. There is not enough here to say for sure what you are doing wrong. We need to see the actual integral you wrote down.
At a guess, you wrote ##\int_{x=0}^z\int_{y=0}^{z-x}1.dxdy##.
One problem with that is, e.g., for z>1 and x<z-1 your y integral range goes above 1.
A more fundamental problem is that it will give you the CDF, not the PDF.
 
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