Let's say Z=X+Y, where X and Y are independent uniform

William Curry 2021-12-14 Answered
Lets
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sonSnubsreose6v
Answered 2021-12-15 Author has 21 answers
If we want to use a convolution, let fX be the full density function of X, and let fY be the full density function of Y. Let Z=X+Y. Then
fz(z)=fX(x)fY(zx)dx.
Now let us apply this general formula to our particular case. We will have fZ(z)=0forz<0, and also for z2. Now we deal with the interval from 0 to 2. It is useful to break this down into two cases (i) 0<z1and(ii)1<z<2.
(i) The product fX(x)fY(zx)is 1 in some places, and 0 elsewhere. We want to make sure we avoid calling it 1 when it is 0. In order to have fY(zx)=1,weedzx0,ti^s,xz. So for (i), we will be integrating from x = 0 to x = z. And easily
0z1dx=z
Thus fZ(z)=zfor0<z1
(ii) Suppose that 1<z<2. In order to have fY(z−x) to be 1, we need zx1,ti^s,weedxz1. So for (ii) we integrate from z−1 to 1. And easily
1z11dx=2z
Thus fZ(z)=2zfor1<z<2.
Another way: (Sketch) We can go after the cdf FZ(z)ofZ, and then differentiate. So we need to find Pr(Zz).
For a few fixed z values, draw the lines with equation x+y=z on an x-y axis plot. Draw the square S with corners (0,0), (1,0), (1,1), and (0,1).
Then Pr(Zz) is the area of the part S that is "below" the line x+y=z. That area can be calculated using basic geometry. For example, when z is 2, the whole square area is under the line so Pr=1. There is a switch in basic shape at z=1.

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ol3i4c5s4hr
Answered 2021-12-16 Author has 48 answers
By the hint of jay-sun, consider this idea, if and only if fX(zy)=1when0zy1. So we get
z1yz
however, z[0,2], the range of y may not be in the range of [0,1] in order to get fX(zy)=1, and the value 1 is a good splitting point. Because z1[1,1].
Consider (i)ifz10then1z10ti^sz[0,1], we get the range of y[0,z]sincez[0,1]. And we get fX(zy)dy=0z1dy=zifz[0,1].
Consider (ii)ifz10ti^sz[1,2], so we get the range of y [z1,1],andfX(zy)dy=z111dy=2zifz[1,2]
To sum up, consider to clip the range in order to get fX(zy)=1.

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