Density Function for Sums of Random Variables

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SUMMARY

The discussion focuses on deriving the probability density function for the sums of random variables Z = X + Y and V = Y - X, given the joint density function f(x,y) = 2 for 0 < x < y < 1. The solution involves integrating f(x, z-x) to find f_Z(z) and determining the support for Z, which is 0 < z < 2. The analysis reveals two cases for the intervals based on the values of z, specifically for 0 < z < 1 and 1 < z < 2, highlighting the significance of bounds in the integration process.

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Homework Statement



Given the joint density, f(x,y), derive the probability density function for Z = X + Y and V = Y - X.

Homework Equations



f(x,y) = 2 for 0 < x < y < 1
f(x,y) = 0 otherwise.

The Attempt at a Solution



For Z = X + Y, I can derive the fact that,

f_Z(z) = \int_{-\infty}^{\infty} f(x,z-x)dx

The support should be 0 < x < z - x < 1? But I am kind of lost from here.

0 < x < 1 and 0 < y < 1, so 0 < z < 2? The book I am using tell me there are two cases but I have no idea how they deduced the two cases. From my very limited understanding, f(x,y) = 2 for all x,y in its support. So why are there two cases?

Thanks!
 
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try thinking about it geometrically, f_{X,Y}(x,y) is uniform on the triangle within the unit square and below y=x

now think about lines of constant z = x+y, and compare with your intgeral
 
i came up with an answer on my own, is the process correct? i looked at the two arguments of f(x, z-x) and try to deduce which bounds are significant.

1) argument "x"

0 < x < z - x.

so x/2 < x < z/2. the z/2 is significant across all z in (0,2) since for z < 2, z/2 < 1. x < z/2 < 1, so the value of z/2 takes precedent over the numerical upper bound of 1.

the lower bound of 0 is significant

2) argument "z - x"

x < z - x < 1

so z - 1 < x < z - x.

the upper bound of z - x is not significant as x < z/2 is equivalent to x < z - x.

0 < z < 2 => -1 < z - 1 < 1, so the lower bound z - 1 < x is significant only for some z.

for 0 < z < 1, -1 < z -1 < 0 is totally not relevant compared to 0 as z-1 < 0 < x anyway. but for 1 < z < 2 we have 0 < z - 1 < 1 which is relevant as 0 < z -1 < x, so z - 1 is the greatest lower bound.

3) Intervals

Hence the intervals are 0 < z < 1 where 0 < x < z/2, and for 1 < z < 2 where z - 1 < x < z/2.

4 ) Algorithm

Basically I followed this self made algo which i hope is correct!

1. figure out the bounds for the two arguments of f(x, z - x)
2. compare it to the bounds for x, e.g. a < x < b
3. find the least upper bound and greatest lower bound. partition z if need be.


Thanks!
 

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