Question.
otherwise
Suppose Z=X+Y. Find the density function of Z.
Answer.
I am having difficulty understanding how to work out the support for Z. I guess I would try to make u and (z-u) non-negative, right?
For example, I know that
Then if X=u then Y=z-u and in this case
Where to from here?... split the z interval into ? (just looking at similar questions, they do split z into several intervals, but I haven't figured out how they decide which ones).
Is that really that simple? that was my first intuitive response. So basically I should be looking for intervals where z is non negative.
Do you not need to evaluate z-u and u separately? Sorry, I am asking because I saw that in similar questions.
For example, this chapter, example 7.3 on page 8
there are two intervals for z in the final density function, one based on interval for y, another based on interval for z-y
http://www.dartmouth.edu/~chance/tea...k/Chapter7.pdf
This is what I don't get, the split intervals in the final density function.
Furthermore, if I want to draw the area of integration for the sum function fz, can I use (x,y) coordinates? and will it help me to see the required area?
I am still not getting it (although closer). First of all, what the graph would look like in 3D? I have the 'triangle' area on XY-plane, bounded by (0,0), (0,1) and (2,0), according to the bounds of X and Y. Then I have the third dimension in the direction Z (z=x+y is a plane right?), which can take values from 0 to 3. But in between these values (of z, from 0 to 3), the function of z=x+y is not the same - ie the probability of z is not determined by one function on the interval - there will be three different density functions on the intervals from 0 to 1, 1 to 2 and 2 to 3.
on 0<z<1, I think I can take z=x+y as the 'top' curve, so the density function for 0<z<1 would be
Further on, for 1<z<2 and 2<z<3, I am still not sure how to proceed...
PS I realise this is now about competing the question, not just about intervals for z, but I guess I need help with the whole thing...
I let W=X and Z=X+Y. The Jacobian is then one.
So the joint density of W and Z is w(z-w).
The region to integrate isn't bad at all.
Draw the z as the "y-axis" and w as the "x-axis".
Then you have the region bounded between 0<w<1 and the two lines z=2+w and z=w.
In the case of 0<z<1, you integrate w with the bounds of 0 and z
In the case of 1<z<2, you integrate w with the bounds of 0 and 1
In the case of 2<z<3, you integrate w with the bounds of z-2 and 1.
Hopefully you obtain a valid density, I'm heading to the gym.
Thank you both Mr Fantastic and Matheagle for your patience! I promise I am not dumb, but I may be having a 'bad convolution' day
I earnestly wanted to try both methods - even though I think they are the same but explained in two different math 'dialects' - and the second one (change of variable language) proved to be easier for me. I did see the three different regions of integration in the W,Z coordinates and the rest was automatic.
0<z<1
which means
1<z<2
therefore density is
2<z<3
(or thereabouts...)
therefore density is
I hope I did get it right this time...
When I tried to do distribution functions method using definition of G(z)=P(Z<z), I did get lost again, and I'd love to get to the bottom of it - Mr Fantastic, if you don't mind, could you let me know if you referred to 3D coordinates (x,y,z) or 2D coordinates (z,x) in your sentence "simple sketch graph of the line y = -x + z over the region defined by the support f_XY"? Also, is it
or