# sum of random variables - help to evaluate support of function

• Jan 28th 2011, 01:31 AM
Volga
sum of random variables - help to evaluate support of function
Question.

$f_{X,Y}(x,y)= xy, 0\leq{x}<1, 0\leq{y}<2$
$0$ otherwise

Suppose Z=X+Y. Find the density function of Z.

$f_Z(z)=\int_{-\infty}^{\infty}f_{X,Y}(u,z-u)du$

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 $0\leq{x}<1, 0\leq{y}<2$

Then if X=u then Y=z-u and in this case

$0\leq{u}<1, 0\leq{z}-u<2$

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).
• Jan 28th 2011, 02:09 AM
CaptainBlack
Quote:

Originally Posted by Volga
Question.

$f_{X,Y}(x,y)= xy, 0\leq{x}<1, 0\leq{y}<2$
$0$ otherwise

Suppose Z=X+Y. Find the density function of Z.

Is it not obvious that:

$f_Z(z) \ne 0$ for $z \in (0,3)$ and $f_Z(z)=0$ otherwise?

So the required support is $(0,3)$

CB
• Jan 28th 2011, 03:41 AM
Volga
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?
• Jan 28th 2011, 05:32 AM
mr fantastic
Quote:

Originally Posted by Volga
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?

A simple sketch graph of the line y = -x + z over the region defined by the support of $f_{XY}$ shows that you need to consider the cases $0 \leq z \leq 1$, $1 \leq z \leq 2$, $2 \leq z \leq 3$.
• Jan 29th 2011, 12:32 AM
Volga
Quote:

Originally Posted by mr fantastic
A simple sketch graph of the line y = -x + z over the region defined by the support of $f_{XY}$ shows that you need to consider the cases $0 \leq z \leq 1$, $1 \leq z \leq 2$, $2 \leq z \leq 3$.

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

$\int_0^zf_{X,Y}(u,z-u)du=\int_0^zu(z-u)du=[z\frac{u^2}{2}-\frac{u^3}{3}]_0^z=\frac{z^3}{6}$

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...
• Jan 29th 2011, 04:53 AM
mr fantastic
Quote:

Originally Posted by Volga
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

$\int_0^zf_{X,Y}(u,z-u)du=\int_0^zu(z-u)du=[z\frac{u^2}{2}-\frac{u^3}{3}]_0^z=\frac{z^3}{6}$

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...

Using the method of distribution functions:

Case 1: z < 0.

$G(z) = \Pr(Z \leq z) = 0$

$\displaystyle \Rightarrow g(z) = \frac{dG}{dz} = 0$.

Case 2: $0 \leq z \leq 1$.

$\displaystyle G(z) = \Pr(Z \leq z) = \Pr(X + Z \leq z) = \int_{x = 0}^{x = z} \int_{y = 0}^{y = -x + z} xy \, dx \, dy = \int_{x = 0}^{x = z} \frac{x}{2} (z - x)^2 \, dx = ....$

$\displaystyle \Rightarrow g(z) = \frac{dG}{dz} = ....$

Case 3: $1 \leq z \leq 2$.

$\displaystyle G(z) = \Pr(Z \leq z) = \Pr(X + Z \leq z) = ....$

$\displaystyle \Rightarrow g(z) = \frac{dG}{dz} = ....$

Case 4: $2 \leq z \leq 3$.

$\displaystyle G(z) = \Pr(Z \leq z) = \Pr(X + Z \leq z) = ....$

$\displaystyle \Rightarrow g(z) = \frac{dG}{dz} = ....$

Case 5: $z > 3$.

$\displaystyle G(z) = \Pr(Z \leq z) = \Pr(X + Z \leq z) = 1$

$\displaystyle \Rightarrow g(z) = \frac{dG}{dz} = 0$.

The terminals for the double integral in cases 2-4 are found by doing what I said in my previous post. "...." denotes work for you to do.
• Jan 29th 2011, 03:56 PM
matheagle
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.
• Jan 29th 2011, 10:50 PM
Volga
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 (Itwasntme)

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

$\int_{z=0}^1\int_{w=0}^z(wz-w^2)dwdz=\int_0^1dz[\frac{w^2}{2}z-\frac{w^3}{3}]_0^z=\int_0^1\frac{z^3}{6}dz$

which means $f_Z(z)=\frac{z^3}{6}, 0

1<z<2

$\int_{z=1}^2\int_{w=0}^1(wz-w^2)dwdz=\int_1^2dz[\frac{w^2z}{2}-\frac{w^3}{3}]_0^1=\int_1^2\frac{z}{2}dz$

therefore density is $\frac{z}{2}, 1

2<z<3

$\int_{z=2}^3\int_{w=z-2}^1(wz-w^2)dwdz=\int_2^3dz[\frac{w^2z}{2}-\frac{w^3}{3}]_{z-2}^{1}=\int_2^3\frac{-z^3+15z-18}{6}dz$ (or thereabouts...)

therefore density is $\frac{-z^3+15z-18}{6}, 2

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

$G(z)=P(X+Z\leq{z})$ or $P(X+Y\leq{z})$
• Jan 29th 2011, 11:04 PM
mr fantastic
Quote:

Originally Posted by Volga
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 (Itwasntme)

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

$\int_{z=0}^1\int_{w=0}^z(wz-w^2)dwdz=\int_0^1dz[\frac{w^2}{2}z-\frac{w^3}{3}]_0^z=\int_0^1\frac{z^3}{6}dz$

which means $f_Z(z)=\frac{z^3}{6}, 0

1<z<2

$\int_{z=1}^2\int_{w=0}^1(wz-w^2)dwdz=\int_1^2dz[\frac{w^2z}{2}-\frac{w^3}{3}]_0^1=\int_1^2\frac{z}{2}dz$

therefore density is $\frac{z}{2}, 1

2<z<3

$\int_{z=2}^3\int_{w=z-2}^1(wz-w^2)dwdz=\int_2^3dz[\frac{w^2z}{2}-\frac{w^3}{3}]_{z-2}^{1}=\int_2^3\frac{-z^3+15z-18}{6}dz$ (or thereabouts...)

therefore density is $\frac{-z^3+15z-18}{6}, 2

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"? Mr F says: It's a sketch of the line y = -x + z in the xy-plane. z is a parameter here not a variable.

Also, is it

$G(z)=P(X+Z\leq{z})$ or $P(X+Y\leq{z})$ Mr F says: By the given definition Z = X + Y so surely it must be the latter ....!

..
• Jan 30th 2011, 07:21 PM
matheagle
you're not dumb and you shouldn't feel that way
(Only cows call people dumb)
• Jan 31st 2011, 12:26 AM
Moo
Dumm, not dumb :D
• Jan 31st 2011, 01:10 AM
mr fantastic
Quote:

Originally Posted by Moo
Dumm, not dumb :D

One of my favorite books was written by a Dumas, I'll have you both know ....

(Probably time to close this thread. Anyone with more mathematics to add can ask me in a pm to open it).