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Math Help - sum of random variables - help to evaluate support of function

  1. #1
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    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.

    Answer.

    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).
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    Quote Originally Posted by Volga View Post
    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
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    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?
    Last edited by Volga; January 28th 2011 at 03:01 AM.
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    Quote Originally Posted by Volga View Post
    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.
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    Quote Originally Posted by mr fantastic View Post
    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...
    Last edited by Volga; January 28th 2011 at 11:46 PM.
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    Quote Originally Posted by Volga View Post
    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.
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    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.
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    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

    \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<z<1

    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<z<2

    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<z<3

    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})
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  9. #9
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    Quote Originally Posted by Volga View Post
    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

    \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<z<1

    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<z<2

    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<z<3

    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 ....!
    ..
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  10. #10
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    you're not dumb and you shouldn't feel that way
    (Only cows call people dumb)
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  11. #11
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    Dumm, not dumb
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  12. #12
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    Quote Originally Posted by Moo View Post
    Dumm, not dumb
    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).
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