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Math Help - Convolution in statistics

  1. #1
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    Post Convolution in statistics

    Dear all,

    I learned, to derive the common probability distribution function f_{Z}(i)
    of e.g. two independent distribution functions f_{X}(i)X and Y
    f_{X}(i), one can use the convolution.

    f_{Z}(i)=\sum_{j=0}^{i}f_{X}(j) f_{Y}(i-j)


    The second way, from my point of view, could be, if e.g. the two distribution have, let's say two states with given probabilities


    f_{X}(i)=[0.6,0.4] with 0.6, for instance, as the probability for an electrical current IX1 to occur and 0.4 for an electrical current IX2 to occur.
    f_{Y}(i)=[0.3,0.7] with 0.3, for instance, as the probability for an electrical current IY1 to occur and 0.7 for an electrical current IY2 to occur.


    The new distribution function of Z is then[tex]
    f_{Z}(i=0)=0.6*0.3=0.18
    f_{Z}(i=1)=0.6*0.7=0.42
    f_{Z}(i=2)=0.4*0.3=0.12
    f_{Z}(i=3)=0.4*0.7=0.28[\MATH]
    with [tex]f_{Z}(i=1)+f_{Z}(i=2)=0.54[\MATH]

    The convolution would give only three stages:

    I) f_{Z}(i=0)=0.6*0.3=0.18
    II) f_{Z}(i=1)=0.6*0.7+0.4*0.3=0.54
    III) f_{Z}(i=2)=0.4*0.7=0.28

    The convolution spitts out two results in one as in II). Basically to me that is fine and right, but I do need the fourth stage as I get it without concolution.

    Is the second way wrong? From my point of view is it even better as it gives more stages in the result.

    So, hopefully one of you guys can help with this.
    Which version is correct? From my point of view both.

    Thanks

    HHsts
    Last edited by HHsts; December 28th 2010 at 12:53 AM. Reason: To clarify questions in reply 1
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  2. #2
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    Convolution is a way of deriving probability distribution for Z = X + Y.

    I'm not sure exactly what you mean by Z(1), Z(2) since no values were given for X and Y.

    Suppose X and Y range over integers 1 through 10 (and X,Y independent)
    Asking the probability that Z = 5 for example is the same thing as
    P{Z = 5}
    = P{X = 1, Y=4} + P{X=3,Y=2} + P{X=3,Y=2} + ...
    = P{X = 1}P{Y=4} + P{X=3}P{Y=2} + P{X=3}P{Y=2} + ...

    Which is exactly the expression for a convolution.
    You can decompose probabilities for P{X=i}P{Y=j}, but adding them gives the same thing as a convolution.
    Last edited by snowtea; December 27th 2010 at 08:10 PM.
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  3. #3
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    Quote Originally Posted by HHsts View Post
    Dear all,

    I learned, to derive the common probability distribution function f_{Z}(i)
    of e.g. two independent distribution functions f_{X}(i)X and Y
    f_{X}(i), one can use the convolution.

    f_{Z}(i)=\sum_{j=0}^{i}f_{X}(j) f_{Y}(i-j)


    The second way, from my point of view, could be, if e.g. the two distribution have, let's say two states with given probabilities

    X=[0.6 0.4];
    Y=[0.3 0.7];

    Z(1)=0.6*0.3=0.18
    Z(2)=0.6*0.7=0.42
    Z(3)=0.4*0.3=0.12
    Z(4)=0.4*0.7=0.28
    with Z(2)+Z(3)=0.54

    The convolution would give:

    0.1800 0.5400 0.2800

    Is the second way wrong? From my point of view is it even better as it gives more stages in the result.

    So, hopefully one of you guys can help with this.
    Which version is correct? From my point of view both.

    Thanks

    HHsts
    Please post the entire original question you are working on, not just your summary of it.
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  4. #4
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    Hello snowtea,

    Thanks for your reply. I have detailed my question (see original post). Hope this does clarify my concern.

    HHsts
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  5. #5
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    f_Z(1) = f_X(1)f_Y(0)+f_X(0)f_Y(1)

    This is the correct probability. For Z=1, we can have either X=1 Y=0 or X=0 Y=1.
    This is the expression given by the convolution. Dividing it into "multiple stages" does not help when you just want probabilities for a particular value of Z.
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  6. #6
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    Snowtea,

    Thanks for your reply.
    I see that Z=1 can happen if X=1 and Y=0 or Y=1 and X=0.
    according to
    X|Y|Z
    0|0|0
    1|0|1
    0|1|1
    1|1|1

    if you mean this and apply binomial distr.

    I still have one thing on my mind.

    I have four current, as above

    System X) Can take IX1=3A with 0.6 probability to occur and IX2=5A with 0.4 prob. to occur
    System Y) can take IY1=1A with 0.3 to occur and IY2=2A with 0.7 prob. to occur

    Now, I can get four current levels.

    1) IX1+IY1=3A+1A=4A
    2) IX1+IY2=3A+2A=5A
    3) IX2+IY1=5A+1A=6A
    4) IX2+IY2=5A+2A=7A

    When I would like to find the prob. for each level to occur I require four results

    1) IX1+IY1=3A+1A=4A=0.6*0.3
    2) IX1+IY2=3A+2A=5A=0.6*0.7
    3) IX2+IY1=5A+1A=6A=0.4*0.3
    4) IX2+IY2=5A+2A=7A=0.4*0.7

    The convolution would give me 1) then 2+3 and then 4.

    But I would like to find all four seperate prob.s
    So what do?

    Thanks again for a short feedback.

    HHsts
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