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Homework Help: Integral and probability

  1. Mar 12, 2010 #1
    1. The problem statement, all variables and given/known data
    The probability is given by:
    p(x,y)=6(1-x-y)
    x>=0
    y>=0
    x+y<=1

    Show that p is normilized.
    calculate <x>


    2. Relevant equations
    I found that p is normilized using the integral:
    untitled.JPG

    and It's equal to 1.
    are the limits of the integral correct?
     
  2. jcsd
  3. Mar 12, 2010 #2

    gabbagabbahey

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    Why are you integrating [itex]y[/itex] from [itex]x[/itex] to [itex]1-x[/itex]? Are you told that [itex]y\geq x[/itex]? If not, you should be integrating from [itex]0[/itex] to [itex]1-x[/itex] instead.
     
  4. Mar 12, 2010 #3
    I know that y>=x-1
    and if I do it for 0 to 1 I don't get in the end that the probabilty =1
     
  5. Mar 12, 2010 #4

    vela

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    You were given [itex]x+y \le 1[/itex], so, moving the x over to the other side, you get [itex]y \le 1-x[/itex], right?
     
  6. Mar 12, 2010 #5
    right... so that's why I chose the integral limits for dy to be from x to 1-x.
    Is it wrong?
     
  7. Mar 12, 2010 #6

    gabbagabbahey

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    Again, why are you using [itex]x[/itex] as your lower limit? You are told that [itex]y\geq0[/itex], not [itex]y\geq x[/itex]
     
  8. Mar 12, 2010 #7

    vela

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    Your response to gabbagabbahey's question was confusing. In your previous post, you answered, it was because you knew [itex]y \ge x-1[/itex]. The "x-1" suggested to me you somehow got that condition from the constraint [itex]x+y \le 1[/itex], and my point was that the constraint can only lead to [itex]y \le 1-x[/itex], the upper limit of the integral. There's no way you can get [itex]y \ge x-1[/itex] from it.

    Even if you did have [itex]y \ge x-1[/itex], your answer didn't make sense. Why would the lower limit of the integral be x? Shouldn't it be x-1 if your inequality were true?
     
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