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Joint probability problems

  1. Apr 11, 2010 #1
    1. The problem statement, all variables and given/known data
    Two players, A and B, try to see who has the fastest reflects. They test this by using a button, and whoever presses the button first when a light goes on, wins. Let's assume both A and B are random variables ~U(0,1) and are independent (time in seconds). May W = "winner's reaction time" and Z = "loser's reaction time" both random variables.

    a) Find the average time for W knowing that the loser took at least 1/2 of a second to press the button.

    b) Find the covariance between Z and W.

    3. The attempt at a solution

    OK, I know that the density for A is fA(a) = 1{0<=a<=1} and the one for B is fB(b) = 1{0<=b<=1}.

    I know that W = min{A,B} and Z=max{A,B}

    For point a), what I'm looking for is the average time for W|Z>1/2, which is the same as
    (min{A,B}|max{A,B}>1/2). Now, I could find E[W|Z>1/2] if I had its distribution function or its density function. If I tried to find the distribution function, I should look for P(W|Z>1/2<=w) = P(min{A,B}|max{A,B}>1/2<=w). I just don't know what to do with that. And I don't know how I could get the density function alone.

    For point b) I think I need the joint probability function for W and Z. And I couldn't find a way to work it out.

    Any ideas?

    Thanks.
     
  2. jcsd
  3. Apr 11, 2010 #2

    vela

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    In the ab-plane, the region where the joint probability function for A and B is non-vanishing is the unit square S=[0,1]x[0,1]. Assume that A wins and that B took at least 1/2 second to press the button. What subset of S does that correspond to? You want to find the average of A over that subset. Then consider the case where B is the one who wins.
     
  4. Apr 12, 2010 #3

    lanedance

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    vela makes some good suggestions, just to add though it was implied

    the random variables are independent, so by definition, if you know the indivdual marginal distributions, the joint distribution is related to them by
    [tex] f_{A,B}(a,b) = f_{A}(a)f_{B}(b)[/tex]

    using that with the geometric picture should be enough to get it done
     
  5. Apr 13, 2010 #4
    OK. If B is the loser and took at least 1/2 of a second, that is S = [0,1]X[1/2,1]. If A is the loser, then S = [1/2,1]X[0,1].

    Now, I don't know what you mean with the average of A or B in that subset. As far as I'm concerned, that is still 1/2 in either case.

    I know this is wrong, I just don't know why.
     
  6. Apr 13, 2010 #5

    lanedance

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    ok, so first starting from the joint pdf of a constant value of one over the unit square S = [0,1]x[0,1], or similary [itex] (a,b):a \in [0,1], b \in [0,1] [/itex]

    for visualistaion, set a on the horizontal axis & b the vertical. Imagine the line a=b, this is when a & b have equal times.

    On the side towards the a axis, we have a<b so this represents the B win region and vice versa. Note the situation is entirely symmetric whether A or B wins...

    so for a) say A wins, then we are restricted to one half of the unit square, the triangle towards the b axis, bounded by the line b=a

    now we also know b>1/2s, so this means we are left with the region bounded by:
    - the b axis
    - the line b = 1/2
    - the line b = a
    - the line b = 1

    the centre of mass of the resulting region will represent the average loser & winner time
     
    Last edited: Apr 13, 2010
  7. Apr 15, 2010 #6
    Right, I got it. But that only covers the case in which A wins. I also have to consider the case in which B wins. If both results are equal (same expected value) then that's the answer I'm looking for, right?

    And what if they weren't equal (I think they will be in this case, but I'm curious now)? Could I use this same strategy merging both results somehow?
     
  8. Apr 15, 2010 #7

    lanedance

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    would depend on the specific
    but if you had define
    <A| A win> - expectation of A given a wins + including any other constaraints
    <B|B win>


    then win
    <W> = P(A win)*<A|A win> + P(B win)*<B|B win>

    in this case due to the symmetry of the distribution
    <A|A win> = <B|B win>
    P(A win) = P (B win)
     
  9. Apr 15, 2010 #8
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