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Joint distribution of functions

  1. Nov 24, 2009 #1
    1. The problem statement, all variables and given/known data
    X1 and X2 are independent~u(0,1)

    Y1=X1+X2

    Y2=X1-X2

    Find the density function of Y1


    2. Relevant equations

    X1=(Y1+Y2)/2
    X2=(Y1-Y2)/2

    0[tex]\leq[/tex]Y1[tex]\leq[/tex]2
    -1[tex]\leq[/tex]Y2[tex]\leq[/tex]1

    0[tex]\leq[/tex]Y1+Y2[tex]\leq[/tex]2
    0[tex]\leq[/tex]Y1-Y2[tex]\leq[/tex]2

    -y1[tex]\leq[/tex]y2[tex]\leq[/tex]2-y1
    -1[tex]\leq[/tex]y2[tex]\leq[/tex]y1

    3. The attempt at a solution
    I don't understand how to set up the upper and lower bounds for these problems. I have spent the last two days wrestling with this, and I just don't get it. My professor spent almost an hour today trying to explain this to me, and I got nothing out of it. I will send a chocolate chip cookie through the mail to whoever can explain this in a way that I will finally understand. And while you're at it, maybe you can tell me if I should have used whomever in the last sentence, because that's another thing beyond my comprehension.
     
  2. jcsd
  3. Nov 24, 2009 #2
    Problem 1.

    I would use the cumulative distribution function. Let F be the cdf for Y1. (Y1 will not be uniform, by the way.)

    Then [tex]F(t)=P(Y_1 \le t)=P(X_1 + X_2 \le t)[/tex]. Now finding [tex]P(X_1 + X_2 \le t)[/tex] is a somewhat straightforward problem. Draw [tex]x_1 x_2[/tex] coordinate axes. Draw the set of points [tex](x_1,x_2)[/tex] where the pdf for X1 and the pdf for X2 are nonzero (should be a square). Draw the line [tex]x_1+x_2=t[/tex] and shade the correct side. Then as a great time-saving trick, instead of integrating, you can find areas using the formula for the area of a triangle (since the joint pdf is constant). There are two cases to consider, depending on what t is (i.e., depending on whether the line [tex]x_1+x_2=t[/tex] is below or above the main diagonal of the square).

    That's how I would do it. That doesn't look close to your notes at all, though.

    Problem 2.

    Determine the case of each pronoun by its use in its own clause. The case is not affected by any word outside the clause.

    The subject of a clause takes the subjective case, even when the whole clause is the object of a verb or preposition.

    In your example, the whole clause is whoever can explain this in a way that I will finally understand.

    Whoever is the subject of can explain, so whoever is the correct pronoun, not whomever.

    The whole clause whoever can explain this in a way that I will finally understand is the object of the preposition to, but that is irrelevant. The case is not affected by the word to outside the clause.
     
  4. Nov 24, 2009 #3
    You had me until you said to draw the line x1+x2=t. What value of do I use t?
     
  5. Nov 24, 2009 #4
    0.6


    Then try it with 1.7


    Then try a generic t value.
     
  6. Nov 25, 2009 #5
    So 0[tex]\leq[/tex]Y1[tex]\leq[/tex]2. The line will have a slope of -1, cutting the square diagonally. It will cut the square in half at t=1. What do I do with this information to get f(y)? F(y)=[tex]\int[/tex]f(y), so I need to find an equation for the area, and differentiate. This will have two parts. First 0[tex]\leq[/tex]Y1[tex]\leq[/tex]1, and then 0[tex]\leq[/tex]Y1[tex]\leq[/tex]2. How do I get the second half? Maybe I'll be able to put this together when I'm not sleepy.
     
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