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Homework Help: Probability integral not converging

  1. Nov 26, 2011 #1
    Probability integral not converging :(

    1. The problem statement, all variables and given/known data

    for the joint probability density function:
    [tex] f(x,y)= \begin{cases}
    y & 0 \leq x, \, y\leq1 \\
    \frac{1}{4} (2-y) & 0\leq x\leq 1, \, 0\leq y\leq 2 \\
    0 & \text{elsewhere} \end{cases} [/tex]
    find the following:
    a) [itex] f_1 (x) [/itex]
    b) [itex] f_2 (y) [/itex]
    c) [itex] F(x,y) [/itex]

    there are others but maybe this will be enough to help me see what the solution to my actual problem is, problem being that my integrals are not converging and I don't know what to do with that.

    2. Relevant equations

    a) [tex] f_1 (x)= \int_{-\infty }^\infty \! f(x,y) \, \mathrm{d}y [/tex]
    Where the integral is integrated over the range space of y.

    b) [tex] f_2 (y)= \int_{-\infty }^\infty \! f(x,y) \, \mathrm{d}x [/tex]
    Where the integral is integrated over the range space of x.

    c) [tex] F(x,y)= \int_{-\infty }^y \int_{-\infty }^x \! f(s,t) \, \mathrm{d} s \mathrm{d} t [/tex]
    Again, the lower bound is replaced by the lower bound of the range space for the appropriate variables.

    3. The attempt at a solution

    a) [tex] f_1 (x) = \int_{-\infty }^1 y \, \mathrm{d}y = \frac{1}{2} y^2 |_{-\infty }^1 = \frac{1}{2} -\frac{1}{2} \infty [/tex] ???

    Lets start there. My time is up on the library computer.
     
    Last edited: Nov 26, 2011
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  3. Nov 26, 2011 #2

    gb7nash

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    Re: Probability integral not converging :(

    I'm not sure if this will help much, but you have contradicting information in your problem statement. You're defining:

    f(x,y) = y for 0 ≤ x, y ≤ 1
    f(x,y) = .25(2-y) for 0 ≤ x ≤ 1, 0 ≤ y ≤ 2.

    These two regions intersect.
     
  4. Nov 26, 2011 #3
    Re: Probability integral not converging :(

    (have I mentioned before how much I hate my erroneous book? Especially pdfs that don't sum to 1. Those are among my favorites...)

    I'm going to email my professor about the bounds. I will post back when he responds.
     
  5. Nov 26, 2011 #4

    I like Serena

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    Re: Probability integral not converging :(

    Hi Arcana!

    I propose you modify the problem to:
    [tex]f(x,y)= \begin{cases}
    y & 0 \leq x, \, y\leq1 \\
    2-y & 0\leq x\leq 1, \, 1\leq y\leq 2 \\
    0 & \text{elsewhere} \end{cases}[/tex]

    I expect that this was intended, and at least that gives you the opportunity to exercise your problem solving skills without delay.
     
  6. Nov 26, 2011 #5
    Re: Probability integral not converging :(

    Did you mean to drop the [itex] \frac{1}{4} [/itex] ? Also, even if I accept these bounds, what should I do about the integral that's not converging? I don't think the new bounds change it.
     
  7. Nov 26, 2011 #6

    I like Serena

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    Re: Probability integral not converging :(

    Yes, I intended to drop the 1/4.
    Perhaps (somewhat later) you can tell me why?


    Perhaps you should change the bounds on your integral?

    Since f(x,y)=0 for negative values of y, it does not seem right that you integrate from -infty.
     
  8. Nov 26, 2011 #7
    Re: Probability integral not converging :(

    Uh, without doing extra math since I have way to much to do already (the lenghtiest problem set for this class, ever), I'm guessing you dropped the 1/4 because when you do then it sums to 1 like a real pdf should. Haha. We don't have many pdfs that actually obey the properties of a pdf in this book.... *grumble grumble* I guess the ability to check your work by using such properties is a luxury we've learned to do without in this class. What a detriment. *sigh*


    Oh yeah. no negative probabilities. That's just silly! :P

    [STRIKE]Tell you what though, I'm going to change the function to the bounds [itex] 0 \leq x, \, y\leq 1 [/itex] and [itex] 0\leq x\leq 1, \, 1\leq y\leq 2 [/itex] but leave the 1/4, for the reason that the bounds are stupid if the overlap, and I have many more things to work out about this distribution that may be really stupid if the bounds are like that, but I'll levae the 1/4 because we work on pdfs that are big fat lies all the time anyway, and the grader will probably have worked it out with the 1/4.[/STRIKE]

    Edit: no, forget it. I'll change it all the way you said. I'll just email the prof again and tell him I did.
     
  9. Nov 26, 2011 #8

    I like Serena

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    Re: Probability integral not converging :(

    Right! It gives you the unique opportunity to really dive into the problems and rectify them.
    When you are able to do that, you will truly understand the material!

    Next time I will just ask you if you can find the mistake in the problem, and perhaps give a hint for that! :wink:


    Yep. And good! :smile:
     
  10. Nov 26, 2011 #9
    Re: Probability integral not converging :(

    Okay, problem. First, I can't tell my prof I'm dropping the 1/4 without justification, and I can't make the math work out to justify.
    To prove that the pdf sums to 1 when I don't include 1/4, I tried to do:

    [tex] \int_0^1 \int_0^{\infty } y \; \mathrm{d} x \, \mathrm{d} y + \int_1^2 \int_0^1 2-y \; \mathrm{d} x \, \mathrm{d} y [/tex]
    But the first integral goes like this:
    [tex] \int_0^1 \int_0^{\infty } y \; \mathrm{d} x \, \mathrm{d} y = \int_0^1 (xy |_0^{\infty } ) \; \mathrm{d} y [/tex] and I'll save myself the tedious typing, you can see the non-convergence problem I'm having. Reversing the order of integration didn't help.


    Also, I solved part (a) by adjusting the bounds, but I'm still stuck on part b), which is essential the problem I just said about the total sum anyway. Basically copy and pasted this from above:
    b) [tex] \int_0^{\infty } y \; \mathrm{d} x = xy |_0^{\infty } [/tex]
     
    Last edited: Nov 26, 2011
  11. Nov 26, 2011 #10

    I like Serena

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    Re: Probability integral not converging :(

    What would f(x,y) be for large values of x? Say x > 1?
     
  12. Nov 26, 2011 #11
    Re: Probability integral not converging :(

    great big values of x, when y is still less than 1, are still only =y. How does that affect the integral?
     
  13. Nov 26, 2011 #12

    I like Serena

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    Re: Probability integral not converging :(

    Nope. They're not.
    Check your definition of f(x,y).
    x and y are simultaneously bound.
     
  14. Nov 26, 2011 #13
    Re: Probability integral not converging :(

    when [itex] x \geq 0 \: , \text{and} \: 0 \leq y \leq 1 \: , f(x,y)=y [/itex]
    Thus, for example, [itex] f(1274689,\frac{1}{2} )=\frac{1}{2} [/itex]

    Isn't it?
     
  15. Nov 26, 2011 #14

    I like Serena

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    Re: Probability integral not converging :(

    That's not what was intended.
    Read it like this:
    0 ≤ (x,y) ≤ 1.

    That is, 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1.
    It's a common shorthand.
     
  16. Nov 26, 2011 #15
    Re: Probability integral not converging :(

    *expletive*

    Okay, I think I can solve the rest of this now that we re-defined and further clarified the *expletive* problem.

    Thank you, you're a life saver, really. You've prevented my suicide at least four times now. j/k :P
     
  17. Nov 26, 2011 #16
    Re: Probability integral not converging :(

    ps,
    can you take a look at my other question? people are using big words at me again.... :uhh:
     
  18. Nov 26, 2011 #17
    Re: Probability integral not converging :(

    alas, further problem with this problem. I'm doing part c.
    first, I do two integrals:
    [tex] \int_0^y \int_0^x t \; \mathrm{d} s \, \mathrm{d} t = \int_0^y xt \; \mathrm{d} t = \frac{1}{2} xy^2 [/tex]
    [tex] \int_1^y \int_0^x 2-y \; \mathrm{d} s \mathrm{d} t = \int_1^y x(2-t) \; \mathrm{d} t = \int_1^y 2x-tx \; \mathrm{d} t = (2xt-\frac{1}{2} xt^2)|_1^y = [/tex]
    [tex] (2xy-\frac{1}{2}xy^2)-(2x-\frac{1}{2} x)=2xy-\frac{1}{2} xy^2-2x+\frac{1}{2} x=2xy-\frac{1}{2} xy^2-\frac{3}{2}x [/tex]
    Now, [tex] F(x,y)= \begin{cases} 0 & x \: \text{or} \: y \leq 0 \\ \frac{1}{2} xy^2 & 0 \leq x,y \leq 1 \\ 2xy-\frac{1}{2} xy^2 -\frac{3}{2} x & 0 \leq x \leq 1 \: \text{and} \: 1\leq y\leq 2 \end{cases} [/tex]

    Now, when one or both variable is above it's bound, you just plug in the maximum value of the variable, right? so the rest of F(x,y) looks like this:
    [tex] \begin{cases} \frac{1}{2} y^2 & x>1 \: \text{and} \: 0\leq y \leq 1 \\ 2y-\frac{1}{2} y^2-\frac{3}{2} & x>1 \: \text{and} \: 1\leq y \leq 2 \\ \frac{1}{2} x & y>2 \: \text{and} \: 0\leq x \leq 1 \\ \frac{1}{2} & y>2 \: \text{and} \: x>2 \end{cases} [/tex]

    But, [itex] F(\infty ,\infty ) [/itex] should equal 1, not 1/2

    What gives?
     
  19. Nov 26, 2011 #18

    I like Serena

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    Re: Probability integral not converging :(

    Haven't checked your entire calculation yet, but I know you'll have to divide the problem up into 4x3 cases, like this:

    [tex]\begin{array}{| c | c | c | c |}
    \hline \\
    & x < 0 & 0 \le x < 1 & x \ge 1 \\
    \hline \\
    y < 0 & & & \\
    0 \le y < 1 & & & \\
    1 \le y < 2 & & & \\
    y \ge 2 & & & \\ \hline \end{array}[/tex]

    Can you calculate F(x,y) for each of the cells?
    Some of them will be the same.
     
  20. Nov 26, 2011 #19
    Re: Probability integral not converging :(

    I think I have that table covered. I have seven cases, because x or y is less than zero is one case, so count 5 cases on your table as one case, and 12-5 is 7. Anyway, the problem lies with the case x>1 and y>2. I get 1/2 but it should be 1.
     
  21. Nov 26, 2011 #20
    Re: Probability integral not converging :(

    should I be adding the integral of the first piece, y, and the integral of the second piece, 2-y, when both are maximum? that would be 1.
     
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