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A Continuous Multivariate Distribution

  1. Apr 9, 2016 #1
    1. The problem statement, all variables and given/known data
    The random variable ##(x,y)## has density ##f(x,y) = ce^{-(ax+by)}## for ##0\leq y\leq x\leq 1##, with given constants ##a > 0##, ##b > 0##.
    1. Compute the constant ##c##.
    2. Find the conditional probability density ##f_y(y|x)##.
    3. Compute the regression curve of ##Y## on ##X## i.e. ##E(Y|X = x)##.
    4. Sketch this regression curve for ##a = 1##, ##b = 2##.

    2. Relevant equations


    3. The attempt at a solution
    For 1. my limits of integration for both ##x## and ##y## is ##0## and ##1##. This gives (after doing the double integration) ##c = \frac{ab}{e^{-(a+b)}-e^{-a}-e^{-b}+1}##. Is this right???
    For 2. the conditional density is given by the joint density divided by the marginal density of ##x##. I just wanna make sure that my ##c## is correct first.
    Not sure on 3. and 4.
     
  2. jcsd
  3. Apr 10, 2016 #2

    vela

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    Did you write the conditions on ##x## and ##y## exactly as they were given to you? If so, the limits on your integral and your answer to #1 are wrong.
     
  4. Apr 10, 2016 #3
    I did ##c\int_0^1 \int_0^1 e^{-(ax+by)} \; dx \; dy = 1## and solved for ##c##. If this is wrong, then my problem is how to interpret ##0\leq y\leq x\leq 1##.
     
  5. Apr 10, 2016 #4

    vela

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    You need to take into account that it says ##y \le x##.
     
  6. Apr 10, 2016 #5

    Ray Vickson

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    In problems involving two-dimensional densities, always sketch the correct (x,y)-region before you start trying to do integration. Your error will stare you in the face if you do that.
     
  7. Apr 10, 2016 #6
    Oh ok so in that case I get ##y## from ##0## to ##x## and ##x## from ##0## to ##1##.
     
  8. Apr 10, 2016 #7
    Using these new limits i get ##c = \frac{ab(a+b)}{e^{-(a+b)}-(a+b)e^{-a}+b}##
     
  9. Apr 10, 2016 #8
    Using these new limits i get ##c = \frac{ab(a+b)}{e^{-(a+b)}-(a+b)e^{-a}+b}##
     
  10. Apr 10, 2016 #9

    vela

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    It looks like you made an algebra mistake somewhere. You should have gotten
    $$c = \frac{ab(a+b)}{a e^{-(a+b)}-(a+b)e^{-a}+b}.$$
     
  11. Apr 10, 2016 #10
    Yes I did get that just forgot to put the ##a## in there haha. Thanks.
     
  12. Apr 10, 2016 #11
    To find the marginal density do I calculate
    $$\begin{split}
    f_X(x) &= \int_{-\infty}^{\infty} f(x,y) \; dy \\
    &= c\int_{0}^{1} e^{-(ax+by)} \; dy???
    \end{split}$$
    or is it
    $$\begin{split}
    f_X(x) &= \int_{-\infty}^{\infty} f(x,y) \; dy \\
    &= c\int_{0}^{x} e^{-(ax+by)} \; dy???
    \end{split}$$
     
  13. Apr 11, 2016 #12
    Still not sure which it is and why. Please help.
     
  14. Apr 11, 2016 #13
    Never mind. Pretty sure it's the second one.
     
  15. Apr 12, 2016 #14
    I'm asked to find the conditional probability function ##f_y(y|x)##. This is
    $$\begin{split}
    f_y(y|x) &= \frac{f(x,y)}{f_x(x)} \\
    &= \frac{ce^{-(ax+by)}}{\frac{c}{b}\left[e^{-ax}-e^{-(ax+bx)}\right]} \\
    &= \frac{be^{-(ax+by)}}{e^{-ax}-e^{-(ax+bx)}}.
    \end{split}$$
    Just wanna make sure I'm good.
    I'm also asked to find the regression curve (which is the one I'm struggling with) ##E(Y|X=x)##. I get
    $$\begin{split}
    E(Y|X=x) &= \int_0^1 y f_y(y|x) \; dy \\
    &= \int_0^1 \frac{bye^{-(ax+by)}}{e^{-ax}-e^{-(ax+bx)}} \\
    &= \frac{be^{-ax}}{e^{-ax}-e^{-(ax+bx)}} \int_0^1 ye^{-by} \; dy.
    \end{split}$$
    We use integration by parts on the integral above. Let ##u=y##, so ##du=dy##. Also ##dv = e^{-by}##, so ##v = -\frac{e^{-by}}{b}##. Hence,
    $$\begin{split}
    E(Y|X=x) &= \frac{be^{-ax}}{e^{-ax}-e^{-(ax+bx)}} \int_0^1 ye^{-by} \; dy \\
    &= \frac{be^{-ax}}{e^{-ax}-e^{-(ax+bx)}} \left(\left[-\frac{ye^{-by}}{b}\right]_0^1+\frac{1}{b}\int_0^1 e^{-by} \; dy\right) \\
    &= \frac{be^{-ax}}{e^{-ax}-e^{-(ax+bx)}} \left(\left[-\frac{ye^{-by}}{b}\right]_0^1-\left[\frac{e^{-by}}{b^2}\right]_0^1\right) \\
    &= \frac{e^{-ax}}{e^{-ax}-e^{-(ax+bx)}}\frac{1-(b+1)e^{-b}}{b^2}.
    \end{split}$$
    My question is are the limits of integration ##0## to ##1## shown or ##0## to ##x## or even another set of limits???
    Thanks again.
     
  16. Apr 13, 2016 #15
    I guess I'm correct then.
     
  17. Apr 14, 2016 #16
    Nope that is wrong.
    $$E(Y|X=x) = \int_0^x \frac{bye^{b(x-y)}}{e^{bx}-1} \; dy = \frac{1}{b}-\frac{x}{e^{bx}-1}$$
    after several lines of working.
    Is this right as my limits of integration are ##0## and ##x##???
     
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