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Joint probability distribution

  1. Nov 21, 2015 #1
    1. The problem statement, all variables and given/known data
    1. The joint probability density function of X and Y is given by

      f(x,y)=(6/7)(x^2+ xy/2) , 0<x<1, 0<y<2.

      (a) Find the pdf of X.
      (b) Find the cdf of X.
      (c) FindP(X<.5).
      (d) Determine the conditional pdf of Y given X = x.
    3. The attempt at a solution
    a) the pdf is what is stated in the probem, that P(x,y) = (6/7)(x^2+ xy/2) , 0<x<1, 0<y<2. and P(x,y) = 0 elsewhere
    b)
    [itex] F(x,y) = \frac{6}{7} \int_0^Y \int_0^X(x^2 + \frac{xy}{2})dxdy [/itex]

    [itex] = \frac{6}{7} \int_0^Y (\frac{X^3}{3} + \frac{X^2y}{4})dy [/itex]

    [itex] = \frac{6}{7} (\frac{YX^3}{3} + \frac{X^2Y^2}{8})[/itex]

    c)
    [itex] \frac{6}{7} \int_0^2 \int_0^.5(x^2 + \frac{xy}{2})dxdy [/itex]

    [itex] = \frac{6}{7} \int_0^2 (\frac{(.5)^3}{3} + \frac{(.5)^2y}{4})dy [/itex]

    [itex] = \frac{6}{7} (\frac{(2)(.5)^3}{3} + \frac{(.5)^2(2)^2}{8})[/itex]

    d) fy(y|x) = f(x,y)/fx(x)

    [itex] fx(x) = \frac{6}{7}\int_0^2(x^2 + \frac{xy}{2})dy [/itex]

    [itex] fx(x) = \frac{6}{7}(2x^2 + x) [/itex]

    fy(y|x) = f(x,y)/fx(x) =

    [itex] \frac{(6/7)(x^2+ xy/2)}{(6/7)(2x^2 + x)} [/itex]

    [itex] = \frac{x(x+ y/2)}{x(2x + 1)} [/itex]

    [itex] = \frac{(x+ y/2)}{(2x + 1)} [/itex]


    Is this correct?
     
  2. jcsd
  3. Nov 21, 2015 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    No: (a) and (b) are wrong. You are given ##f_{XY}(x,y)##, and have been asked to find ##f_X(x)## and ##F_X(x)##, the marginal pdf and cdf of ##X## alone. Anyway, what you wrote in (b) makes no sense: you have numbers ##x## and ##y## on the left in ##F(x,y)## but have random variables ##X## and ##Y## on the right.

    I have not checked the details of (c), but the basic calculations look OK. Your (d) looks OK as well.
     
  4. Nov 21, 2015 #3
    a)
    [itex] fx(x,y) = \frac{6}{7} \int_0^2 (x^2 + \frac{xy}{2})dy [/itex]

    [itex] fx(x,y) = \frac{6}{7} ((2)x^2 + \frac{x(2)^2}{4}) [/itex]

    [itex] fx(x,y) = \frac{6}{7} (2x^2 + x) [/itex]

    b)
    [itex] Fx(x,y) = \frac{6}{7} \int_0^x(2u^2 + u)du [/itex]

    [itex] Fx(x,y) = \frac{6}{7} (\frac{2x^3}{3} + \frac{x^2}{2}) [/itex]
     
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