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Probability: What are p.d.f.'s of x+y and x/y?

  1. Feb 23, 2014 #1
    1. The problem statement, all variables and given/known data

    The probability desity function (p.d.f.) of joint distribution of random variables X and Y is given as

    [itex] f(x,y) = \begin{cases} e^{-(x + y)} \;\; when \;\; x > 0 \\ 0 \;\; \;\;\;\;\;\;\;\;\;\;otherwise \end{cases} [/itex]

    Question 1: What are the p.d.f.'s of X + Y and X/Y ?

    Question 2: Does the expectation of X/Y exist ?

    2. Relevant equations

    Nothing special.

    3. The attempt at a solution

    Answer 1:

    [itex] \begin{cases} u = x \\ v = x + y \end{cases} [/itex]

    [itex] \begin{cases} x = u \\ y = v - u \end{cases} [/itex]

    [itex] Jacobian = \begin{bmatrix} \frac{dx}{du} & \frac{dx}{dv} \\\frac{dy}{du} & \frac{dy}{dv} \end{bmatrix} = \begin{bmatrix}1 & 0 \\{-1} & 1 \end{bmatrix} [/itex]

    [itex] g(u,v)=f(u,v-u)|Jacobian|= e^{-v} [/itex]

    [itex] h(x+y)=h(v) = \int_0^v g(u,v)du = e^{-v} u |_{u=0}^{u=v} = ve^{-v} [/itex]

    [itex]\begin{cases}z = x\\ w = \frac{x}{y} \end{cases} [/itex]

    [itex]\begin{cases} x = z\\ y = \frac{z}{w} \end{cases} [/itex]

    [itex]Jacobian2 = \begin{bmatrix} \frac{dx}{dz} & \frac{dx}{dw} \\\frac{dy}{dz} & \frac{dy}{dw} \end{bmatrix} = \begin{bmatrix}1 & 0 \\{ \frac{1}{w} } & {- \frac{z}{ w^{2}} } \end{bmatrix}[/itex]

    [itex]g2(z,w)=f(z, \frac{z}{w} )|Jacobian2|= e^{-z- \frac{z}{w} } \frac{z}{ w^{2} } [/itex]

    [itex]h2(w)= \int_0^ \infty g2(z,w)dz = \int_0^ \infty e^{-z- \frac{z}{w} } \frac{z}{ w^{2} } dz = -\int_0^ \infty \frac{z}{ w^{2} } {(1+ \frac{1}{w} )}^{-1} d e^{-z(1+ \frac{1}{w})} = 0 + \int_0^\infty \frac{e^{-z(1+ \frac{1}{w})}}{ w^{2} \frac{1}{w}} dz = - \frac{1}{ {w+1}^{2} } e^{-z(1+ \frac{1}{w})} |_{z=0}^{z= \infty }= \frac{1}{{w+1}^{2}} [/itex]

    Answer 2:

    [itex]E( \frac{x}{y} )=E(w)= \int_0^\infty \frac{1}{{w+1}^{2}} dw = -\frac{1}{w+1}|_{w=0}^{w=\infty} = 1[/itex]

    Are the two answers correct? Thank you in advance.
     
  2. jcsd
  3. Feb 23, 2014 #2

    Ray Vickson

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    Homework Helper

    There is something wrong with the question as stated: your joint density ##f(x,y)## blows up as ##y \to -\infty##, and does not have a finite integral over ##y \in (-\infty,\infty)##. You restrict x but not y.

    You never define for us what is meant by ##h(w)## and ##h2(w)##, so we end up having to try to guess---a good way to lose marks on an assignment.

    Also, when you write
    [tex] \frac{1}{w+1^2}[/tex]
    you are writing
    [tex] \frac{1}{w+1}[/tex]
    If you really mean
    [tex] \frac{1}{(w+1)^2}[/tex]
    then use parentheses. BTW: that last form is the correct density of ##X/Y## at ##w \geq 0##.

    Finally, the integral you need for ##E(X/Y)## is not the integral you wrote.
     
  4. Feb 25, 2014 #3
    Thank you very much for your reply, Ray.

    The p.d.f. should be

    [itex] f(x,y) = \begin{cases} e^{-(x + y)} \;\; when \;\; x > 0 \;\; y > 0\\ 0 \;\; \;\;\;\;\;\;\;\;\;\;otherwise \end{cases} [/itex]

    I forgot to write the part [itex]y > 0[/itex], my mistake.

    [itex] h(v)=h(x+y) [/itex] actually is the marginal p.d.f. of [itex] v [/itex]

    Similarly, [itex] h2(w) =h2( \frac{x}{y} ) [/itex] is the marginal p.d.f. of [itex] w [/itex]

    The derivation of h2(w) should be

    [itex]h2(w)= \int_0^ \infty g2(z,w)dz = \int_0^ \infty e^{-z- \frac{z}{w} } \frac{z}{ w^{2} } dz = -\int_0^ \infty \frac{z}{ w^{2} } {(1+ \frac{1}{w} )}^{-1} d e^{-z(1+ \frac{1}{w})} = 0 + \int_0^\infty \frac{e^{-z(1+ \frac{1}{w})}}{ w^{2} \frac{1}{w}} dz = - \frac{1}{ {(w+1)}^{2} } e^{-z(1+ \frac{1}{w})} |_{z=0}^{z= \infty }= \frac{1}{{(w+1)}^{2}} [/itex]

    Yes, as you said, I lost the parentheses when writing [itex] \frac{1}{{(w+1)}^{2}} [/itex]

    The last integral is wrong, it should be

    [itex] E( \frac{x}{y} )=E(w)=\int_0^ \infty w \frac{1}{{(w+1)}^{2}} dw = \int_0^ \infty (\frac{w+1}{{(w+1)}^{2}} - \frac{1}{{(w+1)}^{2}} )dw= \int_0^ \infty \frac{1}{{(w+1)}^{2}} d(w+1)^2 - \int_0^ \infty \frac{1}{{(w+1)}^{2}} dw = ln(w+1)^2|_{w=0}^{w= \infty } + \frac{1}{w+1}|_{w=0}^{w= \infty } = \infty [/itex]

    Hence, [itex] E( \frac{x}{y} ) [/itex] doesn’t exist.
     
  5. Feb 25, 2014 #4

    Ray Vickson

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    You are correct now.
     
  6. Feb 25, 2014 #5
    Thank you again, Ray.
     
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