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Homework Help: Probablity: What's the p.d.f. of the random variable Z = X|X|

  1. Feb 17, 2014 #1
    1. The problem statement, all variables and given/known data

    If the probability density function(p.d.f.) of a random variable X is f(x) = 1/6 * e-|x|/3 where x is lying in (-∞,∞) and |-x| = x if x≥0, then what is the p.d.f. of the random variable Z = XY = X*|X| where Y = |X| ?

    2. Relevant equations

    Nothing special.

    3. The attempt at a solution

    Answer: h(z) = f(x) * g(y) where g(y) = 2 * 1/6 * e-y/3 = 1/3 * e-y/3
    Comment: Because for y=|x|, its range shrinks to half of x, i.e., (0,∞) instead of the oringinal (-∞,∞), consequently, its p.d.f. should be 2*f(x). It seems like if h(z) = f(x) * g(y) then X and Y are independent? I think this answer is wrong, anyone can help?

    Thank you in advance!
    Last edited: Feb 17, 2014
  2. jcsd
  3. Feb 17, 2014 #2

    Ray Vickson

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    X and Y = |X| are not independent; they are about as dependent as you can get.

    In problems of this type, do not try to "guess"; proceed carefully, step-by-step from first principles. In this case it is easiest to first figure out what is ##F(z) = \text{P}(Z \leq z )##, where ##Z = X |X|##. Look separately at the two cases z < 0 and z > 0. I will help you get started on the case when z < 0.

    So, if z < 0 we can write it as z = -|z|. In order to have Z = X|X| ≤ z = -|z| we need X < 0; do you see why? So, in this case we have ##Z = -X^2##, hence we need ##- X^2 \leq - |z|##, or ##\{ |X| \geq \sqrt{|z|}\; \& \; X < 0\}##. This event has the same probability as ##\{X \geq \sqrt{|z|} \}##, because the distribution of X is symmetric about x = 0.

    So, for ##z < 0## we have
    [tex] F(z) \equiv \text{P}(X|X| \leq z) = \text{P} (X \geq \sqrt{|z|}).[/tex]
    You can complete the calculation, and also do it in the other case of z > 0. Then you get the density function ##f(z)## of ##Z## by differentiation of ##F(z)##.

    I would not classify this as a precalculus problem.
    Last edited: Feb 17, 2014
  4. Feb 17, 2014 #3
    If z > 0, then

    [itex]F(z) = p(Z \leq z) = p(x \mid x \mid \leq z) = p(x^{2} \leq z \; and \; x > 0) = p(0 < x \leq \sqrt{z}) = \int_ 0 ^ \sqrt{z} \frac{1}{6} e^{- \frac{x}{3} } dx[/itex]

    [itex]f(z) = \frac{dF(z)}{dz} = \frac{d\int_ 0 ^ \sqrt{z} \frac{1}{6} e^{- \frac{x}{3} } dx}{dz} = \frac{1}{12 \sqrt{z}} e^{- \frac{\sqrt{z}}{3}}[/itex]

    If z < 0, then

    [itex]F(z) = p(Z \leq z) = p(x \mid x \mid \leq z) = p(-x^{2} \leq z \; and \; x < 0) = p(x \leq - \sqrt{-z}) = \int_ {-\infty} ^ {-\sqrt{-z}} \frac{1}{6} e^{ \frac{x}{3} } dx

    f(z) = \frac{dF(z)}{dz} = \frac{d \int_ {-\infty} ^ {-\sqrt{-z}} \frac{1}{6} e^{ \frac{x}{3} } dx}{dz} = \frac{1}{12 \sqrt{-z}} e^{- \frac{\sqrt{-z}}{3}}

    If it is, then
    [itex]h(x) = \frac{1}{6} e^{-\frac{|x|}{3}} [/itex]
    [itex]g(|x|) = \frac{1}{3} e^{-\frac{|x|}{3}} [/itex]
    [itex]f(x|x|) \neq h(x) * g(|x|)[/itex]

    Hence, x and |x| are not independent.

    Is this correct?

    Thank you very much, Ray Vickson.
  5. Feb 17, 2014 #4

    Ray Vickson

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    You are correct that
    [tex] f(z) = \frac{1}{12} \frac{e^{-\frac{\sqrt{|z|}}{3}}}{\sqrt{|z|}},[/tex]
    which form holds for both z > 0 and z < 0.

    I don't know what you are doing after that, or why you bother. Of course, X and |X| are not independent; no calculations are needed to see this, because for any function k(x) the random variables X and k(X) are dependent. After all, if I tell you a value of X you know exactly what is the value of k(X) with no uncertainty whatsoever.
  6. Feb 18, 2014 #5
    Because I was required to compute the covariance of x and |x|, hence I have to figure out z=x|x| for computing E[x|x|], now I have the formula of z, hence

    [itex]E[x|x|] = E[Z] = \int_0^\infty \frac{\sqrt{z}}{12} e^{ -\frac{ \sqrt{z}}{3} } dz + \int_{-\infty}^0 \frac{-\sqrt{z}}{12} e^{ -\frac{ \sqrt{z}}{3} } dz = \int_0^\infty \frac{ u^{2} }{6} e^{ -\frac{u}{3} } du - \int_0^\infty \frac{ v^{2} }{6} e^{ -\frac{v}{3} } dv = 0 [/itex]

    Then E[x] = 0 since f(x) is symmetric, finally

    [itex] Cov(x, |x|) = E[x|x|] - E[x]E[|x|] = 0 [/itex]
  7. Feb 18, 2014 #6

    Ray Vickson

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    Strictly speaking, to compute expectations of functions Z = G(X) you do not need the distribution of Z (although having it does not hurt). Instead, you can use the so-called "Theorem of the Unconcious Statistician", which asserts that
    [tex] E Z = \int f_X(x) G(x) \, dx [/tex]
    Here, ##f_X(x)## is the original density function of the original random variable ##X##. What the theorem is really saying is that
    [tex] \int f_Z(z) z \, dz = \int f_X(x) G(x) \, dx, [/tex]
    where ##f_Z## is the density function of the random variable ##Z = G(X)##. See, eg, http://en.wikipedia.org/wiki/Law_of_the_unconscious_statistician .

    So, in your case, ##EZ \equiv E(X |X|) = \int f_X(x) x |x| \, dx.##
  8. Feb 18, 2014 #7
    Yes, the theorem is correct.

    [itex] E(x|x|)= \int_{- \infty }^{ \infty } \frac{1}{6} e^{ -\frac{|x|}{3} }x|x| dx = \int_{- \infty }^0 \frac{1}{6} e^{ \frac{x}{3} }(- x^{2} )dx + \int_0^{ \infty } \frac{1}{6} e^{ -\frac{x}{3} }x^{2} dx = - \int_0^{ \infty }\frac{1}{6} e^{ -\frac{u}{3} }u^{2} du + \int_0^{ \infty } \frac{1}{6} e^{ -\frac{x}{3} }x^{2} dx = 0 [/itex]

    Thank you again! Ray Vickson
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