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Mean and variance

  1. Aug 1, 2006 #1
    I'm having a problem with my answer for a question involving finding the mean and variance of a function. I'll state the question and list the steps I did so maybe you guys can see where I went wrong.

    The question is:
    Given Y = |tan(X)| and X is uniformly distributed on (-pi/2, pi/2) Find
    a) the pdf of Y
    b) E(Y)
    c) Var(Y)

    for a) I was able to come up with the answer
    [tex] g(y) = \frac{2}{\pi (1 + y^2)} for 0 < y < \infty [/tex]
    0 otherwise

    which I'm pretty sure is correct (by transformation of variables)

    for b)
    I have:

    [tex] E(Y) = \int_{0}^{\infty} y \frac{2}{\pi (1+y^{2})} dy[/tex]
    [tex] = \frac{1}{\pi} \lim_{b\rightarrow\infty} [ ln (1+y^{2}]_{0}^{b} [/tex]
    [tex] = \frac{1}{\pi} [ \infty - 0] [/tex]
    [tex] = \infty [/tex]

    for c)
    I solve first for E[Y^2]
    [tex] E[Y^{2}] = \int_{0}^{\infty} y^{2} \frac{2}{\pi (1+y^{2})} dy[/tex]
    [tex] = \frac{2}{\pi}\int_{0}^{\infty} \frac{y^{2}}{(1+y^{2})} dy[/tex]
    [tex] = \frac{2}{\pi}\int_{0}^{\infty} 1 - \frac{1}{(1+y^{2})} dy[/tex]
    [tex] = \frac{2}{\pi} \lim_{b\rightarrow\infty}[ y - arctan(y)]_{0}^{b}[/tex]
    [tex] = \frac{2}{\pi} [\infty - \frac{\pi}{2} - (0 - 0)][/tex]
    [tex] = \infty [/tex]

    [tex] Var[Y] = E[Y^{2}] - (E[Y])^{2} [/tex]
    [tex] = \infty - (\infty)^{2} [/tex]
    [tex] = undefined ??? [/tex]

    Is this correct?
    my real problem is the mean being equal to infinity, is this possible?
    as for the variance, is it possible to have undefined variance?

    A similar problem was also asked where I am to find
    Given Y = cot(X) and X is uniformly distributed on (0, pi) Find
    a) the pdf of Y
    b) E(Y)
    c) Var(Y)

    Here, I was able to get
    E(Y) = either 0 or undefined
    Var(Y) = either infinity (if mean is 0) or undefined (if mean is undefined)

    are these possible? and which is correct?
    Specifically, I'm not sure if my solution for the mean is correct, which would affect my solution for the variance

    my solution for the pdf, btw, is
    [tex] g(y) = \frac{1}{\pi (1 + y^{2})} for -\infty < y < \infty[/tex]

    here's what I did:
    [tex] E(Y) = \int_{-\infty}^{\infty} y \frac{1}{\pi (1+y^{2})} dy[/tex]
    [tex] = \frac{1}{2\pi} \lim_{b\rightarrow\infty} [ ln (1+y^{2}]_{-b}^{b} [/tex]
    [tex] = \frac{1}{2\pi}[ \lim_{b\rightarrow\infty} ln (1+b^{2}) - \lim_{b\rightarrow\infty} ln (1+(-b)^{2})[/tex]

    now is this
    [tex] = \frac{1}{2\pi} [ \infty - \infty] [/tex]
    = undefined

    [tex] = \frac{1}{2\pi}[0] [/tex]
    [tex] = 0 [/tex]

    since I'm pretty sure that my answer for
    E(Y^2) = infinity
    is correct

    thus var(Y) = either infinity (if mean is 0) or undefined (if mean is undefined)

    are these answers correct (and which ones are the correct answers?)

    Thanks for you help guys
    Last edited: Aug 1, 2006
  2. jcsd
  3. Aug 1, 2006 #2
    I guess you could say that my main problem has more to do with operations of infinity and limits than on mean and variance :)
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