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Math Help - t - distribution

  1. #1
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    t - distribution

    Let X have a t-distribution with r degrees of freedom.

    How would I go about showing:

    E(X) = 0 , r > 1

    and

    Var(X) = r/(r-2) , r > 2
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    Quote Originally Posted by gtaplayr View Post
    Let X have a t-distribution with r degrees of freedom.

    How would I go about showing:

    E(X) = 0 , r > 1

    and

    Var(X) = r/(r-2) , r > 2
    E(X) = \int_{-\infty}^{+ \infty} x f(x) \, dx where f(x) is the pdf of the t-distribution with r degrees of freedom.

    Var(X) = E(X^2) - [E(X)]^2 and E(X^2) = \int_{-\infty}^{+ \infty} x^2 f(x) \, dx where f(x) is the pdf of the t-distribution with r degrees of freedom.
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    The mean is easy because the density is symmetric about zero.
    So xf(x) is an odd function, hence

    \int_{-\infty}^{\infty}xf(x)dx=0
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    Quote Originally Posted by gtaplayr View Post
    Let X have a t-distribution with r degrees of freedom.

    How would I go about showing:

    E(X) = 0 , r > 1

    and

    Var(X) = r/(r-2) , r > 2
    Here is a recipe which I think will allow you to find var(X) with a minimum of effort.

    1. Show that E(X) = 0, so var(X) = E(X^2).

    2. Show that E(1 + X^2 / r) = 1 + E(X^2) / r.

    3. Write down the integral for E(1 + X^2 / r). Make the substitution t = \sqrt{\frac{r}{r-2}}\; x. Notice that the resulting integral is of the form c \int_{-\infty}^{\infty} f(t) \,dt, where c is a constant (depending on r, but independent of t), and f is the pdf of a t distribution with r-2 degrees of freedom.

    4. Use the fact that \int f(t) \, dt = 1 for any pdf f(t) to evaluate the integral in 3.

    5. Substitute into the equation in 2. and solve for E(X^2).
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    So where does E(Z), E(Z^2), E(1/\sqrt{U}), E(1/U) where Z\sim,N(0,1) and U\sim\chi^2_{r}, come into this?
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    Quote Originally Posted by gtaplayr View Post
    So where does E(Z), E(Z^2), E(1/\sqrt{U}), E(1/U) where Z\sim,N(0,1) and U\sim\chi^2_{r}, come into this?
    Why? Unless you have in mind a specific way of solving the original question that involves those things, the question has been solved.
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    Quote Originally Posted by gtaplayr View Post
    So where does E(Z), E(Z^2), E(1/\sqrt{U}), E(1/U) where Z\sim,N(0,1) and U\sim\chi^2_{r}, come into this?
    Like Mr. Fantastic, I find this question confusing. You don't need to know any of those things, provided you know the formula for the pdf of a t distribution. If you don't know the pdf, then that's a different problem.

    One thing I forgot to mention in my original post: you will need the identity \Gamma(x+1) = x \Gamma(x) to get the answer in its final form.
    Last edited by awkward; September 23rd 2009 at 03:46 PM. Reason: forgot something
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    Quote Originally Posted by mr fantastic View Post
    Why? Unless you have in mind a specific way of solving the original question that involves those things, the question has been solved.
    If you correctly answer this question, with full working, 1% will be added to your final mark.

    Let X have a t-distribution with r degrees of freedom. Show that E(X) = 0, r > 1 and V ar(X) = r/(r − 2), r > 2 by first finding E(Z), E(Z^2), E(1/\sqrt{U}), E(1/U) where Z\sim,N(0,1) and U\sim\chi^2_{r}
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    Quote Originally Posted by awkward View Post
    Here is a recipe which I think will allow you to find var(X) with a minimum of effort.

    1. Show that E(X) = 0, so var(X) = E(X^2).

    2. Show that E(1 + X^2 / r) = 1 + E(X^2) / r.

    3. Write down the integral for E(1 + X^2 / r). Make the substitution t = \sqrt{\frac{r}{r-2}}\; x. Notice that the resulting integral is of the form c \int_{-\infty}^{\infty} f(t) \,dt, where c is a constant (depending on r, but independent of t), and f is the pdf of a t distribution with r-2 degrees of freedom.

    4. Use the fact that \int f(t) \, dt = 1 for any pdf f(t) to evaluate the integral in 3.

    5. Substitute into the equation in 2. and solve for E(X^2).
    Hi, does this work for F-distribution too?
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    Quote Originally Posted by noob mathematician View Post
    Hi, does this work for F-distribution too?
    What are you trying to do? Find the mean and variance?
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    Quote Originally Posted by mr fantastic View Post
    What are you trying to do? Find the mean and variance?
    The expected value. Since they have the same result right?
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    Quote Originally Posted by noob mathematician View Post
    Hi, does this work for F-distribution too?
    Quote Originally Posted by mr fantastic View Post
    What are you trying to do? Find the mean and variance?
    Quote Originally Posted by noob mathematician View Post
    The expected value. Since they have the same result right?
    Well, I suppose it works to a point since Y = X^2 ~ F(\nu_1 = 1, \nu_2 = r) and you've been shown how to calculate E(X^2).
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    Quote Originally Posted by mr fantastic View Post
    Why? Unless you have in mind a specific way of solving the original question that involves those things, the question has been solved.
    Ok this is what iv done so far

    Iv used the normal moment generating function:
    \int \exp^{tz} \frac{1}{\sigma\sqrt2\pi} \exp^{-1/2(z-\mu/\sigma)} dz
    M(t)=\exp{\mu.t+1/2\sigma^2t^2}
    To show that E(Z)=M'(0)=0 and E(Z^2)=M''(0)=1

    I used the Chi moment generating function:
    M(t)=(1-2t)^{-r/2}
    To show that E(U)=M'(0)=r
    E(1/U)=1/E(U)=1/r

    This is where i am stuck
    I am unsure how to calculate E(1/\sqrt{U})
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