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Thread: Stuck on evaluating an integral for expect, in order to show that it's non-convergent

  1. #1
    Senior Member dokrbb's Avatar
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    Stuck on evaluating an integral for expect, in order to show that it's non-convergent

    Hi,

    My question is more related to calculus here but it has a probability component so I decided to stick to the main thread

    I have a pdf defined as $\frac{k}{1+y^2}, y \in R$;

    I found $k = \frac{1}{\arctan(y)}$

    Then I need to show that the random variable Y has a Cauchy distribution; for this I tried to find E[X], and here I get stuck:

    $E[X] = \int_{-\infty}^{+\infty} yf(y)dy = \int_{-\infty}^{+\infty} \frac{y}{arctan(y)(1+y^2)}dy = ???$

    - without that $y$ in the numerator I can do some nice substitutions but in this form...

    I would really appreciate your help guys, in evaluating this integral,
    Thank you
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    Re: Stuck on evaluating an integral for expect, in order to show that it's non-conver

    You've messed up.

    In the first part I assume $k$ is just a constant to ensure that the pdf integrates to 1.

    It's not a function of $y$, it's a constant.

    $k = \dfrac{1}{\displaystyle{\int_{-\infty}^{\infty}}~\dfrac{dy}{1+y^2}}$
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    Senior Member dokrbb's Avatar
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    Re: Stuck on evaluating an integral for expect, in order to show that it's non-conver

    yes, that's right, k is a constant
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    Re: Stuck on evaluating an integral for expect, in order to show that it's non-conver

    Quote Originally Posted by dokrbb View Post
    yes, that's right, k is a constant
    and it's value is?
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    Senior Member dokrbb's Avatar
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    Re: Stuck on evaluating an integral for expect, in order to show that it's non-conver

    oh, it's $\frac{1}{\pi}$

    then my integral becomes $ \int_{-infty}^{infty} \frac{y}{\pi(1+y^2)}dy$
    let $u = 1+x^2, du = 2xdx$
    $ = \frac{1}{2\pi}\int_{-infty}^{infty} \frac{1}{u}du = \frac{1}{2\pi} lim_{a \rightarrow \infty}log(u) \bigg]_{-a}^{a} = \frac{1}{2\pi} lim_{a \rightarrow \infty}log(1+x^2) \bigg]_{-a}^{0} + \frac{1}{2\pi} lim_{a \rightarrow \infty}log(1+x^2) \bigg]_{0}^{a}$

    - both parts tend to $\infty$, thus, and expected value does not exist
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    Re: Stuck on evaluating an integral for expect, in order to show that it's non-conver

    well done
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