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Thread: iid Normal

  1. #1
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    iid Normal

    If $\displaystyle X_1 \;,\; X_2$ are iid $\displaystyle N (0, \sigma^2)$ random variables with $\displaystyle Y_1 = {X_1}^2+{X_2}^2 \; and\; Y_2 = \dfrac{X_1}{\sqrt{{X_1}^2+{X_2}^2}}$. The book shows that $\displaystyle Y_1\;and\;Y_2$ are independent. But how would I find the pdf of $\displaystyle \dfrac{X_1}{X_2}$??


    I tried doing $\displaystyle Let\; U = \dfrac{X_1}{X_2} \;and\; V = {X_2} \implies X_1=UV$

    then $\displaystyle h_{X_1}(u,v) = uv \;and\;h_{X_2}(u,v)=v$

    thus, $\displaystyle J = \left|\begin{array}{cc}v&u\\0&1\end{array}\right|\ ;=\;v$

    so, $\displaystyle f(u,v) = \dfrac{1}{\sqrt{2\pi\sigma^2}}e^{\frac{(uv)^2}{2\s igma^2}} \times \dfrac{1}{\sqrt{2\pi\sigma^2}}e^{\frac{(v)^2}{2\si gma^2}} \times v$

    is this correct? thanks.
    Last edited by chutiya; Nov 29th 2010 at 09:12 PM.
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  2. #2
    MHF Contributor harish21's Avatar
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    Since $\displaystyle X_1 \mbox{and} X_2$ are independent:

    $\displaystyle f(x_1,x_2)= \dfrac{1}{2\pi\sigma^2} e^{\frac{{-({x_1}^2+{x_2}^2)}}{2\sigma^2}}$

    since $\displaystyle x_1 = uv\;,\; x_2=v\;,\;and\; |J|=v$

    $\displaystyle f(u,v)= 2f(x_1,x_2) \; |J| = ....$

    Note that you need to find the pdf of $\displaystyle {X_1}/{X_2} = U$. You can do so by finding the marginal pdf of $\displaystyle f(u,v)$
    Last edited by harish21; Nov 30th 2010 at 06:37 PM. Reason: typo
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