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Thread: Bivariate normal distribution

  1. #1
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    Bivariate normal distribution

    Assume that $\displaystyle (X, Y)$ is a bivariate normal with $\displaystyle \mu_X = 1$ and $\displaystyle \sigma_X = .5$ and I want to find $\displaystyle P(X \leq .5)$. Is it true that $\displaystyle f_X \sim N(\mu_X, \sigma_X$)? Then I can just integrate that from$\displaystyle -\infty$ to .5. Is this correct?
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  2. #2
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    Looks good to me.
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  3. #3
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    Using that method, I get

    $\displaystyle f_X = \frac{1}{.5\sqrt{2\pi}}e^{-2(x-1)^{2}}$

    To find $\displaystyle P(X \leq .5)$, I take the integral.

    $\displaystyle \int_{-\infty}^{.5} f_X dx = \frac{1}{.5\sqrt{2\pi}}\int_{-\infty}^{.5}e^{-2(x-1)^{2}}dx$

    How do I do that integral?
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  4. #4
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    You don't, it's intractable (when I said "looks good" I didn't think you would try to integrate analytically) . Standardize X and use a Z-Table or software. The question is essentially asking the probability of X being one standard deviation below the mean, or more; should be, what, 16% or so?
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    Now I want to find $\displaystyle P(X \leq 0 | Y = -.2)$
    Do I find $\displaystyle E(X|Y = -.2$) and $\displaystyle Var(X|Y = -.2)$ and use that in the standard normal?
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  6. #6
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    Quote Originally Posted by BERRY View Post
    Now I want to find $\displaystyle P(X \leq 0 | Y = -.2)$
    Do I find $\displaystyle E(X|Y = -.2$) and $\displaystyle Var(X|Y = -.2)$ and use that in the standard normal?
    $\displaystyle X|Y = y$ is also distributed normally. It should be a result from your text that $\displaystyle (Y|X = x) \sim N(\mu_y + \rho (\sigma_y / \sigma_X)(x - \mu_X), \sigma^2 _Y (1 - \rho^2))$.
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  7. #7
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    It's incorrect...

    Quote Originally Posted by BERRY View Post
    Is it true that $\displaystyle f_X \sim N(\mu_X, \sigma_X$)?
    Corrected: $\displaystyle f_X \sim N(\mu_X, \sigma_X^2$)

    That assumes Cov(X,Y) = 0 which does not imply independence. On the other end, independence implies Cov(X,Y) = 0.
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  8. #8
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    Quote Originally Posted by paolopiace View Post
    Corrected: $\displaystyle f_X \sim N(\mu_X, \sigma_X^2$)

    That assumes Cov(X,Y) = 0 which does not imply independence. On the other end, independence implies Cov(X,Y) = 0.
    What if Cov(X,Y) isn't 0? I was given that $\displaystyle \rho = .6$
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    Quote Originally Posted by BERRY View Post
    What if Cov(X,Y) isn't 0? I was given that $\displaystyle \rho = .6$
    rho = Covariance(X,Y) / (sigma_X * sigma_Y)

    I suggest that you post all the data of the problem. What is sigma_Y ?
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  10. #10
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    Quote Originally Posted by paolopiace View Post
    Corrected: $\displaystyle f_X \sim N(\mu_X, \sigma_X^2$)

    That assumes Cov(X,Y) = 0 which does not imply independence. On the other end, independence implies Cov(X,Y) = 0.
    The marginal distribution of X does not depend on the Covariance. The conditional distribution does. Moreover, for the Bivariate Normal, Cov(X, Y) = 0, DOES imply independence.
    Last edited by theodds; Nov 13th 2009 at 02:01 PM.
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  11. #11
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    Quote Originally Posted by BERRY View Post

    To find $\displaystyle P(X \leq .5)$, I take the integral.

    $\displaystyle \int_{-\infty}^{.5} f_X dx = \frac{1}{.5\sqrt{2\pi}}\int_{-\infty}^{.5}e^{-2(x-1)^{2}}dx$

    How do I do that integral?
    I am sure you meant:

    $\displaystyle \int_{-\infty}^{.5} f_X dx = \frac{1}{.5\sqrt{2\pi}}\int_{-\infty}^{.5}e^{-\frac{1}{2}(\frac{x-1}{.5})^2}dx$
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  12. #12
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    Those are the same
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