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Math Help - product of continuous and discrete pdf

  1. #1
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    product of continuous and discrete pdf

    X~N(0,1) standard normal
    Z is discrete distrbution with -1 and 1 with equal probability.

    What is the distribution of XZ = Y?

    Do I do:

    f(y) = f(x)f(z)
    = -f(y) when y=-1
    and f(y) when y=1

    where f(y) = {1 \over \sqrt{2 \pi} }e^{-0.5y^2}
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  2. #2
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    Quote Originally Posted by chopet View Post
    X~N(0,1) standard normal
    Z is discrete distrbution with -1 and 1 with equal probability.

    What is the distribution of XZ = Y?

    Do I do:

    f(y) = f(x)f(z)
    = -f(y) when y=-1
    and f(y) when y=1

    where f(y) = {1 \over \sqrt{2 \pi} }e^{-0.5y^2}
    No.

    F(y) = \Pr(Y \leq y) = \Pr(XZ \leq y) =

    \Pr\left(X \leq \frac{y}{Z} \right) when Z > 0

    \Pr\left(X \geq \frac{y}{Z} \right) when Z < 0


    Therefore F(y) =

    \Pr(X \leq y) when Z = 1

    \Pr(X \geq -y) when Z = -1



    Therefore:

    F(y) = \frac{1}{2} ( \, \Pr(X \leq y) + \Pr(X \geq -y) \, )

     = \frac{1}{2} \left( \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{y} e^{-t^2/2} \, dt + \frac{1}{\sqrt{2 \pi}} \int_{-y}^{+\infty} e^{-t^2/2} \, dt \right).


    Therefore f(y) = \frac{dF}{dy} = \frac{1}{2 \sqrt{2 \pi} } \left( e^{-y^2/2} + e^{-y^2/2} \right) = \frac{1}{\sqrt{2 \pi} } e^{-y^2/2}.

    This does not seem right but I don't have time now to find my mistake. I'll post again later.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    No.

    F(y) = \Pr(Y \leq y) = \Pr(XZ \leq y) =

    \Pr\left(X \leq \frac{y}{Z} \right) when Z > 0

    \Pr\left(X \geq \frac{y}{Z} \right) when Z < 0


    Therefore F(y) =

    \Pr(X \leq y) when Z = 1

    \Pr(X \geq -y) when Z = -1



    Therefore:

    F(y) = \frac{1}{2} ( \, \Pr(X \leq y) + \Pr(X \geq -y) \, )

     = \frac{1}{2} \left( \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{y} e^{-t^2/2} \, dt + \frac{1}{\sqrt{2 \pi}} \int_{-y}^{+\infty} e^{-t^2/2} \, dt \right).


    Therefore f(y) = \frac{dF}{dy} = \frac{1}{2 \sqrt{2 \pi} } \left( e^{-y^2/2} + e^{-y^2/2} \right) = \frac{1}{\sqrt{2 \pi} } e^{-y^2/2}.

    This does not seem right but I don't have time now to find my mistake. I'll post again later.
    It is right, what you have is a standard normal random variable, with its sign changed at random with probability 0.5, but as the standard normal density is symmetric that leave the distribution unchanged.

    RonL
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  4. #4
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    Quote Originally Posted by CaptainBlack View Post
    It is right, what you have is a standard normal random variable, with its sign changed at random with probability 0.5, but as the standard normal density is symmetric that leave the distribution unchanged.

    RonL
    Yes, that thought occured to me some time later. Thanks for the confirmation.
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