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Math Help - Mixing a uniform distribution with a normal distribution

  1. #1
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    Mixing a uniform distribution with a normal distribution

    Here is my Problem.

    Let X~Uniform(0,2) and Y~Normal(0,1) be independent random variables. Let

    Z ={1 if X > Y
    {2 if X <Y

    Find E[Z] and V ar[Z]. You may leave your answers in terms of de nite integrals.

    My big question is, how can we join a uniform and a normal distribution on the same probability density function.

    Thank you
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  2. #2
    Grand Panjandrum
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    Re: Mixing a uniform distribution with a normal distribution

    Quote Originally Posted by solvj View Post
    Here is my Problem.

    Let X~Uniform(0,2) and Y~Normal(0,1) be independent random variables. Let

    Z ={1 if X > Y
    {2 if X <Y

    Find E[Z] and V ar[Z]. You may leave your answers in terms of denite integrals.

    My big question is, how can we join a uniform and a normal distribution on the same probability density function.

    Thank you
    If you know x you can find P(Y<x), which is the conditional probability P(Z=1|X=x) that Z=1. Then integrate this multiplied by the pdf of X to get the unconditional probability that Z=1.

    CB
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  3. #3
    MHF Contributor chisigma's Avatar
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    Re: Mixing a uniform distribution with a normal distribution

    Quote Originally Posted by solvj View Post
    Here is my Problem.

    Let X~Uniform(0,2) and Y~Normal(0,1) be independent random variables. Let

    Z ={1 if X > Y
    {2 if X <Y

    Find E[Z] and V ar[Z]. You may leave your answers in terms of denite integrals.

    My big question is, how can we join a uniform and a normal distribution on the same probability density function.

    Thank you
    Some cautions have to be taken because X and Y are continuos variables and Z is a discrete variable. The most sure way is to valuate the probability...

    P\{X > Y\}= P\{X > \alpha\}\ P\{Y < \alpha\} (1)

    The first step...

    P\{X > \alpha\} = 1- \frac{\alpha}{2}\ ,\ 0< \alpha<2 ; 1 , \alpha<0 ; 0 , \alpha > 2 (2)

    P\{Y < \alpha\} = \frac{1}{2}\ (1+ \text{erf}\ \frac{\alpha}{\sqrt{2}}) (3)

    Now other steps are...

    a) use (2) and (3) to obtain (1)...

    b) compute the derivative \pi (\alpha) of (1) respect to \alpha ...

    c) compute...

    P\{Z=1\} = \int_{-\infty}^{+ \infty} \pi(\alpha)\ d \alpha (4)

    d) use the result c) to compute P\{Z=2\} , E \{Z\} , V\{Z\}...

    All these steps are left to You...

    Kind regards

    \chi \sigma
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  4. #4
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    Re: Mixing a uniform distribution with a normal distribution

    So if I understand quite good, I should get P(x>Y)=P(Z=1|X=x)

    I need to find P(X>Y)=P(Z=1)=integral from 0 to 2 (P(Z=1|X=x)*P(X=x)dx)
    But what is P(x>Y)??
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  5. #5
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    Re: Mixing a uniform distribution with a normal distribution

    Thanks to both of you, my problem is solved.
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