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    Two Random Variables Normal Distribution



    X1 ~ N(3,2^2), X2 ~ N(-8, 5^2)
    Let U = X1 - 5X2
    Let V = -6X1 + CX2, where C is constant.

    (a) What are the distributions of U and V?
    I know U ~ N(43, 629). I understand the mean, but don't see how the book gets 629. Obviously, it is taking b^2 * 5 * X2\sigma, but I would expect the formula to be \sigmax1 MINUS b^2 * 5 * \sigmax2, or more specifically 4 - 625 for a total of -619.

    I know variance is non-negative, so should I simply add the two variances together, despite the RV U subtracting them. I see the numbers and answer, just not understanding what theorem(s) or formulas convey this.

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    Re: Two Random Variables Normal Distribution

    Quote Originally Posted by simplemts View Post
    X1 ~ N(3,2^2), X2 ~ N(-8, 5^2)
    Let U = X1 - 5X2
    Let V = -6X1 + CX2, where C is constant.

    (a) What are the distributions of U and V?
    I know U ~ N(43, 629). I understand the mean, but don't see how the book gets 629. Obviously, it is taking b^2 * 5 * X2\sigma, but I would expect the formula to be \sigmax1 MINUS b^2 * 5 * \sigmax2, or more specifically 4 - 625 for a total of -619.

    I know variance is non-negative, so should I simply add the two variances together, despite the RV U subtracting them. I see the numbers and answer, just not understanding what theorem(s) or formulas convey this.
    Are you familiar with the Variance Sum Law?

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    Re: Two Random Variables Normal Distribution

    What do you expect to see if you expand (ax - by)^2 ?

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    Re: Two Random Variables Normal Distribution

    Quote Originally Posted by BGM View Post
    What do you expect to see if you expand (ax - by)^2 ?
    ( \hspace{2cm} a \hspace{2cm} x \hspace{2cm} - \hspace{2cm} b \hspace{2cm} y \hspace{2cm} )^2
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    Re: Two Random Variables Normal Distribution

    Seems like a simple definition:
    ""The variance of X plus or minus Y is equal the variance of X plus the variance of Y."

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    Re: Two Random Variables Normal Distribution

    Quote Originally Posted by simplemts View Post
    Seems like a simple definition:
    ""The variance of X plus or minus Y is equal the variance of X plus the variance of Y."
    It's not a definition though - it's a provable fact (assuming X and Y are uncorrelated).
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    Re: Two Random Variables Normal Distribution

    One might even call it a "Law"...

    All joking aside.

    My next challenge is to identify the value of c that makes U and V independent. For these two variables to be independent, I need their Cov[U,V] to equal 0. That I get. Which would mean E[XY] - E[X]E[Y] = 0. I know E[X] and E[Y], but can't see how I would get E[XY] out of the data values I'm given. Any hints?

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    Re: Two Random Variables Normal Distribution

    Write E[UV] = E[(X1 - 5X2)(-6X1 + CX2)] and then expand. Note that since X1 and X2 are independent (I'm assuming) that E[X1X2] = E[X1]E[X2]
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    Re: Two Random Variables Normal Distribution

    I don't see how that expansion turns into anything "clean":
    -6X_1^2 + CX_1X_2 + 30X_1X_2 - 5CX_2^2

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    Re: Two Random Variables Normal Distribution

    But you're taking expected values so all of that should be wrapped in an expectation operator. I didn't check if you expanded it correctly but if you did then all you need to do is take the expectation and then you just have a linear equation in C.
    "His programming is malfunctioning. It begins! Get your weapons, he's going to become a killbot!!!" - bryangoodrich

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    Re: Two Random Variables Normal Distribution

    I've done the math three times now, it just doesn't make sense.
    -6[3^2] + C[3] + C[-8] - 30[3] - 30[-8] - 5C[64]
    E[UV] = 105 - 325C
    then...
    [105-325C] - E[U]E[V]
    [105-325C] - [43]*[-18-8C]
    [105-325C] - [-774 - 344C]
    19C + 879 = 0
    C = 46.xx

    The answer is C = -24/125

    Clearly, I'm doing something VERY wrong, even though this book has like 20% of the problems wrong.. this is not one of them. *Sigh*

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    Re: Two Random Variables Normal Distribution

    How are you getting this: 6[3^2] + C[3] + C[-8] - 30[3] - 30[-8] - 5C[64]
    Because it doesn't look right to me.
    "His programming is malfunctioning. It begins! Get your weapons, he's going to become a killbot!!!" - bryangoodrich

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    Re: Two Random Variables Normal Distribution

    Quote Originally Posted by Dason View Post
    How are you getting this: 6[3^2] + C[3] + C[-8] - 30[3] - 30[-8] - 5C[64]
    Because it doesn't look right to me.
    Known:
    X1 ~ N(3,2^2)
    X2 ~ N(-8, 5^2)
    U = X1 - 5X2
    V = -6X1 + CX2

    E[UV] = E[(X1 - 5X2)(-6X1 + CX2)]

    I'll start with just the first few...
    E[UV] = E[X1 * - 6X1] becomes -6E[X1^2] and E[X1 * CX2] becomes C[EX1] + C[EX2]...

    See where I am going wrong? I am just substituting in the E values for X1 and X2 then.

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    Re: Two Random Variables Normal Distribution


    No I still don't see how you're doing it exactly. When you multiply everything out there only should be 4 terms. Also note that E[X_1^2] \neq (E[X_1])^2.
    "His programming is malfunctioning. It begins! Get your weapons, he's going to become a killbot!!!" - bryangoodrich

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