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Math Help - Std. normal distribution - T / F questions

  1. #1
    Member courteous's Avatar
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    Question Std. normal distribution - T / F questions

    Independent X, Y both have std. normal distribution.
    Which statements are correct?

    1. P(X>0)=P(Y<0)
    2. X+Y is independent of X-Y
    3. X-Y is normally distributed
    4. X+Y is std. normally distributed
    5. P(X>0, Y>0)=\frac{1}{4}
    6. P(X>Y)=P(Y>X)
    7. X+Y is independent of  X
    8. X^2+Y^2=1
    T = true, F = false, blank = "I have (really) no idea."

    1. T: P(X>0)=\frac{1}{2}=P(Y<0)
    2. F: "intuition" ... really, why (if "F" is correct, of course)?
    3. T: solved it on MHF in Normal distribution?
    4. F: follows from 3.?
    5. T: =P(X>0)P(Y>0)=\frac{1}{2}\frac{1}{2}=\frac{1}{4}
    6. T: again, "intuition"
    7. F: I guess X+Y is most strongly dependent on X?
    8. : no clue (at all)

    Help me with argumenting why it is T (or F) ... correct any flawed reasoning (there are many).
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  2. #2
    Moo
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    Quote Originally Posted by courteous View Post
    T = true, F = false, blank = "I have (really) no idea."

    1. T: P(X>0)=\frac{1}{2}=P(Y<0)
    2. F: "intuition" ... really, why (if "F" is correct, of course)?
    3. T: solved it on MHF in Normal distribution?
    4. F: follows from 3.?
    5. T: =P(X>0)P(Y>0)=\frac{1}{2}\frac{1}{2}=\frac{1}{4}
    6. T: again, "intuition"
    7. F: I guess X+Y is most strongly dependent on X?
    8. : no clue (at all)

    Help me with argumenting why it is T (or F) ... correct any flawed reasoning (there are many).
    Hello,

    Unfortunately, I don't know your background and don't know if you've studied gaussian vectors...
    Question 1) is correct. More beautifully, you could say that X and -Y have the same distribution and hence the result.

    Question 2) is T. Indeed, X and Y being independent normal rv's, (X,Y) is a Gaussian vector (any linear combination of X and Y has a normal distribution). So (X+Y,X-Y) is a Gaussian vector too, because any linear combination of X+Y and X-Y will be a linear combination of X and Y.
    And if (Z,T) is a Gaussian vector, there is an equivalence between cov(Z,T)=0 and independence between Z and T. So just check here that cov(X+Y,X-Y)=0 and you'll have the result.
    Given the following questions, try to find the joint pdf of (X+Y,X-Y) by any means you may have learnt.

    Both 3) and 4) are T. Use characteristic functions for instance.

    5) ok.

    6) use the fact that X and -X are identically distributed and Y and -Y also are : P(X>Y)=P(-X>-Y)

    7) cov(X,X+Y)= ? if it's not 0, then it can't be independent

    8) is it an almost sure equality ? It's false since X and Y are continuous rv's and thus it is not possible, almost surely, for them (and for X+Y as well) to take a single value
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  3. #3
    Flow Master
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    Quote Originally Posted by Moo View Post
    [snip]

    8) is it an almost sure equality ? It's false since X and Y are continuous rv's and thus it is not possible, almost surely, for them (and for X+Y as well) to take a single value
    Indeed. Since it is well known that X^2 and Y^2 both follow a chi-squared distribution with 1 df, their sum will follow a chi-squared distribution with 2 df => 8) is clearly false.
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