Thread: Std. normal distribution - T / F questions

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

1. 
2. is independent of
3. is normally distributed
4. is std. normally distributed
5. 
6. 
7. is independent of
8. 

T = true, F = false, blank = "I have (really) no idea."

1. T:
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:
6. T: again, "intuition"
7. F: I guess is most strongly dependent on ?
8. : no clue (at all)

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

Originally Posted by courteous

T = true, F = false, blank = "I have (really) no idea."

1. T:
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:
6. T: again, "intuition"
7. F: I guess is most strongly dependent on ?
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 X²+Y² as well) to take a single value

Originally Posted by Moo

[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 X²+Y² 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.