# Thread: Std. normal distribution - T / F questions

1. ## Std. normal distribution - T / F questions

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

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

1. T: $\displaystyle P(X>0)=\frac{1}{2}=P(Y<0)$
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: $\displaystyle =P(X>0)P(Y>0)=\frac{1}{2}\frac{1}{2}=\frac{1}{4}$
6. T: again, "intuition"
7. F: I guess $\displaystyle X+Y$ is most strongly dependent on $\displaystyle X$?
8. : no clue (at all)

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

2. Originally Posted by courteous
T = true, F = false, blank = "I have (really) no idea."

1. T: $\displaystyle P(X>0)=\frac{1}{2}=P(Y<0)$
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: $\displaystyle =P(X>0)P(Y>0)=\frac{1}{2}\frac{1}{2}=\frac{1}{4}$
6. T: again, "intuition"
7. F: I guess $\displaystyle X+Y$ is most strongly dependent on $\displaystyle X$?
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 X²+Y² as well) to take a single value

3. 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.