Thread: Two Random Variables Normal Distribution

1. 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, but I would expect the formula to be x1 MINUS b^2 * 5 * x2, 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.

2. Re: Two Random Variables Normal Distribution

Originally Posted by simplemts
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, but I would expect the formula to be x1 MINUS b^2 * 5 * x2, 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?

3. Re: Two Random Variables Normal Distribution

What do you expect to see if you expand ?

4. Re: Two Random Variables Normal Distribution

Originally Posted by BGM
What do you expect to see if you expand ?

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BGM (10-19-2012)

6. 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."

7. Re: Two Random Variables Normal Distribution

Originally Posted by simplemts
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).

8. 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?

9. 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]

10. Re: Two Random Variables Normal Distribution

I don't see how that expansion turns into anything "clean":
+ C + 30 - 5C

11. 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.

12. 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*

13. 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.

14. Re: Two Random Variables Normal Distribution

Originally Posted by Dason
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)]

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.

15. 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 .

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simplemts (10-19-2012)

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