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Probability, Bivariate Normal Distribution

  • Thread starter WHB3
  • Start date
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1. The problem statement, all variables and given/known data
Let the probability density function of X and Y be bivariate normal. For what values of a is the variance of aX+Y minimum?


2. Relevant equations
The answer in the book is -p(X,Y)(std dev of X/std dev of Y)


3. The attempt at a solution
I think the equation for Var(aX+Y) is,
Var(aX+Y)=a^2Var(X)+Var(Y)+2aCov(X,Y), but I have no idea how to work this equation to equal the answer in the book.

Any ideas would be much appreciated!!
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
 

statdad

Homework Helper
1,480
29
Yes

[tex]
Var(aX+Y) = a^2 Var(X) + Var(Y) + 2 a Cov(X,Y)
[/tex]

which, as a function of [tex] a [/tex], looks like a quadratic function. How about you? (There's a hint here :) )
 
21
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Thanks, Statdad!! Your hint was just the insight I needed.
 
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I think I spoke too soon, Statdad.

I can see where the equation Var(aX+Y) is a quadratic equation, but I still can't factor the equation to obtain "-p(X,Y)(std dev of Y/std dev of X)" as one of the factors where -p(X,Y) is the negative correlation coefficient.

I am wondering whether I should be working with a different equation.

Any further assistance would be appreciated.
 

EnumaElish

Science Advisor
Homework Helper
2,285
123
Remember p(X,Y) = Cov(X,Y)/sqrt(Var(X)Var(Y)).
 
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When I try to factor this equation using the formula we learned in high school, I get

-2Cov(X,Y) +-radical(4Cov^2(X,Y)-4VarXVarY)divided by 2VarX.

Since everything under the radical goes to zero, I am left with

-2Cov(X,Y)/2VarX = Cov(X,Y)/VarX; this is not the answer I should be coming up with.
 
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I still need an answer to this problem, so if anyone knows what I'm doing wrong here, I would appreciate the help.
 

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