Quants

anbu.edu
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Joined: Mon Feb 04, 2013 3:35 pm

Quants

Postby anbu.edu » Tue Oct 15, 2013 6:12 am

NEW
Consider the following linear regression model: Y = a + b*X + e. Suppose a = 0.05, b = 1.2, Std(Y) = 0.26, Std(e) = 0.1, what is the correlation between X and Y?
Choose one answer.
a. 0.852 Incorrect
b. 0.701 Incorrect
c. 0.923 Correct
d. 0.462 Incorrect
The correct answer is 0.923.
The standard deviation of X can be calculated as follows: std(X) = SQRT( (std(Y)2 - std(e)2) / b2 ) = SQRT( (0.262 - 0.12) / 1.22) = 0.2
Then, the correlation between X and Y can be calculated as follows: Corr(X,Y) = b * std(X) / std(Y) = 1.2 * 0.2 / 0.26 = 0.923


Can anyone explain the concept behind std(X) = SQRT( (std(Y)2 - std(e)2) / b2 ) = SQRT( (0.262 - 0.12) / 1.22) = 0.2

pradeeppdy
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Posts: 258
Joined: Thu Sep 20, 2012 3:42 pm

Quants

Postby pradeeppdy » Tue Oct 15, 2013 8:47 am

From regression equation Y = a + b*X + e take Variances on both sides,
Var(Y)=VaR(a + b*X + e )=VaR(a)+VaR(b*X)+VaR(e)+2Cov(a,b*X)+2Cov(a,e)+2Cov(b*X,e)
Var(a)=0 as vaariance of a constant is 0.
correlation of constant with variable is 0 =>2Cov(a,b*X)=2Cov(a,e)=0
we also assume that correlation of e errors terms with the independent variable is 0 which is one of assumption of regression(not any specific correlation given in question)=>2Cov(b*X,e)=0
after considering all this,
Var(Y)=VaR(a + b*X + e )=0+VaR(b*X)+VaR(e)+0+0+0
Var(Y)=VaR(b*X)+VaR(e)
Var(Y)=b^2*Std(X)^2+Std(e)^2]
[ From regression Y = a + b*X + e take CoVariances on both sides w.r.t X,
Cov(X,Y)=Cov[X,(a + b*X + e )=>Cov(X,Y)=Cov[X,a]+Cov[X, b*X ]+Cov[X,e]
Covariance of constant a with variable X is 0=>Cov[X,a]=0 also from assumption of regression we assume that otherwise not specified oin the Question correlation of independent variable wr.t the error terms is 0=>Cov[X,e]=0
Cov(X,Y)=0+Cov[X, b*X ]+0=>Cov(X,Y)=b*Cov[X, X ]=b*VaR(X,X)=b*Std(X)^2=>b=Cov(X,Y)/Std(X)^2]

Hope it helps


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