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# FRM EXAM 2007—Q 28: expected annual return

#### snigdha.agrawal@gmail.com

##### New Member
Q. Consider 2 stocks, A and B / assume there annual returns are jointly normally distributed , the marginal distribution of each stock has mean 2% AND STD deviation 10%. And correlation is 0.9 . What is the expected annual return of the stock A if the annual return of stock B is 3%?

a. 2%
b. 2.9%
c. 4.7%
d. 1.1%

Answer: The info in this question can be used to construct a regression model of A and B. We have R(A)= 2% + .9(10%/10%) [R(B)-2%] + E
Next replacing R(B) by 3% gives R(A)=2% + .9(3%-2%)=2.9%

#### David Harper CFA FRM

##### David Harper CFA FRM
Staff member
Subscriber
Hi snigdha,

This is classic FRM question, by asking to apply understanding rather than formulas. You can get there another way, might be more intuitive:

What is slope or beta(A with respect to B)?
Slope (A regressed on B) = Beta (A with respect to B) = Covariance(A,B) /Variance(A,B) = correlation(A,B)*StdDev(A)*StdDev(B)/Variance(B) = correlation(A,B)*StdDev(A)/StdDev(B).
Slope (A regressed on B) = 0.9*10%/10% = 0.9

Then, you want to know that all OLS regression lines pass through the point of both averages (mean B, mean A):
... in more familiar terms, average Y = b1 + b2*average X (always!)

A regressed on B: average A = intercept + slope*average B;
(i.e., average A = intercept + beta(A with respect to B)* average B
In this case,
Intercept = average A - slope*average B = 2% - 0.9*2% = 0.2%

Regression (A on B): A = 0.2% intercept + 0.9 slope * B.
E[A|B=3%] = 0.2% + 0.9*3% = 2.9%

...some juicy fundamentals in this simple question. If this makes sense, the given answer is also instructive for using the line to condition from mean of B.

David

#### snigdha.agrawal@gmail.com

##### New Member

1st
average A = intercept + beta(A with respect to B)* average B is used to arrive at calculation for intercept and then
same formula is modified to
Regression (A on B): A = 0.2% intercept + 0.9 slope * B.
and
E[A|B] is calculated??
This is confusing to me. can you please guide.

#### David Harper CFA FRM

##### David Harper CFA FRM
Staff member
Subscriber
Hi snigdha,

The beta(A with respect to B) is just to remind that is the same as the regression slope (A on B); more familiar:
Y = mx + b where m is slope and key fundamentals are:
m (slope) = beta (y with respect to x) = cov(x,y)/var(x) = correlation(x,y)*standard Dev(y)/standard deviation (x)

So we just have: average A = intercept + slope(A on B)* average B; a function of averages

Then using the regular regression

A(i) = intercept + slope*B(i) = 0.2% + 0.9*3%; i.e., this is the regression

It is the same thing to use:
A(i) = E[A|B] = A(i) = intercept + slope*B(i) = 0.2% + 0.9*3%
E[A|B] b/c the dependent/explained (i.e., A) is a conditional mean: what is A conditional on a value of B? … it is a value on the OLS line

David

#### Vaishnevi

##### New Member
Hi David,

Understand the solution can provided using the formula for bivariate normal distribution, wherein
E[A|B]=E[A]+Beta(A,B)*(B-E(B))

Please let me know if im wrong.

Vaishnevi

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#### David Harper CFA FRM

##### David Harper CFA FRM
Staff member
Subscriber
@Vaishnevi Yes, sure totally! I read yours as a simple application of the univariate regression. I solved it with (simple) univariate regression, relying on the property that the OLS must pass through {E(A), E(B)} to solve for the intercept, such that using the regression we can say that E(A|B=3%) = intercept + slope*B = α + β(A, B)*B = 0.2% + 0.9*3% = 2.9%

Your starts with E(A) and "moves up/down the regression line" with Beta(A,B)*(B-E(B)). Yours is equivalent, thanks!

append: in fact, let's prove it:
• E(A|B)= E(A) + β(A,B)*(B-E(B)); i.e., yours
• E(A|B)= E(A) + β(A,B)*B - β(A,B)*E(B)),
• E(A|B)= [E(A) - β(A,B)*E(B))] + β(A,B)*B, but the intercept, α = [E(A) - β(A,B)*E(B))] such that:
• E(A|B)= α + β(A,B)*B, which is mine. I hope that's interesting!

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#### Vaishnevi

##### New Member
In retrospect, apologies I should have given a reference to the chapter from where i picked this formula from.

I was referring to John Hull's 'Correlation and Copulas'. Hull gives an example:
If marginal distributions of V1 and V2 are normal, it is safe to assume joint probability of V1 & V2 are bivariate normal.If we know the correlation between V1 & V2, the expected mean of V2 condition on V1 is given by:
E(V2/V1)=E(V1)+[Correlation(V1,V2)*STD(V2)/SRD(V1)*(V1-E(V1))]

Yes, extremely interesting the simple regression formula matches the above. In this scenario I have a question here, can covariance and correlation (or say simple regression) be applied only when the variables are normally distributed?

Thanks
Vaishnevi

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#### David Harper CFA FRM

##### David Harper CFA FRM
Staff member
Subscriber
@Vaishnevi good question, but no, in order to predict A conditional on B, E(A|B), in the regression we do not need to assume, nor do we require, that either variable is normal. This is so stringent a requirement. Now, the linear regression model does make several assumptions, but none that either variable is normal. Rather, we are merely relying on the implied linear relationship (which is also a feature of correlation and covariance). So, we do assume a linear relationship, but not normality. In this way, the "prediction" of E(A|B) = 2.9% is a conditional expectation assuming the linear relationship. Thanks!