Application of Variance theorems

[sdoshi004]

David

Referring to your 2008 study notes on Quant , you have given the example of Coke and pepsi to explain the application of variance theorems which uses the formula E(X^2)- (E(x))^2. However, I got confused when you used the formula for Variance of X*Y i.e
E(XY)- E(X)E(Y).

Can you please explain

Thanks

Sumit

 

[David Harper CFA FRM]

Hi Sumit:

You are rightly confused: the yellow arrow should say "Covariance of X,Y" rather than "Variance of X*Y." Apologies.
The variance of a product of two random vars can indeed be calculated but finds no role here (or in the assignged Gujarati).

The calc is correct, and the explaing text is correct: E(XY) - E(X)E(Y) = 3.6 is the Covariance(X,Y).

It's a good observation as, IMO, the following equation in Gujarati is especially helpful and useful:

(3.23) cov(x,y) = E(xy) - E(x)E(y)
It is good to have this memorized and further see it specializes, if y=x, to the formula for variance (x)

that is, note the covariance of a variable with itself is the variable's variance:
cov(x,x) = var(x) = E(x^2) - E(x)^2

second, note we can solve for E(xy):
E(xy) = E(x)E(y) + cov(x,y)

Then, in classic credit risk terms, if x = pd & y = lgd, then:

Expected loss (EL) = E (pd*lgd) = E(pd)*E(lgd) + covariance (pd, lgd)

where often the assumption is independence btwn PD & LDG such that cov(pd, lgd) = 0 and
Expected loss (EL) = E (pd*lgd) = E(pd)*E(lgd)

but per 3.23, we see that if pd & lgd are not independent, the EL is understated!

Hope that helps, sorry the label is incorrect, will make sure that is fixed for 2009 notes - David

 

[sdoshi004]

David

Thanks for the explanation. It helped to get more insight.

Also would appreciate your guidance on understanding the calculation and interpretation of chi squared test of significance (Quant Study Notes 2008). I am little confused on the calculation of Confidence interval in the example given on page 60. Also the use of specific value 32.852 and 8.907 in the calculation of lower limit and upper limit.

Please let me know if I am missing any important concept.

Thanks

Sumit

 

[David Harper CFA FRM]

Hi Sumit:

I think it may be the trickiest sample distribution for us. IMO, it's key here to finally see the chi-squared statistic itself as a random variable:
chi^2 r.v. = (df)(sample variance/population variance)

And if you lookup Table A-4 for 95% (i.e, 2.5% in each tail), you get:
.(JavaScript must be enabled to view this email address)% = 32.85
.(JavaScript must be enabled to view this email address)% = 8.9

Okay, so a confusion here is that the right-tail of the chi-square distribution counter-intuitively corresponds to a lower population variance, and the left-tail corresponds to the higher population variance in the confidence interval. I think this takes time to learn! But notice the fraction: a higher population variance implies a lower chi^2 variable.

So we take the formula for chi^2:
chi^2 r.v. = (df)(sample variance/popluation variance)

and solve for population variance:
popluation variance= df*sample variance/(chi^2 r.v)

Now on inference is that the population variance lies within:
popluation variance <= df*sample variance/(chi^2 r.v @ 97.5%). See how this is the left-tail of the chi-square but will correspond to the upper interval ?!
popluation variance >= df*sample variance/(chi^2 r.v @ 2.5%)

In other words,
df*sample variance/(chi^2 r.v @ 2.5%) <= popluation variance <= df*sample variance/(chi^2 r.v @ 97.5%)
9.25 <= pop variance <= 34.13

Hope that helps...I think this requires study...I will put several into a testing/practice XLS...Thanks, David