Covariance of Bernoulli Random Variables

[mikey10011]

David,

I am going through Example 18.8 in Jorian's FRM Handbook (p. 420).

I know that for a Bernoulli random variable
E[X] = p
Var[X] = p (1-p)

Why is E[XY] = Prob[X and Y]?

Apologies that I don't have Gujarati but could you refresh my memory of probability theory on how I should derive that from first principles (e.g., Jorian pp. 55-57)?

 

[David Harper CFA FRM]

Hi Mikey,

Not sure i follow. Here p(a) = 10% and p(b) = 20%. If the assets were independent, p(ab) = (20%)(10%) = 2%. But generally,

E(ab) = E(a)E(b) + COVARIANCE(a,b)

This from Gurati:

COVARIANCE(a,b) = E(ab) - E(a)E(b)

I'm not liking the question for its imprecision; it should say unconditional/marginal probabilities of 10% and 20% (IMO without this qualification, we are struggling to figure exactly what 10% and 20% mean), but...

We can figure the implied default correlation here. From above:

COVARIANCE(a,b) = Correlation(a,b)StdDev(a)StdDev(b) = E(ab) - E(a)E(b), so:
Correlation(a,b) = [E(ab) - E(a)E(b)]/[StdDev(a)StdDev(b)]

and using your variance for a bernouilli above, StdDev(a) = SQRT[a(1-a)], so
Default correlation(a,b) = [3% - (10%)(20%)]/(SQRT[90%*10%]*SQRT[80%]*[20%]) = 0.083

you don't need all that, of course, the question only needs a tiny 2 x 2 "migration matrix:"

0 1
0 73% 17%
1 7% 3%

Let me know if that doesn't answer your question. David

 

[mikey10011]

Sorry David but it didn't answer my question.

For example, where did you get E(ab)=3%? If I read Jorian correctly (p. 425), E(AB)=Prob(A and B) = 0.15%.

Note that your answer of 0.083 is not one choices for answering Jorian's question (p. 420).

 

[David Harper CFA FRM]

Sorry, I was looking at 18.8. Re 18.9, okay, the same idea is used. From Gujarati,

COVARIANCE(a,b) = E(ab) - E(a)E(b), or
E(ab) = E(a)E(b) + COVARIANCE(a,b)

So, per the above, you can substitute Covariance(a,b) with Correlation(a,b)Volatility(a)Volatility(b) and then solve for Correlation.

Here the joint probability E(ab) is given as .15%. If you are asking about E(), given it is a bernoulli, the E(a) = p(default)*1 + p(no default)*0 = p(default)
Maybe that is the issue? for a bernouilli, the E(a) = p(a) if default is defined as 1.

when the question says joint probability of default = 0.15%, it means P(a defaults, b defaults) = P(a=1,b=1) = 0.15%. And E[ab] = P[ab]*(1)(1) = P[ab]. So you raise a good point, that p(a) is not necessarily equal to E(a). Rather E(a) = p(a)*f(a) but for Bernoulli's the f(a) = 1 so it's implicit.

David

 

[mikey10011]

E[ab]=P[ab]*(1)(1). That's it--thanks!

Glad I'm going though Jorian's FRM Handbook. There are so many subtleties from these elementary building block concepts.