What's new

# Quantitative Analysis - Chapter 1- Probabilities - FRM May 2015

#### ShaktiRathore

##### Well-Known Member
Subscriber
Hi,
exp(.95%*5)=(exp(5))^.95%=148.41^.95%
Let x= 148.41^.95%
Take log both sudes
log(x)=log(148.41^.95% )=.95%log148.41
Find rhs from log table,
Then find antilog of obtained rhs value,then this antilog is x the value we yearned for.
Thanks

Hi Brian,

Thanks!
Jayanthi

#### brian.field

##### Well-Known Member
Subscriber
Jayanthi, the method Shakthi is using is a property of exponentials, although it isn't necessary to do it that way on a calculator.

I couldn't figure out how to add parentheses around the X^a on the right hand side, but assume they are there.

#### Dr. Jayanthi Sankaran

##### Well-Known Member
Hi Brian,

I should have mentioned the source - sorry about that! I was just trying to keep it simple....this is with reference to Question 300.3 of the Question set for the Probabilities Chapter. The question was:

If loss severity is given by x, then its pdf can be characterized by f(x) = c/x s.t. 1<x<e^5. What is the 95% VAR given that the losses are expressed in positive value, at what loss severity value (x) is only 5.0% of the distribution greater than x?

Jayanthi

#### ShaktiRathore

##### Well-Known Member
Subscriber
Jayanthi u got .95% incorrect it shud be .95/95% not .95%.
U shud find .95* log(148.41) then find antilog of obtained value.
Thanks

#### Dr. Jayanthi Sankaran

##### Well-Known Member
Yes, thanks for that Shakti. I was also somewhat confused. It's just that in David's Question set page 10 of Probabilities Chapter, there is a typo exp(5*0.95%). It should be exp(5*0.95) instead.

Jayanthi

#### Dr. Jayanthi Sankaran

##### Well-Known Member
Hi Brian,

X^ab = (X^)(a)^(b).......I guess this is it as far as the parenthesis, goes....

Thanks!
Jayanthi

#### ShaktiRathore

##### Well-Known Member
Subscriber
Jayanthi,
Yes Brian is right but he has forgotten to put () creating confusion. Its X^(ab)=(X^a)^b not( X^)a^b this how paranthesis goes. X and a shall be together in ().
E.g1024= 2^10= 2^(2*5)= (2^2)^5=(4)^5=1024=(2^5)^2=32^2=1024. Yours shall become( 2^)2^5=2^32 which is wrong.
Thanks

Last edited:

#### Dr. Jayanthi Sankaran

##### Well-Known Member
Ok thanks, Shakti for your clarification - appreciate it!

Jayanthi

#### Dr. Jayanthi Sankaran

##### Well-Known Member
Hi David,

From observation of numerical examples (specifically, those with probability matrices) in the Probabilities Chapter - Study Notes/Question Set, it appears that the concept of independence where Pr(X=x,Y=y) = Pr(X=x)Pr(Y=y), hardly applies. In other words, Bayes' Theorem does not need statistical independence.....am I right in saying this?

Thanks!
Jayanthi

#### David Harper CFA FRM

##### David Harper CFA FRM
Staff member
Subscriber
Hi @Jayanthi Sankaran Yes, that's a good observation! Bayes says P(A|B) = P(B|A)*P(A)/P(B) but if A and B are independent then P(B|A) = P(B) and P(A|B) = P(A)--these are by definition, right?--such that I don't see how Bayes adds value if A and B are independent because P(A|B) = P(B|A)*P(A)/P(B) then becomes tautological P(A) = P(B)*P(A)/P(B) ... so yes, unless I am missing something, the entire premise of Bayes depends on some kind of dependence (pun intended) and doesn't tell us much under independence. Thanks!

#### brian.field

##### Well-Known Member
Subscriber
Just a quick note - I did mention that the x^a needed to be surrounded by parentheses.

I guess I should have written ((x^a))^b

#### ShaktiRathore

##### Well-Known Member
Subscriber
Jayanthi,
Yes Bayes does not require independence P(A&B)=P(A)*P(B) to hold.Bayes holds both for independent and dependent events A and B. Independence rule is a special case of Bayes. P(A&B)=P(A|B)*P(B) is what Bayes says, now if A and B are independent then P(A|B)=P(A) as David cited above put this condition in Bayes we get P(A&B)=P(A)*P(B) which is the independence rule.Therefore Bayes does not require independence rule otherwise its insignificant as David said but its the vise versa that Bayes theorem is required to hold for independence rule to be true.
Thanks

#### Dr. Jayanthi Sankaran

##### Well-Known Member
Thanks - David and Shakti for the lucid and clear explanation!

Jayanthi

#### brian.field

##### Well-Known Member
Subscriber
To offer some additional thoughts, (which have already been mentioned more or less above,) I always considered them equivalent.

P(A|B) = P(A) iff A and B are independent.

But we also know that P(A|B) = P(A and B) / P(B) and since A and B are independent, we have P(A and B) = P(A)P(B),
so .....
P(A|B) = P(A and B) / P(B) = P(A)P(B) / P(B) and cancelling the P(B)'s, we have P(A|B) = P(A).

Replies
4
Views
4K