Let x= 148.41^.95%
Take log both sudes
Find rhs from log table,
Then find antilog of obtained rhs value,then this antilog is x the value we yearned for.
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.
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?
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.
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?
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!
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.
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),
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).