Lognormal process

notjusttp

New Member
Hi David,

I have 2 questions

1) How did we get this answer?

If X(t) follows a lognormal process then the correlation between X(t) and 1/X(t) is:
Choose one answer.
a. -1
b. 1
c. ½
d. -½

The correct answer is -1

2) What, according to your experience, is the minimum % of marks we should be able to garner to clear FRM exam ( eg 35%,50% etc).

Thanks and Best rgds
Amit
 

David Harper CFA FRM

David Harper CFA FRM
Subscriber
Hi Amit,

1) y = 1/ln(x) so that dy/dx = -(1/(x*ln(x)^2)) so the slope of 1/ln(x) on the y against ln(x) on the x looks negative but also non-constant... so i don't get the answer... maybe i have brain farted to misunderstand the question but -1 correlation requires a constant first derivative; e.g., y = -3x such that dy/dx = -3 and that relationship has a -1 correlation...do you have an answer, can you please share if you do?

2) The history isn't applicable: this is the first time in garp's new format. I sincerely believe the answer is not ex ante knowable, even to GARP.

Thanks, David
 

notjusttp

New Member
Hi David,

Thanks for the clarification, I did not have any explanation. The source had only given answer directly. So sorry i dont have anything additional which i can share.

Thanks and warm rgds..Amit
 

David Harper CFA FRM

David Harper CFA FRM
Subscriber
Hi Amit,

I was thinking maybe the question refers to: Y = 1/X = X^(-1)
...if we take log of both sides: LN(Y) = -1*LN(X); i.e., log-linear transform
...per Gujarati Ch 9, this has a -1 correlation if we graph LN(Y) against LN(X)
...but I feel that would be an unfair way to ask about it

More likely the question confuses the more general "dependence" (i.e., y=1/X has perfect dependence but "dependence" allows for non-lieanr) with correlation, which is a limited type of dependence with flaws (e.g., only a test of linear dependence)

so, it is helpful as a reminder that correlation (coefficient) has a very specific, limited definition that will not cover a much broader class of dependencies

...David
 
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