Question 16 in Quant Round 1

[sridhar]

You've provided the details of the solution here:

http://www.editgrid.com/bt/frm_2008/quant1_16

I am confused about something...Using the terms in this problem --

E(x) = x1*f(x1) + x2*f(x2)..... = sigma(x(i) * f(xi))

This should give you 40b....But you seem to calculate:

E(x) as 40 / (sigma (f(x1) + f(x2) + ....)

I am not clear what the basis is for doing this...I would have thought:

E(x) = 40b and similarly E(x * x) = 130b

But again you are showing E(x * x) as 130 / (sigma (f(x1) + f(x2) + ....)

I am confused...Why do you have the sum of the PDFs in the denominator?

According to me:

VaR(x) should be: 130b - (40b) ** 2

Which cannot be further reduced until we know what b is? I am missing something, but don't know what.

--sridhar

 

[sridhar]

PS to my own post....

I said: Var(x) = 130b - 160b^2

we get the value of b from the fact that sigma(f(x)) = 1, i.e. 15b = 1 and b = 1/15

plugging this into the above, I get VaR(x) = 7.96 and SD(x) = 2.82

I am still persuadable to your answer, if only you tell me why

 

[David Harper CFA FRM]

I may need to look at the question phrasing (i twisted/rephrased an old question for this) but please note: f(x) does not equal p(x). Like density distribution will have y-axis values that exceed 1.0. I am having a bit of trouble quickly following your math, but note that your E(x) appears to exceed 4.0 which is the highest X; the E(x) must be within the [0,4] support.

Your following is not quite true:
E(x) = x1*f(x1) + x2*f(x2)..... = sigma(x(i) * f(xi))

We need:
E(x) = x1*p(x1) + x2*p(x2)..... = sigma(x(i) * p(xi))

For example, p(x1 = 1) = 2/40

Given that, I *think* we will get the same?

(I am headed out to lunch, but if you require more persuasion please do say )

David

 

[fashepard]

I know how to calculate a standard Dev but the working in this question has me confused. I did not know where to start

Frank

 

[fashepard]

David,

I see how you are doing it. Why do we not use means sum of squared deviations in this calculation?

Frank

 

[fashepard]

David,

Can you explain the difference between the variance formulas on page 1 of 79 and page 3 of 79 OF YOUR FORMULA SHEET. I think that if I get that I will be ok

 

[fashepard]

I am sorry David. It is the Variance formulas on page 2 and 4

Frank

 

[David Harper CFA FRM]

Frank,

I modified the "variance of a single die" to illustrate the issue, hopefully. There is no conceptual difference between the two formulas; rather it concerns the type of sample.

The variance is always the expected value of square dispersion around the mean.
Please take a look at this XLS for two different perspectives.
Two variances are calculated (left and right). Both concern the variance of a single six-sided die.
First is "before the roll:" notice I have two/three ways to compute the variance, they all give 2.92; i.e., the variance of a fair die. This is variance against the discrete uniform distribution (i like just "before the roll)"

Second, to the right, is "after the roll:" this simulates six rolls
(you can select F9 to recalculate and give new rolls - new rolls will give a different variance)
In this case, the parametric distribution (i.e., p = 1/6th for each outcome) gives way to an empirical distribution (results will vary)
The variance will give a different result but is conceptually the same idea. Just this time, we are calculating the variance of an actual (historical) sample instead of a probability distribution.

I think this is the issue, let me know...

David