Back testing VAR - Unconditional LR

[kaushik1111]

Hi,

I am not able to calculate LR with the equation as values are gettung too high.

For example, (1-p)^252-n

Calculator is not able to calculate this value

Any suggestions?

Kaushik

050 2422349

[not_a_CFA]

you could try and find an approximate value.

remember the definition of e^x

e^x = sum from k = 0 to infinity of x^k / k!

maybe approximating to the cubic power will be enough. you could add more powers for better accuracy.

for this problem, you have A^B where A = 1 - p and B = 252 - n

on second thoughts, you may need higher powers after all.

for example, when approximating 2³ using the definition of e^x,

approximating to the 3rd power gives me 6.74, which is highly inaccurate.

adding up to the 5th power gives me 7.843, which is somewhat close, but not quite.

adding the 6th power gives me 7.955, which is closer

going up to the 8th power gives me 7.9975 which is very, very close to the true value.

and of course, if you added all the powers all the way to infinity, you would get 8 as the answer.

[S2000magician]

@not_a_CFA

for example, when approximating 2³ using the definition of e^x . . . .

What, exactly, are you using for x here?

It seems that you’re taking a simple calculation and making it (substantially) more complicated.

[not_a_CFA]

@S2000magician

@not_a_CFA

for example, when approximating 2³ using the definition of e^x . . . .

What, exactly, are you using for x here?

It seems that you’re taking a simple calculation and making it (substantially) more complicated.

im just using it as an example. of course, you cant use the approximation directly. since the base is not e, you need to recall that a^b = e^{ln (a^b)} and proceed with the approximation.

i expect that you could use this to calculate powers that are too big for the calculator

(one of my books had a formula, but i think it is based on the definition of e^x)

and its not that complicated. it is just an application of MacLaurin, which is taught in basic calculus courses.

something i would expect frm candidates to know.

[S2000magician]

@not_a_CFA

@S2000magician

@not_a_CFA

for example, when approximating 2³ using the definition of e^x . . . .

What, exactly, are you using for x here?

It seems that you’re taking a simple calculation and making it (substantially) more complicated.

im just using it as an example. of course, you cant use the approximation directly. since the base is not e, you need to recall that a^b = e^{ln (a^b)} and proceed with the approximation.

i expect that you could use this to calculate powers that are too big for the calculator

(one of my books had a formula, but i think it is based on the definition of e^x)

and its not that complicated. it is just an application of MacLaurin, which is taught in basic calculus courses.

something i would expect frm candidates to know.

And, of course, a^b = e^{ln (a^b)} = e^blna, but I’m pretty sure that that won’t solve the problem, and it makes it ridiculously more time-consuming. Not to mention that the author of an FRM problem is not remotely likely to expect a candidate to invoke a Maclaurin series.

I suspect that the problem is simply that (1 − p)^{252 − n} is extremely tiny unless p is really, really small or n is really close to 252. For example, if p = 0.5 and n = 2, (1 − p)^{252 − n} = 0.5²⁵⁰ ≈ 5.53×10⁻⁷⁶; you’ re going to need a lot of terms (i.e., hundreds, which your calculator won’t do) to get an approximation anywhere close to that.

[not_a_CFA]

@S2000magician

@not_a_CFA

@S2000magician

@not_a_CFA

for example, when approximating 2³ using the definition of e^x . . . .

What, exactly, are you using for x here?

It seems that you’re taking a simple calculation and making it (substantially) more complicated.

im just using it as an example. of course, you cant use the approximation directly. since the base is not e, you need to recall that a^b = e^{ln (a^b)} and proceed with the approximation.

i expect that you could use this to calculate powers that are too big for the calculator

(one of my books had a formula, but i think it is based on the definition of e^x)

and its not that complicated. it is just an application of MacLaurin, which is taught in basic calculus courses.

something i would expect frm candidates to know.

And, of course, a^b = e^{ln (a^b)} = e^blna, but I’m pretty sure that that won’t solve the problem, and it makes it ridiculously more time-consuming. Not to mention that the author of an FRM problem is not remotely likely to expect a candidate to invoke a Maclaurin series.

I suspect that the problem is simply that (1 − p)^{252 − n} is extremely tiny unless p is really, really small or n is really close to 252. For example, if p = 0.5 and n = 2, (1 − p)^{252 − n} = 0.5²⁵⁰ ≈ 5.53×10⁻⁷⁶; you’ re going to need a lot of terms (i.e., hundreds, which your calculator won’t do) to get an approximation anywhere close to that.

you may be right, i haven’t considered the need for summing too many terms, thats my bad.

i only now noticed that there is no need for approximation. you can simply use the property of log and change base of logs to find the answer in base 10.

for example, to calculate .5²⁵⁰ (log is in base 10)

.5²⁵⁰ = 10^{[log .5^250]} = 10^{(250 x log .5)}

= 10^{(250 x (ln .5 / ln 10) )} = 10^{(250 x -.301029996)}

= 10^-75.257499 = 5.527 x 10^-76

this means that, for this problem,

(1 - p)^{(252 - n)} = 10^{[(252 - n) x ln (1 - p) / ln 10]}

using the above formula, you can find, for example, that

.15²⁵⁰ = 10^-205.9771852 = 1.05393736 x 10^-206

while the calculator gives 0 as the answer.

[S2000magician]

@not_a_CFA

@S2000magician

@not_a_CFA

@S2000magician

@not_a_CFA

for example, when approximating 2³ using the definition of e^x . . . .

What, exactly, are you using for x here?

It seems that you’re taking a simple calculation and making it (substantially) more complicated.

im just using it as an example. of course, you cant use the approximation directly. since the base is not e, you need to recall that a^b = e^{ln (a^b)} and proceed with the approximation.

i expect that you could use this to calculate powers that are too big for the calculator

(one of my books had a formula, but i think it is based on the definition of e^x)

and its not that complicated. it is just an application of MacLaurin, which is taught in basic calculus courses.

something i would expect frm candidates to know.

And, of course, a^b = e^{ln (a^b)} = e^blna, but I’m pretty sure that that won’t solve the problem, and it makes it ridiculously more time-consuming. Not to mention that the author of an FRM problem is not remotely likely to expect a candidate to invoke a Maclaurin series.

I suspect that the problem is simply that (1 − p)^{252 − n} is extremely tiny unless p is really, really small or n is really close to 252. For example, if p = 0.5 and n = 2, (1 − p)^{252 − n} = 0.5²⁵⁰ ≈ 5.53×10⁻⁷⁶; you’ re going to need a lot of terms (i.e., hundreds, which your calculator won’t do) to get an approximation anywhere close to that.

you may be right, i haven’t considered the need for summing too many terms, thats my bad.

i only now noticed that there is no need for approximation. you can simply use the property of log and change base of logs to find the answer in base 10.

for example, to calculate .5²⁵⁰ (log is in base 10)

.5²⁵⁰ = 10^{[log .5^250]} = 10^{(250 x log .5)}

= 10^{(250 x (ln .5 / ln 10) )} = 10^{(250 x -.301029996)}

= 10^-75.257499 = 5.527 x 10^-76

this means that, for this problem,

(1 - p)^{(252 - n)} = 10^{[(252 - n) x ln (1 - p) / ln 10]}

using the above formula, you can find, for example, that

.15²⁵⁰ = 10^-205.9771852 = 1.05393736 x 10^-206

while the calculator gives 0 as the answer.

For this example, you might as well use zero.

A bigger problem arises when n is, say, 150. Not only do you have to compute 0.15¹⁵⁰ and 0.85¹⁰², you also have to compute ₂₅₂C₁₅₀. Your calculator might not be able to do that.

Overall, the correct reply to the original question is that this isn’t going show up on exam with such huge numbers.

[not_a_CFA]

.15^ 150 and .85^ 102 can be easily found using the formula above.

as for combinatorics, by using the property of logarithms, one can find that

n! = floor (10 ^ [(ln 1 + ln 2 + … + ln n) / ln 10] )

n choose k is equal to n! / (k! (n - k)! )

using the formula above, this simplifies to 10 ^a / (10^b x 10^c)

for some numbers a, b and c one can then just use the property of exponents to arrive at an answer.

the complication here is finding the sum ln 1 + ln 2 + … + ln n in a reasonable amount of time.

it is easy when n = 10. this is not so if n is, say, 200.

[Rasoaisi]

Hi all;

I have just registered to sit for FRM Part I in May 2020. May anyone add me in Whatsapp groups already created for communication and discussions with other candidates.

What i need most is the e books in pdf format. I cannot always have data to read books online.

Your assistance would be highly appreciated.

My Whatapp number is 00266 5062 0799 OR +266 5062 0799

[not_a_CFA]

from my understanding, CFA Institute sells physical copies of the textbook, which you can purchase on sites like Amazon, or directly from the Institute itself.

the same is likely true for FRM