SERVING THE QUANTITATIVE FINANCE COMMUNITY

 
MBVLA
Topic Author
Posts: 1
Joined: January 7th, 2019, 1:50 pm

35th root of 250

January 7th, 2019, 2:08 pm

Hi everyone,

I was watching an interview of Edward Thorpe in which he mentally calculated
his annualized return on stocks over a 35 year horizon by computing the 35th root
of 250 (percentage return) by taking e^(5.6/35) to get approximately 1.17, which
is the right answer. 

Does anyone know where he got the 5.6 from and why the natural log? 
 
User avatar
bearish
Posts: 4153
Joined: February 3rd, 2011, 2:19 pm

Re: 35th root of 250

January 7th, 2019, 6:32 pm

Hi everyone,

I was watching an interview of Edward Thorpe in which he mentally calculated
his annualized return on stocks over a 35 year horizon by computing the 35th root
of 250 (percentage return) by taking e^(5.6/35) to get approximately 1.17, which
is the right answer. 

Does anyone know where he got the 5.6 from and why the natural log? 
OK - you get a point for posting your question in the right forum. e^5.6 is roughly 250 (well, it’s actually 270), and since taking the 35th root is the same as raising to the power of 1/35 , e^(5.6/35) is approximately one plus the annual return. Since 5.6/35 is a little less than 1/6 we can call it .16 and a second order approximation of the exponential gives you 17% and change. Where the rabbit went into the hat was when Thorp memorized tables of logarithms, which makes quick calculations like these relatively straightforward.
 
User avatar
ppauper
Posts: 69846
Joined: November 15th, 2001, 1:29 pm

Re: 35th root of 250

January 7th, 2019, 10:30 pm

more generally, you can write [$]x^k=e^{k\ln x}[$]
you want [$]250^{1/35}=e^{(\ln 250)/35}[$] and as bearish said, Ed Thorpe approximated [$]\ln 250\approx 5.5215[$] by 5.6, It's easy to divide 5.6 by 35
 
 
User avatar
Cuchulainn
Posts: 57641
Joined: July 16th, 2004, 7:38 am
Location: Amsterdam
Contact:

Re: 35th root of 250

January 8th, 2019, 10:27 am

It's probably a 'trick of the trade' Here one to compute [$]e^5[$] (yugely popular thread, btw) from zerdna

I could give a pretty good approximation without using pen and paper, although it's not going to be two decimal digits. I will use the fact that e^3 ~ 20.Knowing that,e^5 ~ 400/e ~ 400/3*(1-0.1) ~ 133.33 *(1 +0.1 +0.01+0.001) ~ 133.33+13.33+1.33+0.13 = 148.12First two additions 133.33+13.33 i could do in a second without pen and paper, so i can easily get 146.66. Beyond that i don't know
ABOUT WILMOTT

PW by JB

Wilmott.com has been "Serving the Quantitative Finance Community" since 2001. Continued...


JOBS BOARD

JOBS BOARD

Looking for a quant job, risk, algo trading,...? Browse jobs here...


GZIP: On