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 Short-Term Actuarial Math Old Exam C Forum

#1
01-13-2015, 08:09 PM
 PrinceNReserve Member CAS SOA Join Date: Aug 2012 Posts: 131
Evaluating binomial coefficients with fractions

Can someone explain to me why 2.5 choose 2 = 1.875? Thanks!
#2
01-13-2015, 08:17 PM
 Abelian Grape Meme-ber                         Meme-ber CAS Join Date: Jul 2014 Favorite beer: Allagash Curieux Posts: 42,069

Where did you come across that? In this exam you should only come across integer binomial coefficients, at least in my experience. In any case,
$\frac{2.5!}{2!\left(2.5-2\right)!}$
You can use the gamma function or just remember that $0.5!=\frac{\sqrt{\pi}}{2}$ and $n!=n\left(n-1\right)!$.
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Quote:
 Originally Posted by Pension.Mathematics what's your problem man?

Last edited by Abelian Grape; 01-13-2015 at 08:23 PM..
#3
01-13-2015, 08:53 PM
 clarinetist Member Non-Actuary Join Date: Aug 2011 Studying for Rcpp, Git Posts: 6,885

Quote:
 Originally Posted by Abelian Grape Where did you come across that? In this exam you should only come across integer binomial coefficients, at least in my experience.
Not necessarily. I assume the OP is going off the ASM convention, where ${x \choose n} = \dfrac{x(x-1)\cdots(x-n+1)}{n!}$. So in this case,
${2.5 \choose 2} = \dfrac{2.5(2.5-1)}{2!} = 1.875$.

I doubt the Gamma function is involved, since the final answer is very "clean."
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#4
01-13-2015, 08:59 PM
 Academic Actuary Member Join Date: Sep 2009 Posts: 8,414

Quote:
 Originally Posted by clarinetist Not necessarily. I assume the OP is going off the ASM convention, where ${x \choose n} = \dfrac{x(x-1)\cdots(x-n+1)}{n!}$. So in this case, ${2.5 \choose 2} = \dfrac{2.5(2.5-1)}{2!} = 1.875$. I doubt the Gamma function is involved, since the final answer is very "clean."
The gamma's cancel in the numerator and denominator.
#5
01-13-2015, 09:12 PM
 clarinetist Member Non-Actuary Join Date: Aug 2011 Studying for Rcpp, Git Posts: 6,885

Quote:
 Originally Posted by Academic Actuary The gamma's cancel in the numerator and denominator.
Lemme see.

${2.5 \choose 2} = \dfrac{\Gamma(3.5)}{\Gamma(3)\Gamma(2.5-2+1)} = \dfrac{\Gamma(3.5)}{2\Gamma(1.5)} = \dfrac{2.5(1.5)\Gamma(1.5)}{\Gamma(1.5)} = \dfrac{2.5(1.5)}{2} = 1.875$.

Okay, so it does.

Although, I would not suggest doing it this way on the exam, since the formula can be easily figured out from the C tables.
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#6
01-13-2015, 09:19 PM
 PrinceNReserve Member CAS SOA Join Date: Aug 2012 Posts: 131

Quote:
 Originally Posted by clarinetist Not necessarily. I assume the OP is going off the ASM convention, where ${x \choose n} = \dfrac{x(x-1)\cdots(x-n+1)}{n!}$. So in this case, ${2.5 \choose 2} = \dfrac{2.5(2.5-1)}{2!} = 1.875$. I doubt the Gamma function is involved, since the final answer is very "clean."
Thanks. I did see it from the ASM. Then, would

${2.5 \choose 3} = \dfrac{2.5(2.5-1)(2.5-2)}{3!} = 0.3125$. ?
#7
01-13-2015, 09:23 PM
 clarinetist Member Non-Actuary Join Date: Aug 2011 Studying for Rcpp, Git Posts: 6,885

Quote:
 Originally Posted by PrinceNReserve Thanks. I did see it from the ASM. Then, would ${2.5 \choose 3} = \dfrac{2.5(2.5-1)(2.5-2)}{3!} = 0.3125$. ?
Yep, you're good there - but I wouldn't bother recognizing that formula. If you need it, 99% chance you would use it for the negative binomial probability function, and it's right in the C tables.
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#8
01-13-2015, 09:29 PM
 Abelian Grape Meme-ber                         Meme-ber CAS Join Date: Jul 2014 Favorite beer: Allagash Curieux Posts: 42,069

Quote:
 Originally Posted by clarinetist Yep, you're good there - but I wouldn't bother recognizing that formula. If you need it, 99% chance you would use it for the negative binomial probability function, and it's right in the C tables.
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Quote:
 Originally Posted by Pension.Mathematics what's your problem man?