Question

Quick question: when using sigma notation, when you shift the lower index by x amount do you also have to add or subtract that amount from the upper index?

Answer 1

nope
but you do have to alter to values inside in sigma

Answer 2

\[\sum_{k=1}^{n} = ((n-1)!/(k-1)!(n-k-1)!)*p^{k-1}(1-p)^{n-k}\]

There we go, if I wanted to shift to k=0 how would it change?

Answer 3

lets see what we do to a simple example

when n=3, 2+n = 5

say we shift the index (n) by -1 but we want to retain the same value

when n=3-1, 2+(n+1) = 5

Answer 4

basically decrease index increase value by same amount inside

usually you can tell it will be good. Look at all those -1's falling away

Answer 5

if we desire to shift the index by "a"; but retain the same value

then n= 3+a; 2+(n-a) = 5

Answer 6

I need to get rid of the n-1's as well, I was hoping simply changing the lower index would do that

Answer 7

That is, they need to be just n's

Answer 8

only your "k" parts are affected; since thats the driving part of the summation

Answer 9

How would I affect the n parts? substitution?

Answer 10

this seems to be part of a larger question; what is the bigger question that you are trying to work out?

Answer 11

but you can change (n-1)! into n!/n-1

Answer 12

the "n"s are just the stopping value; you cant affect them that I can recall

Answer 13

in the same way that you can change (n+1)! into (n+1)!*n!

Answer 14

\[\sum_{k=1}^{n} = \frac{(n-1)!}{(k-1)!!(n-k-1)!}~p^{k-1}(1-p)^{n-k}\]
\[\sum_{k=1-1}^{n} = \frac{(n-1)!}{(k+1-1)!(n-(k+1)-1)!}~p^{k+1-1}(1-p)^{n-(k+1)}\]
\[\sum_{k=0}^{n} = \frac{(n-1)!}{(k)!(n-k-2))!}~p^{k}(1-p)^{n-k-1}\]
hmmm

Answer 15

if the upper bound was "fixed", then you could adjust it; but as is, there is no specific stopping point on it to adjust

Answer 16

I made a mistake

\[\sum_{k=1}^{n} = ((n-1)!/(k-1)!(n-k)!)*p^{k-1}(1-p)^{n-k}\]

thats what it should be

Answer 17

\[\sum_{k=1}^{n} = \frac{(n-1)!}{(k-1)!(n-k)!}~p^{k-1}(1-p)^{n-k}\]
\[\sum_{k=0}^{n} = \frac{(n-1)!}{(k)!(n-k-1)!}~p^{k}(1-p)^{n-k-1}\]
still not working out the way you want it too :/

Answer 18

and trying to get to \[\sum_{k=0}^{n} = (n!/k!(n-k)!)*p^{k}(1-p)^{n-k}\]

Answer 19

which is the binomial distribution

Answer 20

you might have made an error getting it to this point. This is not the start of the question is it???

Answer 21

Im trying to prove E(X) = np for binomial but Ive gotten to this point, so close, just need to show that sum is \[\sum_{k=0}^{n} = (n!/k!(n-k)!)*p^{k}(1-p)^{n-k}\] which is 1 and then Id be done

Answer 22

not sure how to manipulate it further

Answer 23

http://www.math.ubc.ca/~feldman/m302/binomial.pdf

this might be a better read for you then

Answer 24

ah I did need substitution, thanks for that link, Ill see if I can keep going knowing what substitution to make.