Question

Let the discrete random variable X have probability mass function p(x) such that
p(x) =1/n x= a+1, a+2,a+3,...,a+n
=0 elsewhere
what is the mean of this distribution?

Answer 1

the formula for mean is
\[\sum_{a+1}^{a+n}x*p(x)\]
can u find this?

Answer 2

not sure...would i need to try simplify the sum first?

Answer 3

u need to put the value of p(x) first.,then evaluate the summation
ever evaluated the summation??

Answer 4

like i wouldnt know how to evaluate something with\[\sum_{a+1}^{a+n}\]

Answer 5

\[\sum_{a+1}^{a+n} x*1/n ?\]

Answer 6

there are formulae to evaluate the summation,like
\[\sum_{1}^{n}1=n\]
and
\[\sum_{1}^{n}n=n*\frac{(n+1)}{2}\]

Answer 7

u made that substitution correct....
now 'n' is a constant for summation,because u are summing w.r.t x,so u can bring out n like this
\[\frac{1}{n}\sum_{a+1}^{a+n}x\]
got this?

Answer 8

yes i agree. unfortunately the a+1 part of the sum is still a little bit confusing for me...

Answer 9

ok,but till here you understood,right?
i will go further and substitute y=x-a to get rid of a+1
so now lower limit changes from x=a+1 to y=1
got this?if yes,then tell me what would upper limit x=a+n change to?

Answer 10

would it change to just n?

Answer 11

\[like \sum_{y=1}^{n}\]

Answer 12

absolutely correct :)
so now you have
\[\sum_{1}^{n}y-a\]
does it look familiar to any formula i have given??

Answer 13

if you are not comfortable with substitution,then i have another approach also...want that?

Answer 14

Yes please

Answer 15

\[\sum_{1}^{n}y - an\]

Answer 16

ok,lets stick to
\[\sum_{a+1}^{a+n}x\]
this implies the sum starting from x=a+1 to x=a+n
so when substituted,we get
a+1+a+2+a+3+a+4+.......a+n
got this?

Answer 17

yes

Answer 18

good :)
u agree that there will be 'n' times a ---->n*a
and the sum from 1 to n.....like 1+2+3+4+....n

Answer 19

would it be \[\frac{ n(n+1) }{ 2 }-an ?\] for the mean then?

Answer 20

yes

Answer 21

from where did u get the minus??also u forgot to recall about 1/n outside the summation

Answer 22

\[from \sum_{1}^{n}y-a\]

Answer 23

if i bring in the 1/n, then i would have (n+1)/2 +/- an?

Answer 24

lets follow only one method,(forget about substitution for time being)you got a*n because there were 'n' times a
also 1+2+3+4....+n---->n(n+1)/2
so
\[\frac{1}{n}(an+\frac{n(n+1)}{2})\]
will give you a+(n+1)/2
which is the mean

Answer 25

oooh.... i see now thanks. I would follow the same method if find variance?

Answer 26

i made a mistake in substitution
as y=x-a
x=y+a
so summation would be for y+a
and you would get same answer (n+1)/2 +a and not -
yes,for variance there is also the formula like mean......do u need it?
same steps can be followed,but it would be more difficult because it involved square terms.....

Answer 27

yes please if it is not too much trouble?

Answer 28

sure :)
\[\sum_{a+1}^{a+n}(x-MEAN)^{2}f(x)\]

Answer 29

oh sorry p(x) not f(x)

Answer 30

ok so not the usual V(x) = E[X^2] - (E[X])^2 ? because it would be more efficient to find E[X^2]? or is that where it gets complicated?

Answer 31

that usual expression is another representation of variance
you will calculate E[x^2]as
\[\sum_{a+1}^{a+n}x^{2}p(x)\]
and then subtract square of mean
both expression will give same value
although which one will be less complicated will depend on function

Answer 32

Thanks so much for the help! ^_^ would the variance be (2n+1)(n+1)/6 + 2a^2 + a ?

Answer 33

i am not getting that,can u show your work??
also since there is - sign in th formula, 2 'a^2's gets cancelled and does not add up to 2a^2

Answer 34

ok. i started off with \[\frac{ 1 }{ n }\sum_{1}^{n}(y+a )^2\]

Answer 35

is that correct so far?

Answer 36

yes,correct

Answer 37

but for E[x^2]
remember to subtract the square of mean from this for variance

Answer 38

ok, so i just squared it and i have