Question

A) For the uniform discrete distribution with possible values 1,2,3,...n, prove that μx = n+1/2
B) For the binomial distribution with parameters n and 0, prove that
b(0;n,θ)+b(1;n,θ)+....+b(n;n,θ) = 1
C) For the binomial distribution with parameters n and θ prove that
μx = nθ

Answer 1

what can we use for the first one

Answer 2

can we use \[\mu_x=x_1p_1+x_2p_2+\cdots +x_np_n\]

Answer 3

if so the proof should be easy for that one

Answer 4

oh

Answer 5

\[p_i=\frac{1}{n} \text{ for all } i=1,2,3,...,n \\ \text{ and we also know } \\ 1+2+\cdots +n=\frac{n(n+1)}{2}\]

Answer 6

see if you can use those pieces to put your proof together

Answer 7

in the second one what does b(i,n, theta) mean ?

Answer 8

b(0;n,θ)+b(1;n,θ)+....+b(n;n,θ) = 1

Answer 9

is a #1

Answer 10

oh yeah sorry
a is 1
b is 2
c is 3
I forgot they were lettered and not numbered

Answer 11

i guess b(i,n,theta)
just means the probability of i happening given n the number of trials and theta being the probability of success

Answer 12

ohhh

Answer 13

so I'm thinking \[b(i,n, \theta)= \left(\begin{matrix}n \\ i\end{matrix}\right) \theta^i(1-\theta)^{n-i}\]

Answer 14

so what values do i need to plug in there?

Answer 15

you need to show \[\sum_{i=0}^{n} b(i, n, \theta)=1\]

Answer 16

that is you want to show
\[\left(\begin{matrix}n \\ 0\end{matrix}\right) \theta^0(1-\theta)^{n-0}+\left(\begin{matrix}n \\ 1\end{matrix}\right) \theta^1(1-\theta)^{n-1}+\left(\begin{matrix}n \\ 2\end{matrix}\right) \theta^2(1-\theta)^{n-2} + \\ \cdots + +\left(\begin{matrix}n \\n-1\end{matrix}\right)\theta^{n-1}(1-\theta)^{n-(n-1)}+\left(\begin{matrix}n \\ n\end{matrix}\right) \theta^n(1-\theta)^{n-n}=1\]

Answer 17

ohhh i got it now

Answer 18

and yes I know that there is a ++
that was a type-o
should just be +

Answer 19

no problem

Answer 20

like the last one should be similar to the first since they are both discrete

Answer 21

except i do think the third one will be a bit harder to show than the first

Answer 22

what i mean by similar is that
you still want to use
\[\mu_x=x_0p_0+x_1p_1+x_2p_2+\cdots +x_n p_n\]
where
\[p_i=b(i,n, \theta)=\left(\begin{matrix}n \\ i\end{matrix}\right) \theta^{i} (1-\theta)^{n-i}\]

Answer 23

\[\mu_x=0 \cdot p_0+1 \cdot p_1+2 \ \cdot p_2+\cdots +n \cdot p_n \\ \mu_x=1p_1+2p_2+\cdots +np_n\]

Answer 24

oh i see

Answer 25

this might help you
https://proofwiki.org/wiki/Expectation_of_Binomial_Distribution

Answer 26

thanks a lot for everything

Answer 27

as we do see here theta+(1-theta)=1

Answer 28

just like there p+q=1

Answer 29

and no problem

Answer 30

i hope it helped :)

Answer 31

You really did a great job!

Answer 32

thanks :)