Question

How do I proof each problem?

1) For the uniform discrete distribution with possible values 1,2,3,..n, prove that ������x =n+1/2

2) For the binomial distribution with parameters n and ������ prove that ������x = n������

Answer 1

For a uniformly distributed discrete random variable with support \(x=1,...,n\), you have \(P(X=k)=\dfrac{1}{n}\) for \(k=1,2,...,n\), i.e. the probability for any possible outcome is \(\dfrac{1}{n}\).

The mean is given by
\[\large\sum_{x=1}^nx~p(x)=\sum_{x=1}^n\frac{x}{n}\]
Simplify this expression.

Answer 2

For the second question, you have the binomial distribution with parameter \(\theta\) and support \(k=0,1,...,n\), and the probability \(P(X=k)=\dbinom nk \theta^k(1-\theta)^{n-k}\).

The mean here is given by
\[\large\sum_{x=0}^nx~p(x)=\sum_{x=0}^n x\binom nx \theta^x(1-\theta)^{n-x}\]
Rearrange the binomial coefficient:
\[\begin{align*}x\binom nx&=x\frac{n!}{x!(n-x)!}\\\\
&=n\frac{(n-1)!}{(x-1)!(n-1-x+1)!}\\\\
&=n\frac{(n-1)!}{(x-1)!((n-1)-(x-1))!}\\\\
&=n\binom {n-1}{x-1}\end{align*}\]
The mean is then
\[\large\sum_{x=0}^nx~p(x)=\sum_{x=0}^n n\binom {n-1}{x-1} \theta^x(1-\theta)^{n-x}\]
You want to show that the mean is \(n\theta\), which you can do by factoring this out and showing that the remaining summation is equal to 1:
\[\large\sum_{x=0}^nx~p(x)=n\theta\underbrace{\sum_{x=0}^n \binom {n-1}{x-1} \theta^{x-1}(1-\theta)^{n-x}}_{\text{show this }=~1}\]

Answer 3

thanks alot