# Thread: Mean derivation

1. ## Binomial mean derivation

Wikipedia: Algebraic derivation of mean says:
$\displaystyle {E}(X) = \sum_k x_k \cdot Pr(x_k) = \sum_{k=0}^n k \cdot Pr(X=k)$

Sure, $\displaystyle X$ is "a discrete random variable", but does that inevitably mean that $\displaystyle x_k = k$? Couldn't $\displaystyle x_k$ all be, say, multiples of 3 ($\displaystyle x_2 = 6$)?

And in next step they use this to say
The first term of the series (with index k = 0) has value 0 since the first factor, k, is zero.
which is an essential part of proving that if $\displaystyle X\sim(n,p) \Rightarrow E(X)=np$ (which is why I'd like to understand why should $\displaystyle x_k = k$).

2. At the start of their derivation note that the sum is from k=0 to n. That's why they take x sub k = k. The part with x sub k is just a general definition.

After that, the first term is k=0 because they are considering values of k from 0 to n

3. Originally Posted by qmech
At the start of their derivation note that the sum is from k=0 to n. That's why they take x sub k = k. The part with x sub k is just a general definition.
Does from "At the start of their derivation note that the sum is from k=0 to n." always follow "... take xk = k." That is, $\displaystyle \sum_k x_k=\sum_{k=0}^n k$? I'm not saying it doesn't, I'd like to understand why.
Originally Posted by qmech
After that, the first term is k=0 because they are considering values of k from 0 to n
I understood this.

4. Look back at the Wikipedia page. You're focussing on the equations after the line saying "We apply the definition...". Look at the paragraph before this starting with "We derive these quantities from first principles..". The first equation after the text is a sum from 0 to N. That's what's defining this derivation.

5. Originally Posted by qmech
[...] Look at the paragraph before this starting with "We derive these quantities from first principles..". [...]
Paragraph before: $\displaystyle \sum_{k=0}^n Pr(X=k) = \sum_{k=0}^n {n\choose k}p^k(1-p)^{n-k} = 1$.
Let me see, $\displaystyle k$ is here "how many times an event happens": $\displaystyle \underbrace{a_p, a_p, a_p, ..., a_p}_\textrm{k}, \underbrace{a_q, a_q, a_q, ..., a_q}_\textrm{n-k}$ and $\displaystyle {n\choose k}$ gives all arrangements. This is correct, right?

Now, "we apply the definition of the expected value of a discrete random variable to the binomial distribution". ....
... I see now. The case of, say, dice with sides $\displaystyle {2,3,4,5,6,7}$ doesn't apply to binomial distribution (which deals only with "yes/no" types).
Have I got it now?

### mean derivation

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