# Mean derivation

• Jan 17th 2010, 06:15 AM
courteous
Binomial mean derivation
Wikipedia: Algebraic derivation of mean says:
${E}(X) = \sum_k x_k \cdot Pr(x_k) = \sum_{k=0}^n k \cdot Pr(X=k)$

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

And in next step they use this to say
Quote:

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 $X\sim(n,p) \Rightarrow E(X)=np$ (which is why I'd like to understand why should $x_k = k$).
• Jan 19th 2010, 12:29 PM
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.

After that, the first term is k=0 because they are considering values of k from 0 to n
• Jan 20th 2010, 05:20 AM
courteous
Quote:

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, $\sum_k x_k=\sum_{k=0}^n k$? I'm not saying it doesn't, I'd like to understand why.
Quote:

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.
• Jan 20th 2010, 08:57 AM
qmech
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.
• Jan 21st 2010, 02:11 AM
courteous
Quote:

Originally Posted by qmech
[...] Look at the paragraph before this starting with "We derive these quantities from first principles..". [...]

Paragraph before: $\sum_{k=0}^n Pr(X=k) = \sum_{k=0}^n {n\choose k}p^k(1-p)^{n-k} = 1$.
Let me see, $k$ is here "how many times an event happens": $\underbrace{a_p, a_p, a_p, ..., a_p}_\textrm{k}, \underbrace{a_q, a_q, a_q, ..., a_q}_\textrm{n-k}$ and ${n\choose k}$ 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". .... (Wondering)
... I see now. The case of, say, dice with sides ${2,3,4,5,6,7}$ doesn't apply to binomial distribution (which deals only with "yes/no" types).
Have I got it now?