# Thread: Binomial distribution proof of mean

1. ## Binomial distribution proof of mean

Can anyone provide a proof for the variance of binomial distribution?
Show that it is npq without using the Bernoulli distribution and independence way..( which is the typical way of summations or expectations)
thank you so much..

2. Originally Posted by eugene1687
Can anyone provide a proof for the variance of binomial distribution?
Show that it is npq without using the Bernoulli distribution and independence way..( which is the typical way of summations or expectations)
thank you so much..
Here is the solution from my probability book.

$E[X^k] = \sum_{i=0}^n i^k {n\choose i} p^i(1-p)^{n-i} = \sum_{i=1}^n i^k {n\choose i}p^i (1-p)^{n-i}$
Use the identity,
$i{n\choose i} = n{{n-1}\choose {i-1}}$
Thus, letting $j=i-1$,
$E[X^k] = np\sum_{j=0}^{n-1}(j+1)^{k-1}{{n-1}\choose j}p^j(1-p)^{n-1-j}=$ $npE[(Y+1)^{k-1}] \mbox{ where }Y \sim \mbox{bino}(n-1,p)$.

So that means,
$E[X] = np E[(Y+1)^0] = np$
$E[X^2] = np E[(Y+1)^1] = np[(n-1)p+1]$
That means,
$\mbox{Var}[X] = E[X^2] - (E[X])^2 = np(1-p)$

3. Originally Posted by eugene1687
Can anyone provide a proof for the variance of binomial distribution?
Show that it is npq without using the Bernoulli distribution and independence way.
Actually not knowing what text material you have been given, this could be seen as an odd question.
Therefore what I do is just a guess. I expect that your text would have shown the if X is the binomial variable then $E\left( {X^2 } \right) = np\left[ {\left( {n - 1} \right)p + 1} \right]$.
Using the standard definition for variance:
$V(X) = E\left( {X^2 } \right) - E^2 \left( X \right) = np\left[ {\left( {n - 1} \right)p + 1} \right] - \left( {np} \right)^2$.
From which the result follows.

4. ## thank you very much

hey there,
Thank you sooo much
I got the idea on how to show the proof from the help u gave..
Appreciate it..
thanks