# Finding the expected value of x^2 in binomial distribution

• May 8th 2011, 01:28 AM
ohm713
Finding the expected value of x^2 in binomial distribution
Here is the summation (from i=0 to n) of i^2 * (n!/(n-i)!(i!)) * p^i * q^(n-i)

I know that the answer is equal to np[(n-1)p+1].

I am not really sure how to get what is in the bracket of the answer.

Thanks.
• May 8th 2011, 02:17 AM
SpringFan25
There are easier ways to do this than evaluating that sum

You know that
$E(X^2) = Var(X) + E^2(X)$

You also know that by definition the binomial variable X is the sum of n independant bernoulli trials. Its not difficult to compute the mean and variance of the sum of n independant bernoulli trials, then you can use the result above.
• May 8th 2011, 02:25 AM
ohm713

However, I would actually like to know how to get that answer, which is np[(n-1)p+1] using the summation and binomial distribution properties.

Could you please show me the steps?
• May 8th 2011, 04:53 AM
SpringFan25
http://quicklatex.com/cache3/ql_8468...b2406a5_l3.png

Instead of recognising the left hand sum as an expectation, you can use the algebra in the following link to show explicitly that it is (n-1)p.
Mean or Expectation Value

Spoiler:

latex in case i need it later
\begin{eqnarray*}
E(X^2) &=& \sum_{i=0}^n \left( i^2 \frac{n!}{i!(n-i)!} p^{i} (1-p)^{n-i}\right)\\
&& \text{note first term in sum is 0 and rewrite limit}\\
&=& \sum_{i=1}^n \left( i^2 \frac{n!}{i!(n-i)!} p^{i} (1-p)^{n-i}\right)\\
&=& np\sum_{i=1}^n \left( i \frac{(n-1)!}{(i-1)!(n-i)!} p^{i-1} (1-p)^{n-i}\right)\\
&& \text{let } j=i-1 \text{ , so }i=j+1\\
&=& np\sum_{j=0}^{n-1} \left( i \frac{(n-1)!}{(j)!(n-1-j)!} p^{j} (1-p)^{n-1-j}\right)\\
&=& np\sum_{j=0}^{n-1} \left( (j+1) \frac{(n-1)!}{(j)!(n-1-j)!} p^{j} (1-p)^{n-1-j}\right)\\
&=& np\sum_{j=0}^{n-1} \left( (j) \frac{(n-1)!}{(j)!(n-1-j)!} p^{j} (1-p)^{n-1-j}\right)+ np\sum_{j=0}^{n-1} \left( \frac{(n-1)!}{(j)!(n-1-j)!} p^{j} (1-p)^{n-1-j}\right)\\
&& \text{recognise right hand sum as PDF of Bi(n-1,p)}\\
&=& np\sum_{j=0}^{n-1} \left( (j) \frac{(n-1)!}{(j)!(n-1-j)!} p^{j} (1-p)^{n-1-j}\right)+ np \times 1\\
&& \text{recognise left hand sum as Expectation of Bi(n-1,p)}\\
&=& np\left((n-1)p \right)+ np \times 1\\
&=& np\left((n-1)p + 1 \right)

\end{eqnarray*}
• May 8th 2011, 06:30 AM
matheagle
It's easier to find E(X(X-1)) first and then use $E(X(X-1))= E(X^2)-np$