# Thread: Stuck at the therem

1. ## Stuck at the therem

Hi guys i am trying to solve the following theorem

Theorem : If $X_1,X_2,...X_n$ constitute the random sample from an infinite population with the mean $\mu$ and variance $\sigma^2$, then

$E[\bar X] = \mu$ and $var(\bar X) = \frac{\sigma^2}{n}$

Solution: Let $Y = \bar X$
$Y = \sum a_i \ X_i$

$\therefore$ $E[\bar X] = E \left( \sum a_i \ X_i \right)$

Let $a_i = \frac{1}{n}$ and as $E[\bar X] = \mu$ therefore substituting these terms in the above equation we get

$E[\bar X] = \sum_{n=1}^{\infty} \frac{1}{n} \mu$

Now after this in the book it shows this step

$\sum_{n=1}^{\infty} \frac{1}{n} \mu$ = $n. \frac{1}{n} \mu$...............How did he get 'n' instead of $\sum_{n=1}^{\infty}$

2. Because $\bar X={ \sum_{i=1}^n X_i \over n}$

The sample mean is the average of n, not an infinite number of observations.
and it's THEOREM

3. Originally Posted by matheagle
Because $\bar X={ \sum_{i=1}^n X_i \over n}$

The sample mean is the average of n, not an infinite number of observations.
and it's THEOREM
$\sum_{n=1}^{\infty} \frac{1}{n} \mu$ = n $. \frac{1}{n} \mu$
you didnt understand my question . I want to knw from where did the 'n' come from

$\sum_{n=1}^{\infty} \frac{1}{n} \mu$ = n $. \frac{1}{n} \mu$
you didnt understand my question . I want to knw from where did the 'n' come from
Because you're summing n times a 'constant' (which doesn't depend on k)

5. $\sum_{n=1}^3k=k+k+k=3k$

$\sum_{n=1}^{100}k=100k$

$\sum_{n=1}^{100}\mu=100\mu$

$\sum_{n=1}^{\infty} \frac{1}{n} \mu$ = n $. \frac{1}{n} \mu$
you didnt understand my question . I want to knw from where did the 'n' come from
I certainly DID understand your question.
YOU don't have the correct definition of the sample mean.

The sample mean is not an infinite sum

$\bar X={X_1+\cdots + X_n \over n}$

and

$E(\bar X)={E(X_1)+\cdots + (X_n) \over n}$

$={\mu+\cdots + \mu \over n}={n\mu\over n}=\mu$

7. thanks