Query regarding Independent and Identically Distributed random variables

[maverick280857]

Hi

I have a question regarding i.i.d. random variables. Suppose X_1,X_2,\ldots is sequence of independent and identically distributed random variables with probability density function f_{X}(x), mean = \mu and variance = \sigma^2 < \infty.

Define

Y_{n} = \frac{1}{n}\sum_{i=1}^{n}X_{i}

Without knowing the form of f_{X}, how does one prove that var(Y_{n}) = \sigma^2/n?

I suppose this is a standard theorem/result, but any hints/ideas to prove this would be appreciated.

Thanks.

 

[mathman]

var(Yn)=E(Yn²)-E(Yn)²

Plug in the series for Yn and expand, using the fact the E(sum)=sum(E) and E(prod of ind. rv)=prod of E's., it will all work out. Note that all you needed was independence and the fact that the mean and variance was the same for all. The distributions could have been different.

 

[maverick280857]

Thanks mathman

 

[maverick280857]

Can also be done as follows:

If T = \sum_{i=1}^{n}a_{i}X_{i} then Var(T) = \sum_{i=1}^{n}a_{i}^2Var(x_{i}), which gives

Var(Y_{n}) = \sum_{i=1}^{n}\frac{1}{n^2}Var(x_{i}) = \frac{\sigma^2}{n}

Edit: need only the independence of the random variables