If $X_n$ sequence of random variables , equally distributed $EX_n=a, n=1,2,3...$ then $\frac{1}{n}\sum_{k=1}^{n}X_k\to^{P}a$

Question

If $X_n$ sequence of random variables , equally distributed $EX_n=a, n=1,2,3...$ then $\frac{1}{n}\sum_{k=1}^{n}X_k\to^{P}a$ (convergence in probability)

Proof: (using the fact that convergence in distribution to a constant $a$ implies convergence in probability to that constant)

$$f-\text{ characteristic function }\\ f_{\frac{1}{n}\sum_{k=1}^{n}X_k}(t)=\prod_{k=1}^{n}f_{X_k}(\frac{t}{n})\text{ this is clear why but the next two equalities are not}\\ = \prod_{k=1}^{n}(1+\frac{ait}{n}+ \sigma(\frac{t}{n}))\text{ I'm told this is Maclaurin series, I dont see how... }\\ =(1+\frac{ait}{n}+ \sigma(\frac{t}{n}))^n\to_{n\to \infty}**e^{ait}**=f_a(t)$$

I put in stars what I would like to know most..

Answer 1

The first equality that you don't understand is indeed a MacLaurin series expansion of the characteristic function, using the link between the moments of a random variable and the derivatives of the characteristic function. See https://en.wikipedia.org/wiki/Characteristic_function_%28probability_theory%29#Moments .

So Shalop is right, the $\sigma$ is a little o.

Then you use the fact that $\left(1+\frac{x}{n}\right)^n\rightarrow e^x$. You can take the $\ln$ if it's not clear. Finally, we see $e^{ait}$ as the characteristic function of the distribution $\delta_a$, which is the distribution of a random variable constantly equal to $a$.