Convergence in Probability of $X\sim N(0,1/n)$ and $\sqrt{n} X \sim N(0,1)$

Question

I am trying to follow Example 5.3 from Larry Wassermans All of Statistics.

I can follow the argument but I stumble upon: Let $X_{n} \sim N(0,\frac{1}{n})$. Note that $\sqrt{n}X \sim N(0,1)$ I tried comparing the MGF of $X_{n} \sim N(0,\frac{1}{n})$ and the MGF of $\sqrt{n}X \sim N(0,1)$ but it gets ugly.

Why is that? How do you prove it?

Answer 1

Let's use the fundamental transformation theorem:

Let $Y=g(X)$ is a monotone continuous function. Thus

$$f_Y(y)=f_X[g^{-1}(y)]|\frac{d}{dy}g^{-1}(y)|$$

In your example you have:

$$f_X(x)= \frac{\sqrt{n}}{\sqrt{2\pi}}e^{-\frac{n}{2}x^2}$$

$y=\sqrt{n}x$

$x=\frac{y}{\sqrt{n}}$

$x' =\frac{1}{\sqrt{n}}$

$$f_Y(y)=\frac{1}{\sqrt{n}} \frac{\sqrt{n}}{\sqrt{2\pi}}e^{-\frac{n}{2}\frac{y^2}{n}}$$

$$f_Y(y)= \frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}}$$

...we are done.