Equivalence in CLT proof

68 Views Asked by At

I'm studying the proof of CLT in this pdf. My issue is at pag 11.

Let's consider the Moment Generating Function $M(\frac{t}{\sqrt{n}})$. The author applies the logarithm to this function getting $nln(\frac{t}{\sqrt{n}})$.

I'm not able to understand how he came to this result. $M(\frac{t}{\sqrt{n}})$ is defined as $E[e^{\frac{t}{\sqrt{n}}}]$, i don't know how to handle this expected value...

1

There are 1 best solutions below

0
On

As mentioned in the comments, there is no expression involving $n\ln\frac{t}{\sqrt{n}}$ in the linked lecture notes. There is, however, the expression $$ nL\!\left(\frac{t}{\sqrt{n}}\right) $$ on p.11, which is not the same; and indeed, there is nothing abnormal there, as the function $L$ is defined at the bottom of p.10 bt $$ L(u) = \ln M(u)\,. $$ Therefore, indeed $$ \ln \left(M\!\left(\frac{t}{\sqrt{n}}\right)^n\right) = nL\!\left(\frac{t}{\sqrt{n}}\right) $$ by very definition.