Change of probability measure (something relative to Brownian motion) I'm working on a question, it says assume that $X_1, ..., X_n$ are i.i.d. random variables with the finite moment generating function $E[e^{\theta X_1}]$ (under the measure P). Define
$$\kappa(\theta)=\log E[e^{\theta X_1}]$$.
Define the measure Q by
$$dQ= \exp(\theta S_n - \kappa (\theta))dP $$
where $S_n = X_1 +...+X_n$. Assume that $X_1$ has the CDF $F$ under the measure $P$ and let $G$ be the CDF under the measure $Q$. That is, $F(x)=P(X \le x), G(x)=Q(X \le x)$.
The question asks.
a) Let n = 1. If F is standard normal, what is G?
b) Let n = 1. If F is exponential with rate $\lambda$, what is G?
c) Prove that for any measurable function $f:\mathbb R ^n \to \mathbb R$ we have
$$\mathbb E_Q[f(X_1,...,X_n)]=\mathbb E_P [e^{\theta S_n-n\kappa (\theta)}f(X_1,...,X_n)]
$$
I get very confused with most of the symbols here, like the measure Q. I find on textbook
"A measure P is a probability measure on $\Omega$ if, in particular, $0\le P(A) \le 1$ for any "nice" set $A \in \Omega$."
My understanding is that, if F is standard normal, then $X_1$ should be normal as well, since $F(x)=P(X \le x)$, is a presentation of CDF. And I could like the MGF of standard normal in $\kappa (\theta))$, which becomes $\kappa (\theta)= \log e^{t^2/2}=t^2/2$
Where  $dQ = \exp(\theta S_n - \kappa (\theta))dP$ can be repesented as $Q(dx) = \exp(\theta S_n - \kappa (\theta))P (dx)$ (I just copy from book, where I don't know the theorem behind it)
$$\begin{align}
G(X) &= Q(X \le x) = \int_{-\infty}^x Q dx \\\
&= \int_{-\infty}^x \exp(\theta S_n - \kappa (\theta))P (dx)\end{align}$$
and then I can use the $\kappa (\theta)$ idea above. But then the equation becomes very mess.
Can someone give me some help? Or link me to a good textbook, where I can get the idea. Mine is too simply, prof. jumped most of theorem and explaination. Thanks!
 A: The theoretical framework of this exercise is the Radon-Nikodym (RN) theorem and the properties of the RN derivative. Please take a look.
If $F_X$ is standard normal, then $\kappa(\theta)=\ln e^{\frac{\theta^2}{2}}=\frac{\theta^2}{2}$. So $G$ is
$$Q(X_1\leq x)=\int_{\{X_1\in(-\infty,x]\}}dQ=\int_{\{X_1\in (-\infty,x]\}}\frac{dQ}{dP}dP=\int_{(-\infty,x]}e^{\theta a-\frac{\theta^2}{2}}dP_X$$
Now $dP_X=\frac{1}{\sqrt{2\pi}}e^{-\frac{a^2}{2}}da$ thus
$$\int_{(-\infty,x]}\frac{1}{\sqrt{2\pi}}e^{\theta a-\frac{\theta^2}{2}-\frac{a^2}{2}}da=\int_{(-\infty,x]}\frac{1}{\sqrt{2\pi}}e^{\frac{-(a-\theta)^2}{2}}da=\Phi(x-\theta)$$
where $\Phi$ is the standard normal cdf. Similarly: if $F_X$ is exponential with rate $\lambda$, then $\kappa(\theta)=\ln(\frac{\lambda}{\lambda-\theta}),\,\theta<\lambda $. So we have
$$Q(X_1\leq x)=\int_{[0,x]}e^{\theta a-\ln(\frac{\lambda}{\lambda-\theta})}\underbrace{\lambda e^{-\lambda a}da}_{=dP_X}=\int_{[0,x]}(\lambda-\theta)e^{-(\lambda-\theta)a}da=1-e^{-(\lambda-\theta)x}$$
The last one uses the standard machine of Lebesgue integration theory. Let $f\geq 0$ be a measurable function. Then there exists a sequence of simple functions s.t. $f_n \uparrow f$. So
$$\begin{aligned}\int fdQ&=\int \sup_{n \in \mathbb{N}}f_ndQ=\\
&\stackrel{\textrm{MCT}}{=}\sup_{n \in \mathbb{N}}\int f_n dQ=\\
&=\sup_{n \in \mathbb{N}}\sum_{k}\phi_{k,n}Q(E_{k,n})=\\
&=\sup_{n \in \mathbb{N}}\sum_{k}\phi_{k,n}\int_{E_{k,n}}e^{\theta S_j-j\kappa (\theta)}dP=\\
&=\sup_{n \in \mathbb{N}}\int f_n e^{\theta S_j-j\kappa (\theta)}dP=\\
&\stackrel{\textrm{MCT}}{=}\int fe^{\theta S_j-j\kappa (\theta)}dP \end{aligned}$$
and then you can extend it to general measurable and integrable $f$ using the decomposition $\int f dQ=\int f^+ dQ-\int f^-dQ$.
