Here is a problem from an exam I just took several days ago.
Let $(\Omega, \mathcal{F}, P)$ be a probability space.
Let $X:\Omega\to\mathbb R$ be a random variable with $X>0$ a.s. and $EX=1$. Define
$$Q(A)=E[X1_A],\ \ \forall A\in\mathcal{F}.$$
Show that $Q$ is a probability measure on $(\Omega, \mathcal{F})$ and $Q\sim P$, i.e. $Q<<P$ and $P<<Q$.Suppose that $X\sim N(\mu,\sigma^2)$ under $P$ where $\mu\neq0$, $\sigma>0$ and $\sigma\neq1$. Try to construct a probability measure on $(\Omega, \mathcal{F})$ such that $X\sim N(0,1)$ under $Q$.
Suppose that $X\sim Poisson(\lambda)$ under $P$ where $\lambda>0$ and $\lambda\neq1$. Try to construct a probability measure on $(\Omega, \mathcal{F})$ such that $X\sim Poisson(1)$ under $Q$.
The first part is standard and easy for me. But the next two parts stuck me. I have never met and thought those questions before. For the second part, if we introduce $Y=\frac{X-\mu}{\sigma}$, then $Y\sim N(0,1)$ under $P$, which is well-known. But how can I use this and the first part to construct such a probability measure $Q$? I cannot move on the third part, too.
Any help would be appreciated.
Best Answer
For normal case. Let $$ f_P(x)=\dfrac{1}{\sigma \sqrt{2\pi}}e^{-(x-\mu)^2/(2\sigma^2)}, \quad f_Q(x)=\dfrac{1}{\sqrt{2\pi}}e^{-x^2/2} $$ be the given and the desired pdfs of $X$ under $P$ and $Q$ respectively. Then define $$\tag{1}\label{1} Q(A) = \mathbb E_P\left[\frac{f_Q(X)}{f_P(X)}I(A)\right]. $$ Here index $P$ means that we will calculate expectation with respect to initial distribution $P$ of $X$. Firstly check whether positive r.v. $Z=\frac{f_Q(X)}{f_P(X)}$ satisfies conditions from the first descriptive part of the question. We need only to check whether $\mathbb E_P[Z]=1$. $$ \mathbb E_P\left[\frac{f_Q(X)}{f_P(X)}\right]=\int\limits_{\mathbb R} \frac{f_Q(x)}{f_P(x)} \cdot f_P(x)\, dx=\int\limits_{\mathbb R} f_Q(x)\, dx=1. $$ Check if $X$ has standard normal distribution under $Q$. For any Borel set $B$, one get from \eqref{1} $$ Q(X\in B) = \mathbb E_P\left[\frac{f_Q(X)}{f_P(X)}I(X\in B)\right]=\int\limits_B \frac{f_Q(x)}{f_P(x)} \cdot f_P(x)\, dx = \int\limits_{B} f_Q(x)\, dx, $$ so $f_Q(x)$ is indeed (standard normal) pdf of $X$ under probability measure $Q$.
The same you can do with Poisson r.v. $X$: $$ Q(A) = \mathbb E_P\left[I(A) \cdot \frac{\frac{1}{\not{X!}}e^{-1}}{\frac{\lambda^X}{\not{X!}}e^{-\lambda}}\right], $$ after that $$ Q(X=k) = \mathbb E_P\left[I(X=k) \cdot \frac{e^{\lambda-1}}{\lambda^X}\right] = \frac{e^{\lambda-1}}{\lambda^k} P(X=k) = \frac{e^{\lambda-1}}{\lambda^k}\frac{\lambda^k}{k!}e^{-\lambda} = \frac{1}{k!}e^{-1}. $$