Show that $\mathbb{M}$ is a probability measure and that under that measure the random variable $X$ has normal distribution $N(a,1)$

Question

We define a measure $\mathbb{M}$ which is absolutely continuous w.r.t $\mathbb{P}$, and has the Radon–Nikodym derivative $𝑑\mathbb{M}/𝑑\mathbb{P}$ = $𝑒^{𝑎𝑋−𝑎^{2}/2}$. I need to show that $\mathbb{M}$ is a probability measure and that under the measure $\mathbb{M}$ the random variable 𝑋 has normal distribution
𝑁(𝑎,1). I tried to show that $\mathbb{M}(\Omega)=1$, with $\Omega$ the unvivers using the definition from the course :$\mathbb{M}(\Omega)=\int_{\Omega} d\mathbb{M}/d\mathbb{P}~d\mathbb{P}=\int_{\Omega} e^{aX-a^{2}/2} d\mathbb{P}$. But then it seems like I can work with the expectation (I don't know why) and make appear a $1/\sqrt{2\pi}$. I don't know how to pass from $d\mathbb{P}$ to $dx$. Can I have some help please.

Answer 1

For $\ X\ $ to be $\ \mathcal{N}(a,1)$-distributed with respect to $\ \mathbb{M}\ $, you must have \begin{align} \mathbb{M}\big(X\le x\big)&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^xe^{-\frac{x^2}{2}+ax-\frac{a^2}{2}}dx\\ &=\int_{X\le x}d\mathbb{M}\\ &=\int_{X\le x}e^{aX-\frac{a^2}{2}}d\mathbb{P}\\ &=\int_{-\infty}^xe^{ax-\frac{a^2}{2}}dF_\mathbb{P}(x)\ , \end{align} where $\ F_\mathbb{P}\ $ is the distribution of $\ X\ $ with respect to $\ \mathbb{P}\ $. For this to be true for all $\ x\ $, $\ F_\mathbb{P}\ $ must be the standard normal distribution. If you make that assumption you should then be able to prove the required results (and it will be impossible to prove that $\ X\ $ is $\ \mathcal{N}(a,1)$-distributed with respect to $\ \mathbb{M}\ $ without that assumption).