Deduce that $Z_m$ is the Radon-Nikodym-density $Z_m=\frac{dQ_m}{dP}$ of the probability measure $Q_m$, which is equivalent to P

Question

The information given:

Consider an arbitrage-free one-period financial market model $(S^0;S)$. We have a risk free asset $S^0$ with $S_0^0=1,S_1^0=1$ so the risk free rate $r=1$. We have a risky asset $S_0=50$ and $S_1$ uniformly distributed on the interval $(0,100)$ under the probability measure $\mathbb{P}$. It holds for every real valued continuous function $f$ that:

$$\mathbb{E}_\mathbb{P}(f(S_1))=\int_{-\infty}^{\infty}\mathbb{1}_{(0,100)}(x)\frac{f(x)}{100}dx$$

Furthermore for $m\in\mathbb{N}$ we have $\phi(x)=\frac{2m+1}{50^{2m}}(x-50)^{2m}$ and $Z_m(\omega)=\phi_m(S_1(\omega))$

The question:

Show that $Z_m>0$ $\mathbb{P}$-a.s. and $\mathbb{E}_\mathbb{P}(Z_m)=1$. This I have done by showing that $\phi(x)>0$ and calculating the integral respectively.

Deduce that $Z_m$ is the Radon-Nikodym-density $Z_m=\frac{dQ_m}{dP}$ of the probability measure $Q_m$, which is equivalent to $P$. Here is it I'm struggling.

I know the random variable $X$ satisfying: $$\mathbb{E}_{\mathbb{Q}}(\mathbb{1}_A)=\mathbb{Q}(A)=\int_AXd\mathbb{P}=E(1_AX)$$ is called the Radon-Nikodym-density. I don't know how to show that $Z_m$ satisfies this however.

Answer 1

If $\mathbb{Q}_m$ is not defined anywhere, I think the problem is just saying you can define a probability measure by $d\mathbb{Q}_m := Z_m d\mathbb{P}$, i.e. define $\mathbb{Q}_m(A) := \mathbb{E}^{\mathbb{P}}[Z_m 1_A]$ for $A \in \mathcal{F}$. Then $\mathbb{Q}_m(\Omega) = \mathbb{E}^{\mathbb{P}}[Z_m] = 1$, so $\mathbb{Q}_m$ is a probability measure.

We clearly have $\mathbb{Q}_m \ll \mathbb{P}$ because if $\mathbb{P}(A) = 0$ then $\mathbb{E}^{\mathbb{P}}[Z_m 1_A] = 0$. Conversely if $\mathbb{E}^{\mathbb{P}}[Z_m1_A] = 0$ for some $A \in \mathcal F$, the fact that $Z_m > 0$ a.s. implies $\mathbb{P}(A) = 0$, so $\mathbb{P}\ll \mathbb{Q}_m$. Hence we conclude $\mathbb{P}$ and $\mathbb{Q}_m$ are equivalent.

Answer 2

If no more info is given, I agree with @user6247850 , there's just one more thing I want to add to show Q is indeed a probability measure, namely the σ-additivity property:

$\mathbb{Q}(\large{\bigcup\limits_{n=1}^{\infty}}E_{i}) = \int_{{\bigcup\limits_{n=1}^{\infty}}E_{i}}Z_md\mathbb{P} = \large{\int_{\Omega}} \;Z_m1_{{\bigcup\limits_{n=1}^{\infty}}E_{n}}d\mathbb{P} = \int_{\Omega}\sum\limits_{n = 1}^{\infty}{Z_m}{1_{E_n}}d\mathbb{P}. \quad (1)\\ $

Setting $\; \alpha_{n} := \sum\limits_{j = 1}^{n}{Z_m}{1_{A_n}} $ , $\Rightarrow 0\leq \alpha_1 \leq ... \leq \alpha_n \leq \alpha_{n+1} \leq ... \; \\ $

And obviously $ \; \; \lim_{n \to \infty} \alpha_n = \large\sum\limits_{n = 1}^{\infty}{Z_m}{1_{A_n}}$

Hence we can apply the Monotone Convergence Theorem this sequence, and therefore

$\large(1)\Rightarrow \int_{\Omega}\sum\limits_{n = 1}^{\infty}{Z_m}{1_{E_n}}d\mathbb{P} = \sum\limits_{n = 1}^{\infty}\int_{\Omega}{Z_m}{1_{E_n}}d\mathbb{P} = \sum\limits_{n = 1}^{\infty}\int_{E{n}}{Z_m}d\mathbb{P} = \sum\limits_{n = 1}^{\infty}\mathbb{Q}(E_i). $