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tiny clarification
Matthew Gunn
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The risk neutral probability measure $Q$ is the true probability measure $P$ times the stochastic discount factor $M$ (but rescaled so $Q$ sums to 1).

Simple derivation

For maximum simplicity, I'll work in a discrete probability space with $n$ possible outcomes. Everything goes through under measure theory in more general, infinite outcome probability spaces.

Let $\mathbf{x}$ be a vector denoting cashflows in those $n$ states. Let $\mathbf{p}$ be a vector denoting the probabilities of those $n$ states. Let $\mathbf{m}$ be a vector denoting the stochastic discount factor.

If a stochastic discount factor $\mathbf{m}$ exists, today's price of the future cashflow $\mathbf{x}$ is given by:

$$ f(\mathbf{x}) = \sum_i p_i m_i x_i $$

Now define $q_i = \frac{p_i m_i}{\sum_j p_j m_j}$. Also note that risk free rate must satisfy $1 = \sum_i p_i m_i r$. Hence $r = \frac{1}{\sum_i p_i m_i}$. Observe that $\mathbf{q}$ is also a probability vector since it sums to 1. Then: \begin{align*} f(\mathbf{x}) &= \sum_i \underbrace{p_i m_i}_{=\frac{q_i}{ r}} x_i \\ &= \frac{1}{r}\sum_i q_i x_i \end{align*}

So the price of $\mathbf{x}$ is given by the expectation of $\mathbf{x}$ under the probability vector $\mathbf{q}$ discounted by the risk free rate.

The same logic goes through under measure theory (but you have a bit more formal mathematics with a radon-nikodym derivative etc...).

$$ \mathbb{E}^P[MX] = \frac{1}{r} \mathbb{E}^Q[X] \quad \quad \frac{dQ}{dP} = r M \quad \quad r = \frac{1}{\mathbb{E}^P[M]}$$

The whole idea is actually incredibly simple: throw the stochastic discount factor into the probability measure.

Matthew Gunn
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