1
$\begingroup$

Ref: https://en.wikipedia.org/wiki/Forward_measure

I am trying to understand how to move from risk neutral measure $Q$ to T-Forward measure $Q_T$.

It appears we can move from one measure to another using the "Radon-Nikodym derivative $\frac{dQ_T}{dQ}$, i.e

$$P(t,T) = E_Q [ \frac{B(t)}{B(T)} ] = E_{Q_T} [\frac{B(t)}{B(T)} \frac{dQ_T}{dQ} ] $$

What I dont understand is how you deduce what $\frac{dQ_T}{dQ}$ is. Wikipedia states it is the following, but im not sure how?

$$ \frac{dQ_T}{dQ} = \frac{B(t)P(T,T)}{B(T)P(t,T)} = 1$$

In this example P(t,T) is the price of a zero coupon bond at time t for maturity T

$\endgroup$
2
$\begingroup$

I think your statement has a typo. I can't find the statement you made in the article you cite.

The forward measure is the measure induced by using a bond as the numeraire instead of the risk free asset. Letting $H(X_T)$ be the payoff function for an asset $X_t$,

$$ \tilde{\mathbb{E}}\left[\frac{B(t)H(X_T)}{B(T)}\right]=P(t, T)\tilde{\mathbb{E}}\left[\frac{B(t)}{B(T)P(t, T)} H(X_T) \right] $$ $$=P(t, T)\tilde{\mathbb{E}}\left[\frac{B(t)P(T, T)}{B(T)P(t, T)} H(X_T) \right]$$ $\frac{P(s, T)}{B(s)}$ is a martingale under the risk neutral measure,and so the following holds:

$$\tilde{\mathbb{E}}\left[\frac{P(T, T)}{B(T)}\right]=\frac{P(t, T)}{B(t)}$$

Rearranging, it becomes clear that $\frac{B(t)P(T, T)}{B(T)P(t, T)} $ is a martingale with expectation one and is thus mathematically able to be a Radon-Nikodym derivative. Hence the pricing formula can be written as follows: $$g(X_t, t)=P(t, T)\hat{\mathbb{E}}\left[H(X_T) \right]$$

In practice, the dynamics of $X_T$ are often postulated under the T-Forward measure without the intermediary risk-neutral step.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.