T-Forward measure

Question

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

Answer 1

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.