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Why is there no drift adjustment when numeraire is changed from bank account (risk neutral measure) to zero coupon bond who matures at time of payoff (fwd risk neutral measure) ?

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closed as unclear what you're asking by Quantuple, JejeBelfort, g g, Daneel Olivaw, amdopt Nov 3 '17 at 20:06

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    $\begingroup$ I do not know in what context you are saying this. If you assume deterministic interest rates, the the risk-neutral measure and the forward measure are the same. However, with stochastic interest rates, then the drift term will be changed. $\endgroup$ – Gordon Oct 10 '17 at 17:02
  • $\begingroup$ Deterministic interest rates. Why are the 2 measures the same? They are not - they have different values! $\endgroup$ – Randor Oct 10 '17 at 17:50
  • $\begingroup$ Maybe the answer is that the volatility of each numeraire is zero ? $\endgroup$ – Randor Oct 10 '17 at 18:29
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Let $B_t= e^{\int_0^t r_sds}$ be the money-market account value at time $t$, and $P(t, T)$ be the value of the zero-coupon bond with maturity $T$ and unit face amount. Moreover, let $Q$ be the risk-neutral measure and $Q_T$ be the $T$-forward measure. If the interest rate $r_t$ is deterministic, then \begin{align*} P(t, T) &= E\left(e^{-\int_t^T r_s ds} \,|\, \mathcal{F}_t\right)\\ &= e^{-\int_t^T r_s ds} = \frac{B_t}{B_T}, \end{align*} where $E$ is the expectation under the risk-neutral measure $Q$. Moreover, for $0 \le t \le T$, \begin{align*} \frac{dQ}{dQ_T}\big|_t &= \frac{B_t P(0, T)}{P(t, T)}\\ &=\frac{B_t \frac{1}{B_T}}{\frac{B_t}{B_T}}=1. \end{align*} In particular, for any $A\in \mathcal{F}_T$, \begin{align*} Q(A) &= \int_{\Omega} \pmb{1}_A dQ \\ &= \int_{\Omega} \pmb{1}_A \frac{dQ}{dQ_T}\big|_T dQ_T \\ &= \int_{\Omega} \pmb{1}_A dQ_T \\ &=Q_T(A). \end{align*} That is, $Q=Q_T$.

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  • $\begingroup$ Hi gordon in the numeraire of the radon nikodym derivative i think it should just be Bt. Why do you have Bt x P(0,T)? $\endgroup$ – Randor Oct 11 '17 at 5:54
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    $\begingroup$ Let $(V_t)_{t \geq 0}$ denote a self-financing portfolio wealth process adapted to the filtration $\Bbb{F} = \{ (\mathcal{F}_t)_{t \geq 0} \}$. Under the measure $\Bbb{Q}$ associated to the numéraire $B_t$ one should have $$ \frac{V_0}{B_0} = \Bbb{E}_0^\Bbb{Q} \left[ \frac{V_t}{B_t} \right] $$Similarly, under $\Bbb{Q}^T$ associated to $P(t,T)$, $$ \frac{V_0}{P(0,T)} = \Bbb{E}_0^\Bbb{Q} \left[ \frac{V_t}{P(t,T)} \right] $$ $\endgroup$ – Quantuple Oct 11 '17 at 10:02
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    $\begingroup$ Hence indeed: $$ V_0 = \Bbb{E}_0^\Bbb{Q} \left[ V_t \frac{B_0}{B_t} \right] = \Bbb{E}_0^\Bbb{Q^T} \left[ V_t \frac{P(0,T)}{P(t,T)} \right] $$ hence indeed $$ \left. \frac{d\Bbb{Q}}{d\Bbb{Q}^T} \right\vert_{\mathcal{F}_t} = \frac{B_t P(0,T) }{ B_0 P(t,T) } $$ with $B_0=1$ $\endgroup$ – Quantuple Oct 11 '17 at 10:07
  • $\begingroup$ Thanks @Quantuple for your clarification. Indeed, the Radon-NiKodym derivative can be treated as the ratio of the normalized numeraires, that is, $\frac{B_t}{B_0}$ and $\frac{P(t, T)}{P(0, T)}$. $\endgroup$ – Gordon Oct 11 '17 at 12:34
  • $\begingroup$ No problem @Gordon $\endgroup$ – Quantuple Oct 11 '17 at 13:00

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