# Tag Info

3

Let $P(t,T)$ denote the time $t$ price of a zero-coupon bond maturing at time $T$ and $\mathbb{Q}_T$ be the associated equivalent martingale measure which uses $P(t,T)$ as numeraire. Then, for any $\mathcal{F}_T$-measurable payoff $\xi$, the time $t$ value of $\xi$ is given by $$V_t=P(t,T)\cdot\mathbb{E}^{\mathbb{Q}_T} [\xi\mid\mathcal{F}_t].$$ The ...

2

as you have an open question, let me try to give you an example in finance. Let $(M_t)$ be a martingale, i.e. a fair game. Such processes, on average (their expectation), do not increase or decrease, they remain constant. As a consequence, $\mathbb{E}[M_t]=M_0$, i.e. if you ask me right now what value I expect $M$ will have at time $t$, I expect on average ...

0

An expectation is an "average" of a property measured over all the possible outcomes. Suppose you have a die and you take the expectation of a roll. It could be 1-6 with equal chance so the expectation is 3.5. But now give it a conditioning. A conditioning is like a filter, so that you exclude possibilities from the whole. What is the expectation of a die ...

1

Thanks to the Girsanov theorem, we have the following relationship between the forward measure $\mathbb{Q}^{T_i}$ and the historical measure $\mathbb{P}$. \begin{align} \left.\frac{d\mathbb{Q}^{T_i}}{d\mathbb{P}}\right|_{\mathcal{F}_t} &= e^{-\int_t^Tr_sds}\frac{P_t(T_i)}{P_0(T_i)} \\ &= \exp\left(\int^{T_i}_{t}\xi_{i}(s)dB_{s}-\frac{1}{2}\int^{T_i}_{...

Top 50 recent answers are included