10

As @ilovevolatility explains, the seminal reference for this matter is Geman, El Karoui & Rochet (1995). We assume none of the assets are dividend paying, and they are strictly positive. There are two potential options. You are considering a market with only assets $X$ and $N$. Then Assumption 1 of their paper would apply, which is related to the two ...


9

Let $P(t, T)$ be the price at time $t$ of a zero-coupon bond with maturity $T$ and unit face value. Consider the pricing of the caplet with payoff $(L(t_1; t_1, t_2)-K)^+$ at time $t_1$, where $0<t_1 < t_2$ and, for $0\le s \le t_1$, \begin{align*} L(s; t_1, t_2) = \frac{1}{t_2-t_1}\left(\frac{P(s, t_1)}{P(s, t_2)}-1\right) \end{align*} is the forward ...


6

The market is complete iff there is a unique risk-neutral measure: when every contingent claim is attainable, its unique no arbitrage price is the cost of the replicating portfolio. In the case of an incomplete market, you no longer have a unique price for unattainable contingent claim, but rather a range of prices : $\left(-p\left(-G\right), p\left(G\right)\...


5

We work on a probability space $(\Omega,\mathcal{N},\mathfrak{F})$ with filtration $(\mathfrak{F}_t)_{0\leq t\leq T}$ and $\mathfrak{F}_T:=\mathfrak{F}$. Let $\xi$ be a $\mathfrak{F}_T$-mesurable contingent claim, and $N_t$ and $M_t$ two assets with positive prices. We assume the process $M_t/N_t$ is a martingale under the probability measure $\mathcal{N}$. ...


4

$Z(t_0,t_1,t_2)$ is the $t_1$-forward price of the ZC bond with maturity $t_2$, as of $t_0$. We have: $$ Z(t_0,t_1,t_2) = E_{t_0}^{t_1}[Z(t_1,t_2)]\not= Z(t_1,t_2).$$ With a not-trivially stochastic index, there is no way to take out $Z(t_1,t_2)^{-1}$ from under your conditional expectation operator until the running $t_0$ hits $t_1$. It is not $t_0$-...


4

$\frac{1}{S_t}$ is log-normal If $S_t$ is a geometric Brownian motion, so is $\frac{1}{S_t}$ and indeed any power $S_t^\alpha$. Simply use Itô's Lemma and set $f(t,x)=\frac{1}{x}$, \begin{align*} \mathrm{d}f(t,S_t) &= \left(0-\mu S_t\frac{1}{S_t^2}+\frac{1}{2}\sigma^2S_t^2\frac{2}{S_t^3}\right)\mathrm{d}t-\sigma S_t \frac{1}{S_t^2}\mathrm{d}W_t \\ &=-...


4

I would like to add to @DaneelOlivaw answer. Your question: "How does one go about finding the right measure for a product?" Answer: One should choose any measure that will make it easy and convenient to compute the pricing at hand. We are free to use whichever measure we would like. For example, it is possible to derive the process for the Forward ...


4

Consider a financial market with a filtered probability space $\left(\Omega,\mathcal{F},(\mathcal{F}_t),\mathbb P\right)$ satisfying usual conditions equipped with a stock price process $S_t$. Suppose there exists a risk-free asset who is governed by $\mathrm{d}B_t=r_tB_t\mathrm{d}t$. Suppose the market is free of arbitrage, i.e. there exists a probability ...


3

(I might not be answering your question, but I feel this clarification is needed.) A random variable $X$ of $(\Omega, \mathcal{F})$ is a $\mathcal{F}$-measurable function $X : \Omega → \mathbf{R}$. So, $X$ depends on $\Omega$ and $\mathcal{F}$, but does not depend on the probability measure put on $(\Omega, \mathcal{F})$. It is the distribution of $X$ that ...


3

Case I Let us consider a derivative with a payoff $H(L(T_{f},T_{S},T_{E}))$ which is paid at time $T_{p}$. Note that: $T_{f}$ - LIBOR fixing date; $T_{S}$ - LIBOR start date; $T_{E}$ - LIBOR maturity date; $T_{p}$ - derivative payment date. Also, $T_{f}=T_{S}=t_{1}$ and $T_{E}=T_{p}=t_{2}$ in the question. In your first case $H(L(T_{f},T_{S},T_{E}))=(L(T_{...


1

Let $X_{T_1}$ be a random quantity known (fixed) at $T_1$ (measurable wrt $T_1$-information), $B$ be the standard bank account and $P$ standard zero-coupon bond price. From standard pricing, for the second contract: $$E_t\left[B_t B_{T_2}^{-1} \left(X_{T_1} - F_2\right) \right] =0 $$ implies $$ F_2= E_t\left[B_t B_{T_2}^{-1} X_{T_1}\right] P(t,T_2)^{-1},$$ ...


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