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 ...
10
Recall that any traded asset divided by a numéraire is a martingale under the measure associated to that numéraire. For the 3 interest rates you mention, the natural measure (namely the one that makes those processes martingales) is deduced from the structure of the rate.
Always keep in mind that the value at $t_0$ of a cash flow $C$ paid at $T$ is equal to ...
9
As a general principle, I would be wary of economic or financial interpretations of change of measure techniques. Changing numéraires is merely a mathematical tool to ease pricing, see for example the last part of this answer. Nevertheless, here’s my take on your question.
Think of a numéraire as the basic financial asset of your economy, namely a store of ...
9
This is an interesting question that I have asked myself. Below is my take.
Let us consider an economy $(\Omega,\mathcal{F},P)$ equipped with a filtration $(\mathcal{F})_{t \geq 0}$ consisting on a traded asset $S_t$ and a numéraire $N_t$ specified by the following stochastic differential equations:
$$\begin{align}
\text{d}S_t&=\alpha(t,S_t)\text{d}t+\...
6
Let $P(t, T)$ be the price at time $t$ of a zero-coupon bond with maturity $T$ and unit face value. Consider the pricing of the caplet with payoff $(L(t_1; t_1, t_2)-K)^+$ at time $t_1$, where $0<t_1 < t_2$ and, for $0\le s \le t_1$,
\begin{align*}
L(s; t_1, t_2) = \frac{1}{t_2-t_1}\left(\frac{P(s, t_1)}{P(s, t_2)}-1\right)
\end{align*}
is the forward ...
6
The drift is the expectation of the return over an infinitesimal interval. Let $Q$ be the risk-neutral measure and $Q^S$ be measure associated with the stock price numeraire defined by
\begin{align*}
\frac{dQ^S}{dQ}\big|_t = \frac{S_t}{B_t S_0},
\end{align*}
where $B_t=e^{rt}$ is the value at time $t$ of the money-market account. Moreover, let $E$ and $E^S$ ...
6
Proving the existence of a risk neutral measure is the difficult part. Once its existence is established, a simple calculation of conditional expectations allows to go from a numeraire to any other.
Write $\beta$ for the cash numeraire and $Q_\beta$ the corresponding risk neutral measure. Let $N$ be a numeraire (so $N$ is a positive process and $N/\beta$ ...
6
Either $r=0$ in which $B_t$ is constant and is a valid numeraire (as is any multiple of it.)
or $ r \neq 0$ in which case an asset of constant value would give an arbitrage since we could take
$$
B_t - N_t
$$
with $B_0 = N_0$ and get a riskless profit. (or the opposite if $r<0.$) and so it would be a very flawed model.
5
A Numeraire must be a tradeable asset. If you can find a constant tradeable asset, then yes a constant can be used as a numeraire.
5
We work on a probability space $(\Omega,\mathcal{N},\mathfrak{F})$ with filtration $(\mathfrak{F}_t)_{0\leq t\leq T}$ and $\mathfrak{F}_T:=\mathfrak{F}$. Let $\xi$ be a $\mathfrak{F}_T$-mesurable contingent claim, and $N_t$ and $M_t$ two assets with positive prices. We assume the process $M_t/N_t$ is a martingale under the probability measure $\mathcal{N}$. ...
4
$\frac{1}{S_t}$ is log-normal
If $S_t$ is a geometric Brownian motion, so is $\frac{1}{S_t}$ and indeed any power $S_t^\alpha$. Simply use Itô's Lemma and set $f(t,x)=\frac{1}{x}$,
\begin{align*}
\mathrm{d}f(t,S_t) &= \left(0-\mu S_t\frac{1}{S_t^2}+\frac{1}{2}\sigma^2S_t^2\frac{2}{S_t^3}\right)\mathrm{d}t-\sigma S_t \frac{1}{S_t^2}\mathrm{d}W_t \\
&=-...
4
The confusion is that you think that we define the numeraire as this exponential function... It is not the case. We give the numeraire properties to $N$, then we model it. Similar to any other model.
All we know is that $N$ is positive, and we have $$\frac{V_t}{N_t}=E^{N}\left[\frac{V_T}{N_T}|\mathbb{F_t}\right]$$ where $V_t$ is a tradable asset.
$N$ can ...
4
Well, consider using $S_t$ as the numeraire and let the asset be the reinvested stock $S_te^{qt}$. Then this ratio equals $e^{qt}$ so can never be a martingale.
4
We consider a financial market with three assets: a zero-coupon bond of maturity $T_1$, a second one with maturity $T_2$ and the money market account $B_t$. Assuming the market's risk-free rate $r_t$ is normally-distributed, the spot dynamics of the assets under the risk-neutral measure $Q$ are given by:
$$\begin{align}
\frac{\text{d}B_t}{B_t}&=r_t\text{...
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
An obvious example is using the maturity $T$ zero coupon as numeraire, and a European option with premium paid at time $T$ hedged with maturity $T$ forward contracts. You do not need to trade the zero coupon, in fact you don't even really need to know its value in terms of \$ prior to $T$, because all settlements will occur on $T$.
As stated by @Matthew ...
2
Actually, all investments, retirement accounts, mutual fund accounts, utility bills, supermarket price listings are reported or stated in the Constant Numeraire, which may also be called Dollar-kept-under-the-mattress Numeraire
It is the most widely (indeed the only) Numeraire used in real life.
How nice it would be if my retirement account or mutual fund ...
2
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
$$ F_2= E_t\left[B_t B_{T_2}^{-1} X_{T_1}\right] P(t,T_2)^{-1},$$
...
1
Via a combination of the Cameron-Martin-Girsanov Theorom and the Martingale Representation Theorom you can find the equivalent martingale measure.
1
I have a take on the intuition part of the question. Isn't it a simple consequence of Jensen's inequality? Thus, assuming $r=0$ for simplicity, we have in the money market measure: $E(S_T)=S_t$, but then $E(1/S_T)>1/S_t$ by Jensen since $1/x$ is convex. Now in the stock measure, we must force $E_S (1/S_T)=1/S_t$ to create the correct martingale, but ...
1
I am not familiar with the deep mathematical intricacies of advanced no-arbitrage theory, an extremely technical subject. However, from reading literature reviews, I suspect this is an historical legacy of the research path that led to the most general versions of no-arbitrage theory.
If you consider dividend-paying assets whose dividends are not ...
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