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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 ...

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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 ...

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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+\... 11 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 ... 8 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. 6 Consider a coin independently tossed 10 times. Assume under the measure P, Pr(H) > 0.5 but not equal to 1. Let a risk neutral person be iteratively given the gamble between getting atleast n heads versus atleast n tails. Clearly the person always chooses getting atleast n heads. Let X represent number of heads and Y the number of tails. Thus ... 6 \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 \\ &=-... 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}. ... 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 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 ... 4 When you try and discount everything by the stock, every price process now has extra gamma PnL naturally just due to the presence of the stochastic discount factor (due to it's quadratic variation). Every payoff now is a non linear function of this stock and thus after delta hedging earns this extra PnL. Simply applying Ito's lemma on the discounted payoff ... 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 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{...

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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 ... 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$Z(t_0,t_1,t_2)$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 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_{...

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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 ...

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Under the risk-neutral measure both stocks follow the GBMs \begin{align} S^{(i)}_t=S^{(i)}_0\exp\left((r-q_i)t+\sigma_iW^{(i)}_t-\frac{\sigma_i^2t}{2}\right)\,,\quad i=1,2\,, \end{align} where the constant $r$ is the riskless interest rate and the constant $q_i$ is stock $S^{(i)}$'s dividend rate. The Margrabe formula says that the value of the option to ...

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The numéraire changing formula tells us that for a tradeable asset $X$, and two numéraires $N$ and $M$ we can write (by $\mathbb{E}^N$ I denote the expectation under the martingale measure associated to numéraire $N$): $$N_t\times\mathbb{E}^N_t \left[\frac{X_{T_1}}{N_{T_1}} \right] = M_t\times\mathbb{E}^M_t \left[\frac{X_{T_1}}{M_{T_1}} \right]$$ This ...

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When you're taking the SOFR OIS and building the SOFR curve, you are really using the swaps to bootstrap a curve that is the expected fwd rates in a "natural" measure. So you can just bootstrap without worrying about convexity at this step. Now you might also use SOFR futures and there you may need a futures convexity adjustment. See here https://...

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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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Via a combination of the Cameron-Martin-Girsanov Theorom and the Martingale Representation Theorom you can find the equivalent martingale measure.

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