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+\...


7

Let $dB_t = rB_t dt$. Now \begin{equation} d\Big(\frac{1}{B_t S_t}\Big) = -\frac{dS_t}{B_t S_t^2} -\frac{dB_t}{B_t^2S_t} +\frac{2}{2}\frac{(dS_t)^2}{B_t S_t^3} = (-\mu-r+\sigma^2)\frac{1}{B_tS_t}dt-\sigma\frac{1}{B_tS_t} dW_t \end{equation} Using the EMM given by $dW_t = \frac{r-\mu}{\sigma}dt +dW_t^\mathbb{Q}$ we get the $\mathbb{Q}$-dynamics \begin{...


6

Per @SKRX's suggestion, another solution is provided below. For simplicity, we assume that the stock price process $\{S_t \mid t \geq 0\}$ follows an SDE, under the risk-neutral measure $\mathbb{Q}$, of the form \begin{align*} \frac{dS_t}{S_t} = r dt + \sigma dW_t, \end{align*} where $r$ is the constant interest rate, $\sigma$ is the constant volatility, ...


5

Let $$I= \mathbb{E}_t^\mathbb{Q}\left[\text{exp}(-2\sigma W_{T-t}) \cdot\mathbb{1}_{S_T\ge K}\right] = \frac{1}{\sqrt{2\pi}} \int_{\hat{d}_2}^{\infty} e^{-2\sigma x} e^{-x^2/2} dx.$$ So $$ I = \frac{1}{\sqrt{2\pi}} \int_{\hat{d}_2}^{\infty} e^{-(x-2\sigma)^2/2} dx \, e^{2\sigma^2}. $$ Change variables $y = x-2\sigma$ and you are done.


5

Note: your previous question assumed log-normality instead of normality. By Cholesky decomposition, we assume that, under measure $P$, \begin{align*} \frac{dR_t}{R_t} &= \mu_{R,t} dt + \sigma_{R,t}\, dW^1_t\\ \frac{dA_t}{A_t} &= \mu_{A,t} dt + \sigma_{A,t}\, d\left(\rho W^1_t + \sqrt{1-\rho^2} W^2_t\right), \end{align*} where $W^1$ and $W^2$ are ...


5

If you are interested in the proof of the Baye's Rule for conditional expectations you can find it here The sake of completeness: The Baye's rule for conditional expectations states $$ E^Q[X|\mathcal{F}]E^P[f|\mathcal{F}]=E^P[Xf|\mathcal{F}] $$ With $f=dQ/dP$ - thus being the Radon-Nikodyn derivative and $X$ being some random variable and $\...


5

Is this the proof you are looking for? -- from Shreve, S. E.'s book "Stochastic calculus for finance II, continuous-time Models", chapter 5.


5

Let $P$ be the risk-neutral measure. We define the measure $P_S$ such that \begin{align*} \frac{dP_S}{dP}\big|_t &=\frac{S_t}{e^{rt}S_0}\\ &=e^{-\frac{1}{2}\sigma^2 t+\sigma W_t}. \end{align*} Then $\{\widehat{W}_t \mid t \ge 0\}$, where \begin{align*} \widehat{W}_t = W_t -\sigma t, \end{align*} is a standard Brownian motion under the measure $P_S$. ...


4

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 $N/\beta$ ...


4

For the buyer of a forward contract the payoff is $S_T - K$ at time $T$ since at this date he pays $K$ and gets the underlying in exchange. Consider the following strategy: buy the stock $S$ and sell $K$ zero-coupon bonds with maturity $T$. At any time $t$, your portfolio's value is $$ \Pi_t = S_t - KB(t,T) $$ In particular at time $T$, it is $S_T - KB(T,...


4

You miss the cross-derivative term in the Ito formula you use to express $d\left ( \frac {C_t}{S_t} \right)$. More specifically (see [Remark] below), $$d\left ( \frac {C_t}{S_t} \right) = \frac {1}{S_t} dC_t - \frac {C_t}{S_t^2} dS_t + \frac {C_t}{S_t^3} d\langle S_t, S_t \rangle {\color{green}{- \frac {1}{S_t^2} d\langle C_t, S_t \rangle}}$$ This last ...


4

The payoff you mention writes: $$ V_T = N \left( 1 + \frac{1}{S_0}(S_T - K_1)^+ - \frac{1}{S_0}(S_T - K_2)^+ \right) $$ with $K_1=S_0=2078.36 < K_2 = 3034.4056$ and $N=2000$ Thus taking a risk-neutral discounted expectation of the payoff yields the price at time $t$: $$ V_t = N B(t,T) + \frac{N}{S_0} C(K_1, T-t) - \frac{N}{S_0}C(K_2, T-t) $$ hence $N$ ...


4

Under the stock numeraire measure, $\frac{B_t}{S_t}$ is a Martingale. We can compute $$d\frac{B_t}{S_t}= \frac{1}{S_t}dB_t -\frac{1}{S_t^2}B_tdS_t+\frac{1}{S_t^3}B_t\sigma^2S_t^2dt\\=\frac{B_t}{S_t}\left(rdt -\mu dt -\sigma dW_t +\sigma^2dt\right)$$ so the growth rate $\mu$ that makes this a Martingale is $$ \mu = r+\sigma^2.$$ So the growth rate of the ...


4

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.


4

Let $N_t$ be the numeraire for the real world measure $P$. Let $r$ be the risk free rate and $P^*$ be the risk neutral measure. Then from the standard change of measure results one must have $ \frac{dP^*}{dP} = \frac{e^{rt} N_0}{N_t} $ so that we get the general result $$ \boxed{N_t = N_0 e^{rt} \frac{dP}{dP^*}} $$ Now in the particular 1 dimensional ...


3

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.


3

I believe that the dynamics for $V$, under $Y$, is not the form you provided. In particular, a measure change will change the drift of $V$. Specifically, the dynamics of $V$ is typically of the form \begin{align*} \frac{dV}{V} = -\sigma_V\sigma_W \rho dt + \sigma_V d W_V. \end{align*} You can now check that the final result holds. Addendum We assume ...


3

Theory Define a single currency economy with $N+1$ tradable assets. Further assume that individually investing in each of these assets constitutes a self-financing strategy. Simply put, this amounts to considering that the assets of our model economy do not distribute capital in the form of dividends or coupons. Let $S_1, \dots, S_N$ represent $N$ risky ...


3

Your questions are nicely addressed in this short paper by Fabrice Rouah: The T-forward measure More specifically, using your notations and noting that $B(T,T)=1$ by definition, the change of measure between the $T$-forward ($\mathbb{Q}^B$) and risk-neutral ($\mathbb{Q}^M$) measures is characterised by the following Radon-Nikodym derivative: $$\left. \frac{...


3

By definition, under the $T+\delta$-forward measure, the price of any tradable asset relative to the bond price $B(t, T+\delta)$ is a martingale. Since \begin{align*} Lb_t(T, \delta) = \frac{B(t, T)}{B(t, T+\delta)}, \end{align*} it is a martingale by definition. Regarding $L_t(T, \delta)$, since \begin{align*} L_t(T, \delta) = \frac{1}{\delta}\left(\frac{B(...


3

We assume that, under the risk-neutral measure $Q$, \begin{align*} dP(t, T) = P(t, T)(r_t + \sigma(t, T)dW_t), \end{align*} where $\{W_t, \, t \ge 0\}$ is a standard Brownian motion. Then \begin{align*} dL(t) &= \frac{1}{T-S}\bigg(\frac{dP(t, S)}{P(t, T)} -\frac{dP(t, S)}{P^2(t, T)}dP(t, T) \\ &\qquad + \frac{dP(t, S)}{P^3(t, T)} \langle dP(t, T), \,...


3

To build on Antoine's answer (which covers the case where the market consists only of a stock $S$ and a risk free asset $r$). In the general case, if the real world measure $\mathbb{P}$ numéraire depends on a stock $S$, then this means each asset will have its own real world measure... which is clearly not the case. Here, one needs to resort to a framework ...


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 ...


3

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{...


3

For the instantaneous forward, please see the last page of this note: T-Forward Measure by Fabrice Douglas Rouah (http://www.frouah.com/finance%20notes/The%20T-Forward%20Measure.pdf). For the simple forward, you know the relationship between the price of the zero coupon and the simple forward: $ \frac{P \left(t,T_{n}\right)}{P \left(t,T_{n+1}\right) }=1+\...


2

First lets analyse the claim that $\frac{(S_t - F(t,T)}{S_{0}}$ is a martingale under a given risk neutral measure $P^{*}$. Recall that the crucial property of a martingale is that at some point in time $t$, a process $\tilde{S}_{t}$ is a martingale iff for some time $t+\Delta$, the expected value of $\tilde{S}_{t+\Delta}$ is $S_t$. So lets start, assume we ...


2

How to use the stock as Numeraire: $$\mathbb{\tilde{E}}[e^{-rT}(S_T-K)^+]=\mathbb{\tilde{E}}\left[e^{-rT}S_T\left(1-\frac{K}{S_T}\right)^+\right]$$ $$=S_0\mathbb{\tilde{E}}\left[\frac{e^{-rT}S_T}{S_0} \left(1-\frac{K}{S_T}\right)^+\right]$$ $$=S_0\mathbb{\hat{E}}\left[\left(1-\frac{K}{S_T}\right)^+\right]$$ Where under $\mathbb{\hat{P}}$ the stock follows $...


2

for a large class of models that is ones where $\log S_T - \log S_0$ has distribution independent of level, it is possible to show that the delta is $$ \mathbb{P}_S(S_T>K) $$ and this is a martingale in the stock measure. (For a proof see More Mathematical Finance by me)


2

I think you are right. The SDE does not attempt to describe the dynamics of the spot exchange rate with respect to random changes in interest rates. Rather, it describes the evolution of the FX rate as a drift term proportional to the rate differential, plus a random term. Specifically, it says that if domestic rates go up, the rate at which the foreign ...


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 ...


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