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

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 $P_S$. ...


6

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 $\...


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

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


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 ...


4

Let $\mathbb{Q}$ be the risk-neutral probability measure which uses the risk-free bank account $(B_t)$ as numeraire. In general, $\mathrm{d}B_t=r_tB_t\mathrm{d}t$. In the Black-Scholes setting, $r_t\equiv r$, we have $B_t=e^{rt}$. The stock measure $\mathbb{Q}_S$ uses the compounded stock price $S_te^{qt}$ as numeraire and is defined via the Radon Nikodym ...


3

Let $(V_t)_{t \geq 0}$ denote a self-financing wealth process in foreign currency units. In the absence of arbitrage, the former process should emerge as a martingale when expressed in the foreign money market numéraire i.e. $$ V_0 = \Bbb{E}^{\Bbb{Q}^f} \left[ \frac{B_0^f}{B_T^f} V_T \right] \tag{1} $$ Still by absence of arbitrage, the value of that same ...


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)


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