# Tag Info

22

Life Without a Risk-Neutral Measure How would we price assets without the measure $\mathbb Q$? Well, we would start with some version of the Euler equation $P_t=\mathbb{E}_t[M_{t+1}P_{t+1}]$, where $M$ is the stochastic discount factor (SDF). This equation holds under very weak assumptions (law of one price) and uses real-world probabilities. So, we take the ...

15

I think to understand the martingale/local martingale distinction, it helps to bring in a third class of processes, the uniformly integrable martingale. I would argue that the local martingale and the non-uniformly integrable (true) martingale are actually fairly similar. The key property that a uniformly integrable martingale has is the so-called closure ...

15

In the integral $$\int_0^t S_u dW^{*}_u \, ,$$ $dW^{*}_u \equiv W^{*}_{u+du} - W^{*}_u$ is independent from the integrand $S_u$. So, $\mathbb{E}\left[ \int_0^t S_u dW^{*}_u\middle\vert \mathcal{F}_0\right] = \int_0^t \mathbb{E}\left[S_u \middle\vert \mathcal{F}_0\right]\mathbb{E}\left[dW^{*}_u\middle\vert \mathcal{F}_0\right] = 0$, since $\mathbb{E}\... 12 We assume that, under the probability measure$Q, \begin{align*} dS_t &= S_t\big(r_t dt + \sigma dW_s(t)\big),\\ dr_t &= -k\, r_t dt + \alpha dW_r(t),\tag{1} \end{align*} whered\langle W_s(t), W_r(t)\rangle_t = \rho dt$. From$(1)$, for$s\ge t, \begin{align*} r_s = e^{-k(s-t)}r_t + \alpha\int_t^s e^{-k(s-u)} dW_r(u). \end{align*} Then, forT\ge ...

11

Whether it's called volatility pumping, rebalancing premium, or Shannon's Demon it would just be a form of replicating a short gamma option strategy (eg. selling straddles). Intuitively, you are systematically selling at higher levels and buying at lower levels. The payoff for continuously rebalancing an equity/cash portfolio without friction when the ...

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 ...

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

I aim to give a careful mathematical treatment to this answer, whilst following the fantastic book "Basic Stochastic Processes" by Brzezniak and Zastawniak. The reason I am putting this answer on is twofold: first, to compliment @ William S. Wong's answer by adding greater mathematical intricacy for other users of the website, and secondly to confirm that ...

9

Roughly speaking, we can express the difference between a Markov process and a martingale as follows: A Markov process is one for which conditioning its future value on its history is the same as conditioning its future value on its present value, so that $E(h(X_t)\,|\,X_u,\,u\leq s)=E(h(X_t)\,|\,X_s)$, for any appropriate function $h$; A martingale is a ...

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 ...

7

Under a Black-Scholes framework, the dynamics of the stock price under the risk-neutral measure $\mathbb{Q}$ are given by ... $$S_t = r S_tdt +\sigma S_tdW^{\mathbb{Q}}_t$$ ... and those of the risk-free bond by: \begin{align} dB_t = rB_tdt \end{align} Let us define the derivative value as $V(t,S_t)$, which only depends on the time $t$ and the ...

7

Let $(W_t)$ be a standard Brownian motion and $a>0$. We define $X_t=e^{aW_t-\frac{1}{2}a^2t}$. Then, the process $(X_t)$ is adapted and integrable which are the first two conditions of being a martingale. Finally, for any $s<t$, \begin{align*} \mathbb{E}\left[ X_t\mid\mathcal{F}_s\right] &= \mathbb{E}\left[ e^{aW_t-\frac{1}{2}a^2t}\mid\mathcal{F}...

7

Intro: Great answer given by KeSchn above. I would like to contribute an additional perspective. My experience with and my understanding of the Risk Neutral measure is entirely based on "no arbitrage" and "replication / hedging" arguments. The way I would like to explain this view is via the following three-step construction: (i) First, I ...

6

Measure theory helps us overcome some of the drawbacks of constructing measures (measure of probability when ranged at $[0,1]$). Classic probability theory is effective for probability models whose sample space $\Omega$ is a finite or countable set ($P: 2^{\Omega} \to [0,1]$). But for uncountable sets, such as $\mathbb{R}$ (which is uncountable) the ...

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

Note merely that $B_t=B_s+(B_t-B_s)$ which is the sum of independent normally distributed random variables. In particular, $B_s$ is $\mathbb{F}_s$-measurable and $B_{t-s}$ is independent of $\mathbb{F}_s$. Thus, \begin{align*} \mathbb{E}_s[Z_t] &= \mathbb{E}_s\left[\exp\left(-\frac{1}{2}\theta^2t+\theta B_t\right)\right] \\ &= \mathbb{E}_s\left[\exp\...

6

The way I understand it is: In equation 2 $x_{j, t + 1}$ is defined as the change in of $p_j$ over the period $t$ to $t + 1$. The formula says that the expectation of the change is zero which is the same as saying that the expectation of the original variable at $t+1$ is equal to its current value.

6

European Contracts It's a really important question and as @noob2 commented, the FTAP is normally applied to European-style derivatives, even if they are (strongly) path-dependent, including barrier options and Asian options. The idea is always the same, $V_t=B_t\mathbb{E}^\mathbb{Q}\left[\frac{\xi_T}{B_T}\Big|\mathcal{F}_t\right]$, that is the derivative's ...

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 : $\left(-p\left(-G\right), p\left(G\right)\... 6 Consider the Heston (1993) model under the real world measure ($\mathbb{P}) \begin{align*} \mathrm{d}S_t&=\mu^\mathbb{P} S_t\mathrm{d}t+\sqrt{v_t}S_t\mathrm{d}B_{S,t}^\mathbb{P}, \\ \mathrm{d}v_t&=\kappa^\mathbb{P} (\bar{v}^\mathbb{P}-v_t)\mathrm{d}t+\sigma^\mathbb{P}\sqrt{v_t}\mathrm{d}B_{v,t}^\mathbb{P}, \end{align*} where\mathrm{d}B_{S,t}^\...

5

Let's consider a random process $X$. If $X$ is an adapted process, then we know, without any uncertainty, what its value is at the present time. This idea is formalized with measure theory. For $X$ to be a martingale, it needs to have the following property: at any given time, our best estimate of the value at some point in the future (i.e. forecast), is ...

5

The martingale representation theorem says that for any martingale $M$, there exists a unique stochastic process $\nu_t$ such that \begin{align*} M_t = \mathbb{E}(M_0) + \int_0^t\nu_sdW_s. \end{align*} See, for example, this book. Since \begin{align*} C_t &= \Bbb{E} \left( \int_t^T f(r_s, \theta_s, W_s) ds \mid \mathcal{F}_t \right)\\ &= -\int_0^t ...

5

We decree that $D_t$ has a certain process which makes it a martingale. In particular, we let $$D_t = \mathbb{E} ( D_T \, | \, \mathcal{F}_t)$$ This is trivially a martingale by the tower law. Since the discounted stock price and discounted bond price are martingales, we have made everything a martingale and that ensures that there is no arbitrage in the ...

5

You may find the following paper worthwhile. It addresses most of the above points (and many more) in a systematic way: Dubikovsky, Vladislav and Susinno, Gabriele, Demystifying Rebalancing Premium and Extending Portfolio Theory in the Process (May 20, 2015). Available at SSRN: https://ssrn.com/abstract=2927791 or http://dx.doi.org/10.2139/ssrn.2927791 ...

5

Assume that: $$S_0^1(1+r)\leq a,b$$ Arbitrage for a portfolio $V_t$ is defined as: $$V_0\leq0, \quad P(V_1\geq0)=1, \quad P(V_1>0)>0$$ Consider borrowing at rate $r$ to buy the risky asset such that $V_0=0$. Then, assuming $a\not= b$: \begin{align} \min_{\omega}V_1(\omega)=a-S_0^1(1+r)\geq 0 \\ \max_{\omega}V_1(\omega)=b-S_0^1(1+r)> 0 \end{... 5 By definition,Fo(t,T)=E^T[S_T|F_t]$$Note that expectation is taken under T-forward measure. Now, evaluating at s<T:$$E^T[Fo(t,T)|F_s] = E^T[E^T[S_T|F_t]|F_s] = E^T[S_T|F_s] = Fo(s,T)(using the tower property of expectations). Hence Forwards rate is a martingale under the T-forward measure. 5 The Snell envelope is the smallest super-martingale that is greater than X. Since \tau \le N, it is obvious that A_N^{\tau} = A_{N\wedge \tau} = A_{\tau}. For part (b), note that, from the Doob decomposition, M is a martingale, A is increasing, M_0=Z_0, and A_0=0. If Z^{\tau}= \{Z_n^{\tau}\}_{n=1}^N is also a martingale, then \begin{align*}... 5 You do not really need the dynamics of S_t^2. You can simply apply your standard technique from risk-neutral pricing. The time zero price of a European-style contract with payoff X is given byV_0=e^{-rT}\mathbb{E}^\mathbb{Q}[X\mid\mathcal{F}_0]. Thus, \begin{align*} V_0 &= e^{-rT}\mathbb{E}^\mathbb{Q}[\mathbb{1}_{\{S_T^2\geq K\}}] \\ &= e^{-...

4

Most of the time, when you have a simple SDE without a drift, it's a martingale because the Wiener process itself is a martingale. In your example, you have a constant with the Wiener process, therefore the whole process must also be a martingale because the expectation is clearly X(t). However, we can't conclude a driftless SDE is always a martingale. ...

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