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37 votes

Explaining the Risk Neutral Measure

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 $M$...
Kevin's user avatar
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Explaining the Risk Neutral Measure

Intro: Great answer given by Kevin. 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&...
Jan Stuller's user avatar
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Convexity Adjustment for Futures

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*} where $d\langle W_s(...
Gordon's user avatar
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13 votes

Intuitive Explanation for Shannon's Demon?

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 ...
RRL's user avatar
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How do we determine the "correct measure"?

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 ...
Daneel Olivaw's user avatar
12 votes

Heston stochastic volatility, Girsanov theorem

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&...
Kevin's user avatar
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Intuition for Stock Price Numeraire Drift

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 ...
Daneel Olivaw's user avatar
10 votes

Intuition for Stock Price Numeraire Drift

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*} \...
Gordon's user avatar
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Numeraire correlated to the traded asset

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 ...
Daneel Olivaw's user avatar
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Change of measure and Girsanov's Theorem: Do the following models admit arbitrage and are they complete?

First, let's check if these models are abritrage free. The first fundamental theorem of asset pricing says that if there exists an equivalent probability measure under which $\frac{S_t}{\beta_t} = e^{...
user6247850's user avatar
9 votes

Why aren't american put options martingales?

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 ...
Kevin's user avatar
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Why discounted derivative price is a martingale?

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 ...
Daneel Olivaw's user avatar
7 votes

Intuitive Explanation for Shannon's Demon?

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 ...
vonjd's user avatar
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Measure theory in quantitative finance

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 ...
alexbougias's user avatar
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Proof that $\exp(aW(t)-0.5a^2t)$ is a martingale

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 ...
Kevin's user avatar
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Intuition for Stock Price Numeraire Drift

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: $E(S_T)=S_t$, ...
dm63's user avatar
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6 votes

Martingale representation theorem

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*}...
Gordon's user avatar
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Steven Shreve: Stochastic Calculus and Finance

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}...
Kevin's user avatar
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Pricing call option using risk-neutral martingale approach with squared stock price boundary?

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 by $$...
Kevin's user avatar
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Explaining the Risk Neutral Measure

I believe the other answers are nearly exhaustive; but here's a bit of intuition I'd like to add: Think of the decision (= equilibrium price) of a market as: Decision = f(probabilities, risk aversion) ...
Arshdeep's user avatar
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Fama: Efficient Capital Markets: A Review of Theory and Empirical Work - are martingales incorrect?

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 ...
Bob Jansen's user avatar
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If there is a $T$-forward measure and a risk neutral measure, then markets are not complete?

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 ...
siou0107's user avatar
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Proving that a stochastic process is a martingale using Ito's Lemma

$$ d Y \left(t\right) := d \left[\int_0^t{a \left(s\right)\mathrm{d}W_s}\right] = a \left(t\right) dW_t $$ Note that since $Y$ is a driftless process, it is a local martingale, and because $a$ is ...
siou0107's user avatar
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6 votes

Discounted price of an option

The process $Y_t:=(S_t-K)^+$ cannot be the price of a traded asset because of Jensen's inequality. Instead, it is the price of the option which is a martingale. In the Black-Scholes model, the ...
Daneel Olivaw's user avatar
5 votes

Why discounted derivative price is a martingale?

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. ...
Mark Joshi's user avatar
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Equivalent martingale measure exists if and only if $a < S_0^1(1+r)< b$

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 ...
Daneel Olivaw's user avatar
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How to prove martingality of forward rate under T-forward measure

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 ...
Magic is in the chain's user avatar
5 votes

How to prove martingality of forward rate under T-forward measure

By definition, $$Fo(t,T)=E^T[S_T|F_t]$$ Note that expectation is taken under $T$-forward measure. Now, evaluating at $s<T$: $$E^T[Fo(t,T)|F_s] = E^T[E^T[S_T|F_t]|F_s] = E^T[S_T|F_s] = Fo(s,T)$$ (...
Prabhnoor Duggal's user avatar
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I am trying to solve this question about optimal stopping theory. I don't know how to get started. Any hints would be very helpful

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 ...
Gordon's user avatar
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